chapter 5 thomas 9 ed f 2008

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Applicationsof Integrals

OYERVIEW Many things we want to know can be calculated with integrals: theareas between curves, the volumes and sudace areas of solids, the lengths of curves,the amount of work it takes to pump liquids fiom belowground, the forces againsrfloodgates, the coordinates of the points where solid objects will balance. We defineall of these as limits of Riemann sums of continuous functions on closed intervals.that is, as integrals, and evaluate these limits with calculus.

There is a pattern to how we define the integrals in applications, a patternthat, once learned, enables us to define new integrals when we need them. We lookat specific applications 0rst, then examine the pattern and show how it leads tointegrals in new situations.

Areas Between CurvesThis section shows how to lind the areas of regions in the coordinate plane bvintegrating the functions that define the regions, boundaries.

The Basic Formula as a Limit of Riemann SumsSuppose we want to find the area of a region that is bounded above by the curve-y =.l(r), below by the curve ) = g(.r), and on the left and right by the l ines.r : 4 and r : D (Fig. 5.1). The region might accidentally have a shape whose areawe cou ld f inL l w i rh geomet r l . bur i l / and g a re i t rb i l ra r ) conr inuou, tuncr ronr , eusuaily have to flnd the area with an integral.

To see what the integml should be, we lirst approximate the region with,?vertical rectangles based on a parlit ion p : {.r0, rr, ... , r,,} of [ri, D] (Fig. 5.2, onthe following page). The area of the tth rectangle (Fig. 5.3, on rhe followrng page;ls

AA1 : tsjcfi1 x width : Ll (c.t) g(ci)l A4.

We then approximate the area of the region by adding the areas of the ,? rectanqles:

A * f ^A , - I t r , , . r g r r . , r lA , : , .

As ll P | -+ 0 the sums on rhe right approach the limit /j t/(r) _ g(,r)l d,r because

365

' The region between y: f(x) andy : g ( x ) and t he l i nes x : a and x = b .

356 Chapter 5: Applications of Integrals

::.: We approximate the region withrectangles perpendicular to the x-axis.

, ' .]]. The region in Example'l with atypical approximating rectangle.

f(t) - s(ck)

Stct l l

AAk = area of kth rectangle, f(c*) - g(cr) = height, lxk = width

/and g are continuous. We take the area of the region to be the value of this integral'

That is,

A - l i m ) - l / ( c , ) - g t c r t l A x r - f o r f , r , - S t x t ] d xn P o - - J "

DefinitionIfl ancl g are continuous with /(r) > g(x) throughout [4, b], ihen the area

of the region between the curves y : /(*) and y : g(t) ftom a to b is the

integral o[ [/ - 8 | l iom a to D:rb

A= J" U(x) - g(x)ldx.

To apply Eq. (1) we take the following steps.

How to Find the Area Between Two Curves

l. Graph the cunes and draw a tepresentative rectangle. Thts rcveals

which curve is / (upper curve) and which is g (1ower curye). It also

helps find the limits of integration if you do not aheady know them.

2. Find the limits of integration.3. Write a formula for f(x) - g(x). Simplify it if you can.

4. lntegrate [f(x) - gQ, fron a to b Tl1e number you get is the area.

T

Find the area between lv = sec2 x and y : 3i1;5 from 0 to z/4'

( 1 )

EXAMPLE 1

Solution

Step t; We sketch the cutves and a vertical rectangle (Fig. 5.4). The upper curve

is the gmph of /(;r) = secz;r: the lower is the graph of g(;) : sip"I'

Sfep 2; The limits of integration are already given: a :0' b = r /4'

SteP 3: f (x) - 8(x) = sec2;r - sin't

",s'I ,=sinrH I

--''-

5.1 Areas Between Curves 367

Step 4:

(-{,,f(-r)

( 1 , 1 )

Step 4

o: l"' rf ,.,> - e(x)ldx :

lo"'o {r""' *- sinx) r/x : ftan -r + cos x]i/a

f'.+l ,to+\:+L - t f,

Curves That IntersectWhen a region is detemined by curves that intersect, the intersection points givethe limits of integration.

EXAMPLE 2 Find the area of the region enclosed by the parabola ! :2 - x2and the line y : -y.

Solution

Sfep t; Sketch the curves and a vertical rectangle (Fig. 5.5). Identifying the upperand the lower curyes, we take /(x) :2- xz arLd g(r): -x. The ,{-coordinatesof the intersecdon points are the l imits of integration.

Step 2; We find the limits of integation by solving | :2 - xz and y : -1 ri-multaneously for .r.'

2 - x z : - x [ q r l L t r / 1 r ) . u r d . q ( f r

x 2 - x - 2 : 0 R . \ r i r . .

( x + l ) ( x - 2 ) : 0 F x r r . '

x : - 1 , x : 2 . s o t \ .

The region runs from x : - 1 to.r : 2. The limits of integration are a : - l, b : 2.

Step 3:

5.5 The region in Example 2 with atypical approximating rectangle.

t 2 f - 2 - t 1 2

I t z+x -xz tdx= l z r+ " - , IJ - t I z J J r

/ 4 8 \ / | 1 \(o*r-r , l (- '* l*r /

? q qA t

-- ' 2

3 2 l

Technofogy The Intersection of Two Graphs One of the difficult and some-times frustrating parts of integmtion applications is finding the limits of inte-gration. To do this you often have to find the zeroes of a function or theintenection points of two curues.

To solve the equation /(x): g(r) using a graphing utility, you enter

(a s(r))

yr : f@) and y2: g@)

368 Chapter 5: Applications of lntegrals

5.6 When the formula for a bounding curvechanges, the area integral changes to match(Example 3).

and use the grapher routine to find the Points of intersection' Altematively' you

lun solve *tJ "quation

/(:r) - g(T ) : 0 with a root finder' Try both procedures

with

f ( x ) : l n x a n d g ( x ) : 3 - x '

When points of intersection arc not clearly revealed or you suspect hidden

behavior, additional work with the graphing utility or further use of calculus

may be necessary,

a) The intersecting curves y1 : Inx and lz:3 - x' using a built- in fundion

to find the intersectionO) Using a built- in root f inder to find the zero of f(x) : Inx - 3 + x

Boundaries with Changing Formulas

If the formula for a bounding curve changes at one or mole points' we partition

th","gionintosubregionsthatconespondtotheformulachangesandapplyEq.(1) to each subregion.

EXAMPLE 3 Find the area of the region in the fint quadrant that is bounded

above by y : .r/i and below by the t-axis and the line y : x - 2'

solution

Step t.' The sketch (Fig. 5.6) shows that the region's upper boundary is the graph of

f(; : JV. The lowei boundary changes fromg(;r):0for0 S r <2to g(x) =

" -' Z tot Z : t : 4 (there is agreement at r : 2)' We partition the region at r : 2

into subregiJns A ani f and tket"h a representative rectangle for each subregion'

ISECT =.79205996845

step 2: 'f\e limits of integration for region A are a : O and b :,2'

li#t for region B rs a = 2. To find the right-hand limit, we solveThe lefchandthe equations

5.1 Areas Between Curves 369

y : rG and y : x - 2 simultaneously for r.'

v ^ - ^ - z

a : ( x . - 2 )2 : x .2 -4x+4

x2 -5x +4 :o

( . r - 1 ) ( " r - 4 ) : 0

- _ 1 - _ i

Only the value.x =4 satisfies the equation Ji:x-2. The value "r:1is anextraneous root introduced by squaring. The right-hand limit is b:4.Step 3. 'For0 5 x = 2: f (x) - S@):

"E -0:

"qF o r 2 a x a 4 : f ( x ) - e ( x ) : , 6 - ( x - 2 ) : J i - x + 2Step 4.' We add the area of subregions A and B to find the total area:

f ) - f 4To ta f a rea = | J i dx I l t Jx -x+2 tdx

Jo Jz\ - \ - \ - - \ - /arca of A

Eqratc I ( .r) rnd. 9 ( 1 1SqLrafe both

Re$'r'ite.

FacL(r'.

Solve.

area of B

lntegration with Respect to yIf a region's bounding curves are described by functions of y, the approximatingrectangles are horizontal instead of vertical and the basic formula has y in placeof r.

f ) ' l 2

f 1 - 2 1 4

fi""l.+l1*'' '-v+2,1,?ef,, _o_ (1,,0,',, _8 .' 8) _ /2 - \r \ r , \ r {zt ' ' ' -z++1?,t , - , : ! . ,3 3

In Eq. (2),/always denotes the right-handcurve and I the left-hand curve, sofO) - e(y) is nonnegative.

t

1IL*,,,

e: l. ' vrrt - stytldy.

use tJte formula

(2)

(c()), ).)x : y + 2


5.7 lt takes two integrations to find thearea of this region if we integrate withrespect to x It takes only one if weintegrate with respect to y (Example 4).

EXAMPLE 4 Find the area of the region in Example 3 by integrating withrcspect to ).

Solution

Step t; We sketch the region and a typical horiz,ontal rectangle based on a panitionof an interval ofy values (Fig. 5.7). The region's right-hand boundary is the line x :

y f 2, so l (y) : y + 2. The left-hand boundary is the curve ir : y2, so g(y) : y2.

Step 2: The lower limit of integration is l = 0. We find the upper lirnit by solvingI : y *2 and r: )2 simultaneously for y;

l + t : l

, t - ) 2 = o

( )+ 1 ) ( ) 2 ) : 0

_ t 1 ' - )

The upper limit of integmtion is D : 2. (The valuesectlor\ below the r-axis.)

Step 3:

/ ( ) ) - s ( ) ) = ) + 2 y 2 = 2 i y ) ?

Step 4:

12+y -y ' ) l dy

f r,2: L2Y

+;

. 4-o+t-

This is the result of Example 3, fbund with less work.

Combining Integrals with Formulas from GeometryThe fastest way to find an area may be to combine calculus and geometry.

EXAMPLE 5 The Area of the Region in Example 3 Found the Fastesiway

Find the area of the region in Example 3.

So/ution The area we want is the area between the curye I : Ji,O = x S 4, andthe x-axis, minus the area of a triangle with base 2 and height 2 (Fig. 5.8):

xru = fo r ta , l r2r r2r

2 " , " 1 1J .lo

:3tt l-o - 'z:+' ,J J

I u r t , ,

S , i l \ .

y : -1 gives a point of inter-

e: 1,,'' v,rt - srt,rat : l,'

r t l '1 l3 l nl 03

83

1 : r f

2. 2

/' . 2

5.8 The area of the blue region is thearea under the parabola y = 1& minusthe area of the triangle.

-

Exercises 5.1 371

Moral of Examples 3-5 It is sometimes easier to lind the area between twocurves by integrating with respect to y instead of ,r. Also, it may help to combinegeometry and calculus. After sketching the region, take a moment to determine thebest wav to oroceed.

Exercises 5.1

Find the areas of the shaded regions in Exercises 1-8.

l. 2.


In Exercises 9-12, find the total shaded area.

2x3 - "2

5x

]

I

+.

9,

( -3 ,5 )

10.

16.

t7.

18.

19.

20,

21.

y : x 2 - 2 x a n d Y = r

) : 12 and y = -tt2 14x

y = l - l x ' a n o ] : x - 1 4

y = x 4 - 4 x 2 + 4 a n d Y = a 2

y = r n @ = 7 , a > 0 , a n d y : 0

y = {fl nd 5y = x * 6 (How many intersection points are_ , _ , 2

I

there?)

2 2 , y : l x 2 - 4 1 ' a n d ) : ( x 2 1 2 ) + 4

Find the areas of the regions enclosed by the lines and curves

Exercises 23-30.

23 . x :2y2 , : r :0 , and ) :3

24. x = y2 u1t4 a: y 12

25. y2 - 4x:4 and' 4x - Y = 16

26, x - y2 :0 arld x +2Y2 =3

2 7 , x * y 2 : O ^ n d r + 3 Y 2 : 2

2 8 . x - y 2 / t : 0 a n d , r * ) a : 2

2 9 . x = y 2 - l a n d x = ' y J l y

30 . r= ) l3 -y2 and x=2y

Find the areas of the regions enclosed by the curves in Exercises

31'34.

3 1 . 4 x 2 + y : 4 a n d . x a - Y : l

32 . x3 y :O and 3x2 - y :4

3 3 . x * 4 y 2 : 4 a n d : r * ) a : 1 , f o r x Z 0

3 4 , x a y z : 3 a n d 4 . t * y 2 : 0

Find the arcas of the regions enclosed by the lines and curves in

Exercises 35-42.

35 . y=2s1nx and y = s in2 . r , 0 : .x57 t

3 6 . ) : 8 c o s . r a n d y : 5 e 9 2 . r , - 7 t 1 3 = x S r 1 3

37. y :cos( rx /2 ) and 1 :1 . t2

38. y = sin(nx/2) and 1t:r

39 , l : sec2 . r , y= tar fx , x : - t r /4 , and ' x : t t /4

4O. x : ta#y and .x : - tan2) , -7 r14=r 3n /4

4 1 . r : 3 s i n l / i o s y a n d - x = 0 , 0 = y < 7 t / 2

42. y :sss2 11 taJ31 ar rd y : r1 /1 , - l= i r : l

43. Find the area of the propeller-shaped rcgion enclosed by the curvet - y3 : 0 and the line.r - Y : 0.

44. Find the area of the propeller-shaped region enclosed by the

clrues x -yll3 : 0 andr - )l/5 = 0.

45, Find the area of the region in the first quadrant bounded by theline I :1, the line,t :2, the curve 1 : 1/-r'?, and the -t-aris.

-3)

(-2,4\

( 3 , 5 )

(3 ,6 )

12.

- 2 - l

III,

)

I6v

- ' 3 )r(3, 1)

Find the areas of the regions enclosed byExercises 13-22.

1 3 . y : a 2 - 2 a r ' d y : 2

14. y =2x - x2 arLd Y: -J

15 . y :1+ and ) :8 r

the lines and curves in

46.

47,

Find the area of the "tdangular" region in the first quadrantbounded on the left by the y-axis and on the right by the curves

) = sin.r and ): cos,r.

The region bounded below by the pambola ) : ,r2 and above bythe line ) : 4 is to be parlitioned into two subsections of equalarea by cutting across it with the horizontal line ), = c.

a) Sketch the region and draw a line ) : c across it that looksabout right. In terms of c, what are the coordinates of thepoints where the line and parabola intersect? Add them toyour figure.

b) Find c by integrating with respect to ),. (This puts c in thelimits of integration.)

c) Find c by integnting with respect to.x. (This puts c into theintegrand as well.)

Find the area of the region between the curve )):3 ''2 andthe line y = 1 by integrating with respect to (a) t, (b) ).

Find the area of the region in the first quadrant bounded on theleft by the ),-axis, below by the line y = t /4, above left by thecurve y = I * .,[, and above right by he (i.jtr..te y : 2/ Ji,

Find the area of the region in the first quadrant bounded on theleft by the y-axis, below by the curve r : 2/t, above left bythe curye r : () - l)'?, and above right by the line i :3 - ).

The figure here shows triangle AOC inscribed in the region cut

ftom the parabola ) - .r2 by the line y: a2. Find the limit of

the latio of the area of the triangle to the area of the parabolic

region as 4 approaches zerc.

Exercises 5.1 373

Suppose the area of the region between the graph of a positivecontinuous function / and the .x-axis from x : 4 to x : b is 4square units. Find the area between the curves )r: /(.r) andy - 2 f ( x ) f t o m x : a t o x - b .

Which of the following integals, if either calculates the area ofthe shaded region shown here? Give reasons for your answer.

f L f )a l I ( x - l - x ) \ d : ( = | 2 x d x

J - t J I

r o t '

t l

b t / r - x - ( r ) ) d x - | - z x a xJ - t J I

54. True, sometimes tue, or never tue? The area of the regionbetween the graphs of the continuous functions ): /(i) and

) : g(r) and the vertical lines r : a and x: b (a < b) rs

rbI lf @) - s(x)ld)c.

J "

Give reasons for your answer.

S CAS Explorations and ProiectsIn Exercises 55-58, you will find the area between curves in the plane

when you cannot find their points of intersection using simple algebra.

Use a CAS to perform the following steps:

a) Plot the curves together to see what they look like and how manypoints of inrersection (hey ha\e.

b) Use the numerical equation solver in your CAS to find all thepoints of intersection.

c) Integate "f(r) - g(r)l over consecutive pairs of intersection

values.d) Sum together the integrals found in part (c).

5 5 . / 1 x 1 : ^ - ^ - 2 , I . g 1 x . 1 : x - J' 3 2 3

t 45 6 . / t x t : i - t r ' - 1 0 . g ( . r r : 8 - l 2 r

57. /("r) =x *sin(2tc), g(x) : 13

58. /(.x) : -r2 sqsir, g(tc) : x1 - x

49.

50.


lf the area A(x) of the cross sedionR(x) is a continuous function of x, we canfind the volume of the solid bYintegrating A(x) from a to b.

Finding Volumes by Sl ic ingFrom the areas of regions with curved boundaries, we can calculate the volumes of

cylinders with curved bases by multiplying base area by height. From the volumes

of such cylinders, we can calculate the volumes of other solids.

5l ic ingSuppose we want to find the volume of a solid like the one shown in Fig. 5.9. At

each point r in the closed inteNal la, rl the cross section of the solid is a region

R(x) whose area is A(r). This makes A a real-valued function of .r' If it is also a

continuous lunction of .r, we can use it to define and calculate the volume of the

solid as an integral in the following way.

We partition the interval [4, ]l along the r-a{is in the usual manner and slice

the solid, as we would a loaf of bread, by planes petpendicular to the -r-axis at

the partition points. The tth slice, the one between the planes at xk r and tn, has

approximately the same volume as the cylinder between these two planes based on

the region R(,rr) (Fig. 5.10). The volume of this cylinder is

Vr : base area x height

: A(rr) x (distance between the planes at -rr r and xr)

: A(x r )Axr .

The volume of the solid is therelbre approximated by the cylinder volume sum

\ - r , " . r , r - .

?-,"'^""'^This is a Riemann sum for the function A(r) on [c, Dl. We expect the approxlmahons

from these sums to improve as the norm of the partition of la, bl goes to zero, so

we define their limiting integral to be the volume of the solid.

Plane at.r[ IApproxrmatlngcl l inder based plane nr r ,o n ] ( ( r t ]

-'i.---------' -,

The cylinder's baseis the region R(,v*).

NOT TO SCAI-E

. Enlarged view of the slice of the solid between theplanes at xk r and xk and its approximating cylinder'

Cross section R(jr). Its area is A(ir).

5.2 F inding Volumes by Sl ic ing 375

DefinitionThe volume of a solid of known integrable cross-section area A(x) fromx = a to x: D is the integral ofA from a to r.

v : fo e61a,. ( j)J"

To app l l Eq . { l ) . we rake the fo l low ing : rep . .

How to Find Volumes by the Method of Slicing

1 Sketch the solid and a typical cross section.2. Find a formula for A(r).3. Find the limits of integration.4. Integrate A(x) to find the volume.

EXAMPLE 7 A pyramid 3 m high has a square base that is 3 mon a side.The cross section of the pyramid perpendicular to the altitude r m down from theve ex is a square r m on a side. Find the volume of the pyramid.

Solution

Step 1: A sketcll. We draw the pyramid with its altitude along the r-axis and itsvertex at the origin and include a typical cross section (Fig. 5.1 1).

Typical cfosssection

--r$€t-=_, , . ,

The cross sections of the pyramid in Example 1are 5quare5.

Step 2: A.fonnula for A(x). The cross section at r is a square x meters on a side,so lts area ls

A( r ) : 12 '

Step 3: Tlle lintits of integration. The squares go from r :0 to;r :3

Step 4: The volwne.

l b t ' - r . ' l '

v = I A t ^ t r l t - | t : d r - " , | = 9J.. J,, r .l ,

The volume is 9 mr.


Bonaventura Cavalieri ('l 598-1647)

Cavalied, a student of Galileo's, discoveredthat if two plane rcgions can be afianged tolie over the same interval of the i-axis 1nsuch a way that they have identical verticalcross sections at every point, then the rcgionshave the sarne area. The theorem (and a letterof recommendation from Galileo) wereenough to win Cavalied a chai at theUnivelsity of Bologna in 1629. The solidgeometry version in Example 3, whichCavalieri never proved, was given his nameby later geometen.

EXAMPLE 2 A curved wedge is cut from a cylinder of radius 3 by two planes.One plane is perpendicular to the axis of the cylinder. The second plane crossesthe nrst plane at a 45' angle at the center of the cylinder. Find the volume of thewedge.

Solution

Step 1: A sketch. We draw the wedge and sketch a typical cross section perpen-dicular to the:r-axis (Fig. 5.12).

5-t2 The wedge of Example 2. sliced perpendicularto the x-axis. The cross sections are redangles.

Step 2: The formula for A(x). The cross section at r is a rectangle of area

Crcss sections havethe same longth atevery point in [d, b]

I

,f

? b ^ 3

v : l A \ x ) d x : | 2 t c J 9 - x 2 d xJ a J O

1 1 3: -;e - xr3/'

)o

: o +?e),/'J

- 1 9

EXAMPLE 3 Cavalieri's Theorerrr

Cavalieri's theorem says that solids with equal altitudes and identical parallel cross-section areas have the same volume (Fig. 5.13). We can see this imrnediately fromEq. (1) because t}Ie cross-section area function A(.r) is the same in each case. f

1! = l f r1 r . nncs fa lc -

.Lnd subs t i lUre brck .

tr

5.13 Cavalieri's theorem. These solidshave the same volume. You can il lustratethis yourself with stacks of coins.

2lt -7

A(') : (heishtxwidth) : (r) (zJe - 'r)

Step 3: The linxits of integration.

Step 4: The volune.

^ t : ,: z x \ t Y - x ' .

The rectangles run from r :0 to.r :3.

Exercises 5.2 377

Exercises 5.2

Cross-Section AreasIn Exercises I and 2, find a fomula for the arca A(r) of the crosssections of the solid perpendicular to the l-axrs.

1. The solid lies between planes perpendicular to the r-axis at i =-l and,r:1. In each case, the cross sections perpendicularto the r-axis between these planes run from the semicircle y =-^"4 - ,r2 to the semicircle y : 4[ - ,2.

a) The cross sections me circular disks with diameters in thexy-plane.

b) The crcss sections are squares with bases in the r),-plane.

The solid lies between planes perpendicular to the r-axis at.r : 0and l : 4. The cross sections perpendicular to the -t-axis betweenthese planes run from the parabola y = -aE to the parabola

r: Ji .a) The cross sections are circular disks with diameters in the

r)-plane.

b) The cross sections are squares with bases in the r)-plane.

c) The cross sections are squares with diagonals in the r)-plane.d) The cross sections are equilateral triangles with bases in the

rJ-plane.

Volumes by SlicingFind the volumes of the solids in Exercises 3-12.

3. The solid lies between planes perpendicular to the ,-axis at .r : 0and "r :4. The cross sections perpendicular to the axis on theinterval 0 5 i :4 are squares whose diagonals run ftom theparabola 1 : -.vE to the parabola f : "Jtr.

c) The cross sections arc squarcs with diagonals in the xy-plane. (The length of a square's diagonal is .,4 times thelength of its sides.)

The cross sections arc equilateral triangles with bases in thery-plane.

d)

374 chapter 5: Applications of Integrals

4, The solid lies between planes perpendicular to the r-axis at -x :- 1 and r : I . The cross sections perpendicular to the ir axis arecircular disks whose diamerers run from lhe parabola y - x7 tothe Darabola t :2 - t2.

5. The solid lies between planes perpendicular to the r-axis at"r : - I and r : 1. The cross sections perpendicular to the axisbetween these planes are vertical squares whose base edgesrun from the semicircle J - - 1(1 - rt lo rhe semicircle y -!7-7.

6. The solid lies between planes perpendicular to the r-axis at r =I and n: l. The cross sections perpendicular to the ,r-axis

between these planes are squares whose diagonals run frcm thesemicircle ) : -{ -7 to the semicircle y: JT=V. Qn"length of a square's diagonal is r/Z times the length of its sides.)

7. t he base of the solid is the reSion berween lhe curve ) : 2 vfi;and the interval [0, r] on the.r-axis. The qoss sections perpen-dicular to the r-axis are

a) veftical equilateral triangles with bases running from theI-axrs to me culve:

b) vertical squares with bases running from the r-axis to thecutve.

8. The solid lies between planes pe.pendicular to the r-axis at -x =-r /3 and x : xr /3. The crcss sections perpendicular to the "r-axis are

Cavalieri's Theoremll. A twisted solid. A square of side length s lies in a plane per-

pendicular to a line L One vertex of the square lies on t. As thissqu&e moves a distance l? along l, the square tums one revo-lution about Z to generate a corkscrewlike column with squarecross sections.

a) Find the volume of the column.b) What will the volume be if the square tums twice instead

of once? Give rcasons for your answer

12. A solid lies between planesperpendicular to tbe -{-axis at-r = 0 ard .r : 12. The crosssections by planes perpendicularto the r-axis are circular diskswhose diameters run from the line) : -r/2 to the line ) : -r.Explain why the solid has thesame volume as a right circularcone with base radius 3 andheight 12.

Cavalieri's original theorem. Prove Cavalieri's original theorem (marginal note, page 376), assuming that each region isbounded above and below by the graphs of continuous functions.

The volume of a hemisphere (a classical application af Cav-alieri's theorem). Derive the formula V : (2/3)zR3 for thevolume of a hemisphere of radius R by comparing its cross sec-tions with the cross sections of a solid right circular cylinder ofmdius R and height R from which a solid right circular cone ofbase radius R and heisht R has been removed.

13.

t4.

circular disks with diameters running from the curve y :

tan,r to the curve ): secr;verlical squares whose base edges run from the curve y :

tanr to the cuNe ): SeCr.

9. The solid lies between planes perpendicular to the y-axis at y : 0and ) : 2. The cross sections perpendicular to the ]-axis are cir-cular disks with diameters running from they-axis to the parabola

10. The base of the solid is the disk j'? + ),2 5 1. The cross sectionsby planes perpendicular to the )-axis between ], - I and y - Iare isosceles dght triangles with one leg in the disk.

a)

b)

5.3 Volumes of Solids of Revolution-Disks and Washers 379

E n rn::rt sotids of Revorution-Disks

The most common applicarion of the method of slicing is to solids of revolution.Solids of revolution are solids whose shapes can be generated by revoJving planeregions about axes. Thread spools are solids of revolution; so are hand weightsand billiard balls. Solids of revolutign sometimes have volumes we can find withformulas liom geometry, as in the case of a billiard ball. But when we want tofind the volume of a blimp or to predict the weight of a part we are going to haveturned on a lathe, formulas from geometry are of little help and we tum to calculuslor lhe un:* ers.

revolution

6revolution

#Axis of

s

If we can arange for the region to be the region between the graph of a contin-uous function I = R(r), a 5 x < ,, and the x-axis, and for the axis of revolutionto be the x-axis (Fig. 5.14), we can find the solid's volume in the following way.

The typical cross section of the solid perpendicular to the axis of revolution isa disk of radius R(x) and area

A(.r) : 71136iut1: : rlR(x)12.

The solid's volume, being the integral ofA from.r - a to x : b, is the integral ofzfR(x)1'zfrom a to D.

Volume of a Solid of Revolution (Rotation About the .r-axis)

The volume of the solid generated by revolving about the.{-axis the regionbetween the -r-axis and the graph of the continuous function ) : R(r), a S:r 5 b, is

v = [" , ftua;ur1t a* : ['J. J,,

r lR1x) l2 z l r . (1 )

EXAMPLE I The region between the curye y :.,rE,O <.{ S 4, and the r-axisis revolved about the .x axis to generate a solid. Find its volume.

Cross section perpendjcular tothe axis at r is a disk ofarea

,4(r) = 7r(radius)2 = tt (R(,r))2

and the x-axis from a to b about thex-axis.


:l,

(b)

5.15 The region (a) and solid (b) inExample 1 .

(b)

5.16 The region (a) and solid (b) inExample 2.

Solution We draw ligures showing the region, a typical radius, and the generatedsolid (Fig. 5.15). The volume is

f bv : I trlR(x)\'dx

f 1: I r l r tTaxJ o -

F - q r l r

/ l 1 f t \ i

f 4 t 2 l a ( 4 t 2= t r l x d x - u . ,

1 = z - : 8 n .J o L J o l

How to Find Volumes Using Eq. (1)

1, Draw the region and identify the radius function R(x).2. Square R(r) and multiply by z.3. Intesrate to find the volume.

^ . 4l x - ^ t \ t I t n

: T l ^ 2 ' : x - + x l : -

L z J l r o

To find the volume of a solid generated by revolving

)-axis and a curve x : R()), c < l : d, about the y-axis,

replaced by ).

The axis of revolution in the next example is not the .x-axis, but the rule forcalculating the volume is the same: Integmte lr(radius)2 between appropriate limits.

EXAMPLE 2 Find the volume of the solid generated by revolving the regionbounded by y : .rf and the lines ) : l, x:4 about the line y : 1.

Solution We draw figures showing the region, a typical radius, and the generatedsolid (Fig. 5.16). The volume is

FV : I r l R ( x ) 1 z d r f ( l r r

J , '

r 1: I n l Jx_ t ) - dx

J t '

f 4:o I l , z"E + r)a,J t

'

-j

a region between thewe use Eq. (i) with "r

Volume of a Solid of Revolution (Rotation About the y-axis)

, , . f

-

5.3 Volumes of Solids of Revolution_Disks and Washers 391

EXAMPLE 3 Find the volume of the solid generated by revolving Lne reglonbetween the y-axis and the curve x:2/y,I S l,:4, about the y-axis.

Solution We draw figures showing the region, a typical radius, and the generatedsolid (Fig. 5.17). The volume is

v : l,o

:zln6)1'z ay

t 4 / ' r \ 2= J, "\;) o,

pD: I rl2 - y212 dy

/1: r I L 4 - 4 y 2 + y 4 l d y

| 4 . , ,51 f: 7 r l 4y - ; - y ' + a I

I J ) l ^

64rJ2t5

\z

0

-.'t

l . r r : )

/ 1 1 \ r : 1 r f - l l

l _ r l ( l r

/11 r I

r ^ 4 T r ' 1 . l 3 l- t I - t l t : qr | - : | : 4n r . tJ t v L ) l r L 4 J

EXAMPLE 4 Find the volume of the solid generated by revolving the regionbetween the parabola I :12 + I and the l ine I :3 about the l ine; :3.

so/ut on We draw ligures showing the region, a typical radius, and the generatedsolid (Fig. 5.18). The volume is

N1, :

J n r lR (y ) \ ' ?dy

.l

(

(b)



I r : t - + r

(b )

l

) R ( ] ) = 3 - ( ) ' ? + 1 )

(a.)


\,

) = R(-r)

5. ?9 The cross sections of the solid ofrevolution generated here are washers,not disks, so the integral f Ak) dx leadsto a slightly different formula.

The Washer MethodIf the region we revolve to generate a solid does not border on or cross the axis of

revolution, the solid has a hole in it (Fig. 5.19). The cross sections perpendicular

to the axis of revolution ate washers instead of disks. The dimensions of a typicalwasher are

Outer radius:

Inner radius:

R(-r)

r\x )

The washer's area is

A(x) : Tt lR(;) lz - rVG)12: o ( tn{") l ' - t t ( r ) l t ) .

Notice that the function integrated in Eq. (3) is t(Rz - r2), not T (R - r)'z. Also

notice that Eq. (3) gives the disk method fomula if /(jr) is zero throughout [a, r].Thus. the disk method is a soecial case of the washer method.

EXAMPLE 5 The region bounded by the curve y : xz + | and the line y :-x t 3 is revolved about the -r-axis to generate a solid. Find the volume of thesolid.

Solution

Sfep t; Draw the region and sketch a line segment across it perpendicular to thea;ds of revolution (the red segment in Fig. 5.20).

Step 2,' Find the limits of integration by finding the .r-coordinates of the intersectionpoints.

x 2 + l : - x + 3

, ' + * -2 :o(x+2) (x - l ) :0

, - _ 1 . - l

Step 3; Find the outer and inner radii of the washer that would be swept out bythe line segment if it were revolved about the x-axis along with the region. ftVedrew the washer in Fig. 5.21, but in your own work you need not do that.) Theseradii are the distances of the ends of the line sesment from the axis of revolution.

R(x) : -1 -1- 3

r ( x ) : a 2 4 1

Outer radius:

Inner radius:

The Washer Fonnula for Finding Volumesf b

v : I " ( [Rrx ) | ' : . l r t x ) f )dx

ra | |t l

outer innermdius radiussquarcd squarcd

(3)

5.3 Volumes of Solids of Revolution-Disks and Washers 383

( 1 , 2 )

lntegratlon

j . . i The reg ion in Example 5 spanned bya line segment perpendicular to the axisof revolution. when the region isrevolved about the x-axis, the l inesegment wil l generate a washer.

Washer cmss sectionOuterradius: R(r) = r+3Inner radius: r(.r) = 12 + I

Step 4; Evaluate the volume integral.

5.21 The inner and outer radii of thewasher swept out by the l ine segment inF i9 .5 .20 .

rb, =

J,, " ( [Rrx)l ' - l r(x)] ' ) dx

l t ^=

/ _ r t r ( { x | 3 ) ' t x ' I l t ' t d t

f l= / z (S 6x x2 xa ;dx

J-2

t ^ r ' r ' l ' t l l r- ' l * i s l , s

F( t 1 - i l

I ltn!1 l

LI

How to Find Volumes by the Washer Method

l. Draw the region and sketch a line segment across it perpend.icular tothe q,\is of revolution. When the region is revolved, this segment willgenerate a typical washer cross section of the generated solid.Find the limils of integration.Find the outer qnd inner rqdii of the washer swept out by the linesegment.Integrate to flnd the volume.

To find the volume of a solid generated by revolving a region about the )-axis,we use the steps listed above but integrate with respect to l instead of ,r.

EXAMPLE 6 The region bounded by the parabola I : x2 and the l ine y : 2yin the first quadrant is revolved about the )-axis to generate a solid. Find the volumeof the solid.

2.

4.

+-1


t22 The region, l imits of integration,and radii in Example 6.

524 The region, l imits of integration,and radii in Example 7.

Solution

Step t.' Draw the region and sketch a line segment across it perpendicular to theaxis of revolution, in this case the ),-axis (Fig. 5.22).

Step 2: The line and parabola intersect at ) : 0 and l : 4, so the limits of inte-gration are c :0 and d :4.

Step 3.' The radii of the washer swept out by the line segment are R(y) : a/ry,r(y) : y/2 (Figs. 5.22 and 5.23).

523 The washer swept out by the l inesegment in Fig. 5.22.

Step 4:

Eq. (-l ) \\'irh i

sleps I .rnd -l

?

, : 1""

, (tnci)l, - tr(y)t2) dy

: l,^" (ltl' -ltrJ')r,:" L'(,-';)*:"l+-i]:,8

3

5 o l

E r

EXAMPLE 7 The region in the first quadrant enclosed by the parab ola y :2e2,the y-axis, and the line ) : I is revolved about the line,r : 3/2 to generate a solid.Find the volume of the solid.

Solution

Step 1.' Draw the region and sketch a line segment across it perpendicular to theaxis of revolution, in this case the line ; :3/2 (Fig. 5.24).

Step 2: Tbe limits of integration are ) : 0 to l : 1.

Step 3.' The radii of the washer swept out by the line segment arc R(y) :3/2,r(y) : (3/2) - -/1, (Fiss. 5.24 and 5.25).

) = * 2 o r * : ft"

4! .= ; - t i

Exercises 5.3 385

5.-25 The washer swept out by the l ine segment inFig.5.24.

Step 4:

(Ln{r) l ' - t ' (y)1'?)dy= J, n

f t:1""

l

( [ ; ] ' - [ ; - "0) ' ) , ,

(3{y-y)dy =nlzftz +l:"f l

: t I3tr2

Exercises 5.3

Volumes by the Disk Methodln Exercises 1-4, find the volume of the solid generated by revolving

the shaded region about the given axis.

l. About the r-axis

4. About the r-axis

Find the volumes of the solids generated by revolving the regionsbounded by the lines and cu es in Exercises 5-10 about the r-axis.

5 , , y : n 2 , y : Q , x : 2 6 . y : t l , ) : 0 , r : 2

i . J - Je 7 . r : o 8 . y : - r - r 2 , - v : o

9 . y : v € o s i r , 0 S x = n / 2 , ) : 0 , j r : 0

1 0 . I - 5 s g a . 1 = 0 , r : - n / 4 , x : r / 4

In Exercises 1l and 12, lind the volume of the solid generated byrevolving the region about the given line.

11. The region in the iirst quadrant bounded above by the line y :J1, b"lo* by the curve ) : secr tanr, and on the left by the

-r-axis, about the line y = 1i

12. The region in the first quadrant bounded above by the line y : 2,below by the curve l:2sin.t.0l r : r/2, and on the left bythe ;y-axis, about the line 1 : 2

Find the volumes of the solids generated by revolving the regionsbounded by the lines and cuNes in Exercises 13-18 about the )-axis.

1 3 . x : J 5 y 2 , . v = 0 , y : - 1 , t : I

1 4 , x : ! 3 / 2 , a = e , y : 2

2. About the y-axis

3ab Chapter 5: App)icatlons ol )ntegra)s

1 5 . x - \ D i n T , o = y = r 1 2 , x : 0

16, x = Jcodtr y l -$ , 2=j i0 , , r :o

1 7 . x : 2 l O + 1 ) , x : 0 , ) : 0 , ) : 31 8 . x : J z . y l b , 2 + 1 ) , r c : 0 , 1 : 1

Volumes by the Washer MethodFind the volumes of the solids generated by revolving the shaded

regions in Exercises 19 and 20 about the indicated axes.

19, The,r-axis

) :,r'?, below by the i-axis, and on the right by the line, = 1,

about the line i - I

36. The region in the second quadrant bounded above by the curve

1, : -r3, below by the x-axis, and on the left by the line r : - 1,about the line r : -2

Volumes of Solids of Revolution37, Find the volume of the solid generated by revolving the region

bounded by y : ^vE and the lines ) - 2 and r : 0 about

a) the r-axis;b) the )-axis;c) the l ine ) :21

d) the line i - 4.

Find the volume of the solid generated by revolving the tdalgular

region bounded by the lines ) :2x, y :0, and t - I about

a) the l ine;r = 1;

b) the l ine;Y :2.

Find the volume of the solid generated by revolving the region

bounded by the parabola;l :,r2 and the line ]l: I about

a) the line 1, : l;b ) t he l i ne y :2 ;

c) the l ine ): -1.

40. By integration, find the volume of the solid genemted by re-volving the triangular region with vertices (0, 0), (r, 0), (0, r)about

a) the,t-axis;b) the )-axis.

R 41. oesigning a wok. You are designing a wok frying pan that willbe shaped like a spherical bowl with handles. A bit of expe -

mentation at home pe6uades you that you can get one that holdsabout 3 L if you make it 9 cm deep and give the sphere a radiusof 16 cm. To be sure, you picture the wok as a solid of revolu-tion, as shown here, and calculate its volume with an integral.To the nearest cubic centimeter, what volume do you really get?(1 L : 1000 cm3)

20. The y-axis

38.

Find the volumes of the solids generated by revolving the regions

bounded by the lines and curves in Exercises 21 28 about the t-axis.

2 1 . y : a , y : 1 , ; r : 0 2 2 . y : ) ; s , l : j r , " r : 1

z J . J = t J x , l : 2 , x - u

24. y = J i , y : -2 , x -o

2 5 . y : x 2 ] 1 , ) : r + 3

2 6 . ! : 4 - x 2 , y : 2 - r

27 , ) :5gsa, y : .15 , - t r 141x 3n /4

28. y : sec,r, l : tant, "Y :0, ;r : I

In Exercises 29-34, Iind the volume of the solid generated by revolv-ing each region about the )-axis.

29. The region enclosed by the triangle with vertices (1, 0), (2, l),and (1, 1)

30. The region enclosed by the triangle with vertices (0, l), (1, 0),and (1, l)

31. The region in the first quadrant bounded above by the parabola

],: 12, below by the r-axis, and on the right by the line .I :2

32. The region bounded above by the curve y: ^,6 and below byt h e l i n e Y : 1

33, The region in the nrst quadrant bounded on the left by the circlex2 + y2 - 3, on the dght by the line .t : r./3, and above by theline y : .7t

34, The region bounded on the left by the line i : 4 and on the rightbY the clrcle r'+ )' - l)

In Exercises 35 and 36, find the volume of the solid generated byrevol\ ing each region about the given axis.

35. The region in the frrst quadrant bounded above by the cuNe

_' 9 cm deep

Designing a plumb bob. Having been asked to design a brassplumb bob that will weigh in the neighborhood of 190 g, youdecide to shape it like the solid of rcvolution shown here. Find

rc2 : 256

) (cm)

r ' ( c m )

the plunb bob's volume. If you specify a brass that weighs 8'5

g/cml, how much will the plumb bob weigh (to the nearest gram)?

r (cm)

) - sinn, 0 : r = 'r, is revolved about the line -v : ',

1, to genemte the solid in Fig. 5 26.

5.4 Cylindrical Shells 387

f,l d GRAPHER Graph the solid's volume as a function ofc, lirst

for 0: c: I and then on a larger domain What happens

to the volume of the solid as . moves away ftom [0, 1]l

Does this make sense physically? Give reasons for your

answer'

B 44. ln,r.'xri';rf irr.r/ trrll. YoLl are designing an auxiliary fuel tank

that will 1lt under a helicopter's fuselage to extend its range

After some expedmentation at your drawing board, you decide to

shape the tank like the surface generated by revolving the curve

) : I (r2116). -4 = r 5 4. about the i t-axis (dimensions in

feet).

a) How many cubic I'eet of fuel will the tank hold (to the

nearest cubic foot)'i

b) A cubic foot holds 7 481 gal lf the helicopter gets 2 mi to

the gallon, how many additional niles will the helicopter be

able to fly once the tank is installed (to the nearest mile)?

45. Tl t e vo l u me ctf a.o/ trJ The disk,r2 + l2 : d2 is revolved about

1h" 1;ns,1 : b (b > d) to generate a solid shaped like a doughnut

and called a loru.t. Find its volume. @irt, I'1,, uQll 'f d:- -

ra2 f2, since it is the area of a semicircle of radius tl )

46. a) A hemispherical bowl of radius a contains water to a depth

,. Find the volume of waier in the bowl'

b) (Related rates) Water runs inlo a sunken concrete hemispher

ical bowl of radius 5 m at the rate of 0 2 mr/ sec How fastjs the water level in the bowl rising when the water is 4 m

deeP?

41. l t . t in91 ihe .on5i5le/ l .J/ of i / l e 'al . r i t l \ ' i ' i ini i iau al ' /al !1ni ' :

The volume lbrmulas in this secfion are all consistent with the

standard formulas from geometrY.

a) As a case jn point, show that if you tevolve the region

enclosed by the semicircle )'- '/rA 7 and the r-axis

about the x-axis to generate a solid sphere' the disk formula

for volume (Eq l) will give (4/3)r dr just as it should

b) Use calculus to find the volume of a right circular cone of

he igh t , and ba 'e r r d i u . ,

The arch0 5 c =

: . : . ' , : r Exerc ise 43 asks for theva lue o f c tha t min imizesthe volume of this solid.

Find the value ol c that minimizes the volume of the solid'

What is the minimum value?

What value of c in [0, 1l maximizes the volume of the solid?

a)

b)

: j . .

. : : : : . . .

. . : : . . .

: . : . . : . :

Cyl indr ical Shel lsWhen we need to find the volume of a solid of revolution' cylindrical shells some-

times work better than washers (Fig. 5 27, on the following page) In part' the

reil)on is lhat the lormulil lhe) leud lo does not require squaring

The Shell FormulaSuppose we revolve the tinted region in Fig 5.28 (on the following page) about the

)-axis to genemte a solid. To estimate the volume of the solid, we can approxlmate

ihe region with rectangles based on a partition P of the interval fc, Dl over which the

region stands. The typical approximating rectangle is AJrr units wide by I (c1) units

high, where c1 is the midpoint of the rcctangle's base. A formula ftom geometry tells


Rectangleheight =/(cr)

5.27 A, solid of revolution approx-imated by cylindrical shells.

5.28 The shell swept out by the kthrectangle.

us that the volume of the shell swept out by the rectangle is

LVy : )v x average shell radius x shell height x thickness,

which in our case is

LVL :21T x ck x f(c) x Lxk.

We approximate the volume of the solid by adding the volumes of the shells swept

out by the r? rectangles based on P.

v ,= \ - , tY/ r i : ) r z J t c l J t | t t A x ! \ r '

The limit of this sum as llPll + 0 gives the volume of the solid:n f b

v - l i m f 2 z c k . f l c ^ r l x k : | 2 n x l t x t d x .tp .o?a J"

Interval of integration

5.29 The region, shell dimensions,interval of integration in Example

anql .

The Shell Formula for Revolution About the J-axis

The volume of the solid generated by revolving the region betw,

"r-axis and the graph of a continuous function ) : /(x) > 0,0 < a <

about the y-axis is. h , , r ! \ / , t a 1 t l b

v = I z t I sn : r r l l . t l t l ' l a r : I 2n r f t x td t .J . \ radrus / \he 'gh t / l "

EXAMPLE 1 The region bounded by the cuwe y: $, the r-axis, and theline x : 4 is revolved about the,-axis to generate a solid. Find the volume of thesolid.

Solution

Step t: Sketch the region and draw a line segment across it parallel to the axisof revolution (Fig. 5.29). Label the segmenfs height (shell height) and distance from

between the

: d = x = b ,

( 1 )

One way to remember Eq. (1) is to imaginecutting and umolling a cylindrical shell to geta (nearly) flat rectangular solid.

Inner circumference = 2rt

5.4 Cylindrical Shells 389

the axis of revolution (shell radius). The width of the segment is the shell thicknessdr. (We drew the shell in Fig. 5.30, but you need not do that.)

) shell radius

4

5.30 The shell swept out by the l ine segment in Fig. 5.29.

Step 2i Find the limits of integration: r runs from a :0 to D :4.

Alnosl a rectangular solidY= lenglh x height x thickness

: 21tx - f(x\ . dx

v : l.' z,(;fli) (,,i"l!il,) *:

fo" zo1,;1.tri a,

__.LT

d,Y = tbicknessEq . ( l )

Values from stcps1 a n d 2

f 4 f ) l a 1 1 9 -: 2t I x3t2 dx = 2n l'= xs/z | :

'":"

.,/o L) lo J E

Equation (1) is for vertical axes of revolution. For horizontal axes, we replace the.r's with y's.

The Shell Formula for Revolution About the r-axis

v: [" z, ( '"i"fl) (J:l].)o : fo znyystayJr. \ raolus // \nelgnr/ J.

(for /(y) > 0 and0 < c < y < d)

(2)

(4,2)

Shellthickness = d)

EXAMPLE 2 The region bounded by the curve y: Ji, the,r-axis, and theline x : 4 is revolved about the r-axis to senerate a solid. Find the volume of thesolid.

Solution

Step t.' Sketch the region and draw a line segment across it parallel to the axis ofrevolution (Fig. 5.31). Label the segment's length (shell height) and distance ftomthe axis of revolution (shell radius). The width of the segment is the shell thicknessdy. (We drew the shell in Fig. 5.32, shown on the following page, but you need notdo that.)

5.31 The region, shell dimensions, andinterval of integration in Example 2.

4 t '. Shell height


,..:.: The shell swept out by thesegment in F ig . 5 .31 .

Shell radius

I : : The reg ion , she l l d imens lons , andinterval of integration in Example 3.

Step 2; Identify the l imits of integration: lr runs from c:0 to d:2.

Step 3.' lntegrate to lind the volume.

I J / . h e l l \ / s h e l l \v - I 2n | , . 11 . ' . - , l d vJ , l rac t tus 7 \nergn l / /

l ): I 2r (g@ _ 12) d)-J,,

^ t^ .'' 'l'- ) n l ) f . l =8 r

L a l o

This agrees with the disk method of calculation in Section 5.3, Example 1.

How to Use the Shell Method

Regardless of the position of the axis of revolution (horizontal or vertical),the steps tbr implementing the shell method are these:

l. Draw the region arul sketch a line segment across it parallel to theaxis of revolution. Label the segment's height or length (shell height),distance from the axis of revolution (shell radius), and width (shellthickness).

2. Find the limits of irtegration.3. Integrate the product 2z (shell radius) (shell height) with respect to the

appropriate variable (,t or l) to find the volume.

ln the next example, the axis of revolution is the vertical line -{ : 2.

EXAMPLE 3 The region in the first quadrant bounded by the parabola y :.r2,the )-axis, and the line l' : 1 is revolved about the line x :2 to generate a solid.Find the volurne of the solid.

Solution

Step 1; Draw a line segment across the region parallel to the axis of revolution (theline r : 2) (Fig. 5.33). Label rhe segment's height (shell height), distance tuom rheaxis of revolution (shell radius), and width (in this case, dx). (We drew the shell inFig. 5.34, but you need not do that.)

Step 2: The limits of integration: r runs from a :0 to b = 1.

Step 3:

i

l r ne

Shellradius

I i n e , = 2 S h e l l h e i r h t

" l

.;

InteI.''aL ofrntegratron

i . i .,r The shell swept out by the l inesegment in F ig .5 .33 .

, = [ ' z , | , \he l i

) / . ' l . l l ) a ,^Jd \ raorus / \nergn t /

- lo ' r r t r -x) ( r -x2)c t r

=2, fo ' {z - ,

-2x '+x3)dx

13ir

6

Shell

lnterval of integf ation

. r + l + 2 - v

5.4 Cy l indr ica l She l ls 391

Table 5.1 summarizes the washer and shell methods for the solid generated byrevolving the region bounded bY ) : x and ) : 12 aboui the coordinate axes. Forthis particular region, both methods work well for both axes of revolution. But thisis not always the case. When a region is revolved about the y-axis, for example,and washers are used, we must integrate with respect to l. However, it may not bepossible to express the integrand in terms of ). In such a case, the shell methodallows us to integrate with respect to x instead.

The washer and shell methods for calculating volumes of solids of revolutionalways agree. In Section 6.1 (Exercise 52), we will be able to prove the equivalencefor a broad class of solids.

Table 5.1 Washers vs. shells

The region bounded by

or

GENERATING SEGMENT -L TO AXIS: WASHERS.

vGENERATING SEGMENT ll TO AXIS: SHELLS.

]

,/

\>' : l ' : R ; ' r

l. t .

J" - -sh" l ih " ishr= f i -nt -

\ , - - - \ ,

N:f=."tt'l' t.'l'l^\ \_ 2z

\ l s

Fr-I

zrlx)[x xt),t^: !


Exercises 5.4In Exercises 1-6, use the shell method to find the volumes of the

.otiO. g"n"rut"O ty revolving the shaded region about the indicated

axis.

]

6. The )-axls

Use the shell method to find the volumes of the solids generated by

revolving the regions bounded by the curves and lines in Exercises

7-14 about the )-axis.

7 . y : x , Y : - x 1 2 ' x : 2

8 . y : 2 x , Y = t l 2 , x : I

9 . , : t ' , Y : 2 - x , . r : 0 , f o r ' r Z 0

1 0 , ) : 2 - x 2 , Y = f , r : 0

l l . y : 1 Q , y : Q , : r : 4

1 2 . y : 2 x - 1 , l : J i , t = O

7 3 . y : l l x , ) : 0 , . x : 1 1 2 , x = 2

r t . y : 3 1 1 2 5 1 , ! : 0 , x : r , x : 4

Use the shell method to find the volumes of the solids generated by

revolving the regions bounded by the curves ald lines in Exercises

15-22 about the t-axis

1 5 . j { = . / t , a - - Y ' Y : 2

1 6 . x : ; l 2 , a = - Y , Y : 2

l J . 7 : ) y - y 2 , x : 0 1 8 . x : 2 Y - Y 2 , x : Y

1 9 . y : 1 . x , y : 1 2 0 ' Y : a , Y : 2 x ' Y : 2

21 . r - J r , Y=O, Y=x -222 . y=aE , y :O , ! : 2 - x

In Exercises 23 and 24' use the shell method to find the volumes ofthe

solids generated by revolving the shaded regions about the indicated

at(es,

23. a) The r-axisb) The line Y: tc) The tine )' : 8/5d) The line Y: -2l5

24. a) The i-axisb) The l ine ) :2c) The line Y = 5d) The line 1: -5l8

9,I'V r' + 9

In Exercises 25-32, find the volumes of the solids generated by re-

volving the regions about the given a'{es If you thin} it would be

better;o use disks or washers in any given instance' feel free to do

so.

25. The triangle with vertices (1, 1), (1, 2), and (2, 2) about (a) the

x-axis; Oi the )-aris; (c) the line t - l0/3; (d) the line;v: I

-

26. The region in the lirst quadrant bounded by the curve.r - I - y3and the ]-a{is abour (a) the r-axis; (b) the line l: 1

27. The region in the first quadrant bounded by .r : y - y3,.r : I,and )' : I about (a) rhe x-axis; (b) the y-axis; (c) the line jr : l;(d) the line I - I

28, The triangular region bounded by the lines 2J : x l 4, ! : x,and j = 0 abour (a) the r-axis; (b) the y-axis; (c) the line x : {;(d) the line y : 8

29. The region in the first quadrant bounded by y :;r and ) = 4.rabout (a) the i-axis; (b) the line y : 8

30. The region bounded by y: uE and ) = x2/8 abour (a) the _raxis; (b) the ]-axis

31. The region bounded by )-:2x - x2 and 1:r n5eu1 1uy,5"J,-axis; (b) the line r : 1

32. The region bounded by y : ^,tr, y :2, t: 0 about (a) rhe .r-axis; (b) the y-axis; (c) the tine r - 4; (d) the line ) :2

33. The region in the first quadrant that is bounded above by thecurve ), : 1/rrl4, on the left by rhe line r : 1/16, and below bythe line ): l, is revolved about the _r-axis to generate a solid.Find the volume of the solid by (a) rhe washer method; (b) rheshell method.

34. The region in the first quadrant that is bounded above by thec\$ve y : I/rE, on rhe left by the lirle y : y14, and below bythe line ) - I is revolved about the ),axis to generate a solid.Find the volume of the solid by (a) the washer methodl (b) theshell method.

3 5 . L e r / r . r r - | G i n I t 7 * o ' ' r ^ 1 E

l r . x : u .

a) Show that f(.r) : sin r,0: r : r i.b) Find the volume of the solid generated by revolving the

shaded region about the l-axis.

5.5 Lengths of Plane Curves 393

0 < r = T / 4

a) Show that r8(.x): (tanx)2,0 <x =n14.b) Find the volume of the solid generated by revolving the

shaded region about the )-axis.

The region shown here is to be revolved about the,-axis togenerate a solid. Which ofrhe methods (disk, washer, shell) couldyou use to nnd the volume of the solid? How many integralswould be required in each case? Explatn.

]

The region shown here is to be revolved about the y_axis togenerate a solid. Which ofthe merhods (disk, washer, shell) couldyou use to find the volume of the solid? How many integralswould be rcquired in each case? Give reasons for vour answers.

36. Let s(x) - {;'"*l 'z'.

Suppose that the function /(r) is nonnegative and continuous forx - 0. Suppose also that, for every positive number ,, revolvingthe region enclosed by the graph ofJ the coordinate axes, andthe line r - , about the )-axis generates a solid of volume 2n03.Find /(.x).

Lengths of Plane CurvesWe approximate the length of a curved path in the plane the way we use a ruler toestimate the length of a curved road on a map, by measuring from point to pointwith straight-line segments and adding the results. There is a limit to the accumcyof such an estimate, howevet imposed in part by how accurately we measure andrn part by how many line segments we use.


With calculus we can usually do a better job because we can imagine uslng

straighlline segments as short as we please, each set of segments making a poly-

gona-l path that fits the curve more tightly than before When we proceed this way'

iuitt u .-oottl cuNe, the lengths of the polygonal paths approach a ljmit we can

calculate with an integral.

The Basic FormulaSuppose we want to find the length of the curve y: /(;) from. x = a to x: b'

We pafition [4, b] in the usual way and connect the conesponding points on the

curve with line segments to form a polygonal path that approximates the cuNe (Fig'

5.35). lf we can find a formula for the length of the path' we will have a formula

for approximating the length of the curve.

5.35 A typical segment PQ of a polygonal path approximating the curve AB'

The length of a typical line segment P0 (see the figure) is

The lensth of the curve is therefore approximated by the sum

\-.44'-)' + (^r,-t. (1)/ - \

We expect the approximation to improve as the partition of [4, b] -becomes

finer'

and we would liie to show that the sums in (l) approach a calculable limit as the

norm of the partition goes to zero. To show this, we rewrite the sum in (1) in a

form to which we can apply the Integral Existence Theorem from Chapter 4' Our

starting point is the Mean Value Theorem for derivatives'

DefinitionA function with a continuous first derivative is said to be smooth and its

graph is called a smooth curve'

If/is smooth, by the Mean Value Theorem there is a point (ct' /(cr)) on the

curve between P and O where the tangent is parallel to the segment P0 (Fig 5 36)

At this point

Ayr.f '@n: ^x , ,

(Ax1)2 * (Ayr

Tangentparallelto chord

Gt, fGr))

r l t c t t *

[-- Aju+l

5.36 Enlargement of the arc PQ in Fig. or Ay1 : / '(c1)A,r1.

5.5 Lengths of Plain Curves 395

With this substitution for Alr, the sums in (l) take the form

\ - , ^ " t r - , / ' , - . , / ' \ - - : - - - - - - - - - - - - - \ l ( ( r r r ' r r r l

L \ ' - . . , , . J . . ̂ , 1 - \ r ) ' - L \ /

| | ( J ( c i ) ) ' A x a .

Because ./1 a 1y1x112 is continuous on [4, D], the limit of the sums on the rightas the norm of the partition goes to zero it I: Jl+ U\iYrh. We define thelength of the curve to be the value of this integral.

DefinitionIflis smooth on [a, D], the length of the culve ] : /(.r) from a to b is

L: J" Jr + c\,ly ay. (2)

EXAMPLE I Find the length of the curve

4av : - - x ) ' ' - 1 . 0 < x 1 1 .' . t

Solution We use Eq. (2) with a :0, b: 1, and

4J2 .,,,,' _ 3 "

d y 4 A 3 , , ^ / ^ , )E = I

. Z . *

: z r z x ' -

. , , 2(o r \ : ( z , f zx t , z \ , : s^ .\ d x / \ /

The length of the curve from x : 0 to .r : I is

ffi,,:1"'

, ra r : I

l r 1 1

t " o

I _I

"4+ u a" Eq. ( l ) wi lh

lntcgrirlc- itnd

l r s f . D

Deaf ing with Discontinuities in dyldxAt a point on a curve where dy / d x fails to exist, dx f dy may exist and we may beable to find the curve's length by expressing r as a function of y and applying thefollowing analogue of Eq. (2):

Formula for the Length of a Smooth Curve x = g(J), c 1y I d

Jr + rcoy ay. (3)

] . ] , t * r - , , , , ' ]


5.37 The graph o l y = (x l2 l '1 f rom-x=0tox=2 is ; l so the qaph o I x =2Y11 'f r o m y : 0 t o Y : 1 .

EXAMPLE 2 Find the length of the curve y: (xl2)2/3 fromx:0tox:2'

Solution The derivativeav -2 - (^ \ - ' " / l \ | / 2 \ ' "a-= i \z) \z / : : \ ; /

is not defined at .r :0, so we cannot find the cuwe's length with Eq (2)'

We therefore rewrite the equation to express x in terms of )'/ \ 2/3

} : l ; ,

. / - 2

- - ) , ,312

Fromthisweseethatthecuwewhoselengthwewantisalsothegraphoft:21312f r o m y : g t o y : 1 ( F i g . 5 . 3 7 ) .

The derivatived x , / ' l \

; : ' \ i )Y "z - 3Y ' / z

is continuous on [0, l]. We may therefore use Eq. (3) to find the cuve's length:

lo lhe po$er 3/1.

Sol \e lo r .

[ q . ( ] ) \ \ i l l r

L e t l l : l + 9 . r .

subst i tL(c brck.

L: J.(4)dydx

are often written with differentials instead of derivatives. This is done formally by

thinking of the derivatives as quotients of differentials and bringing the dx ar.,d dy

insicle the radicals to cancel the denominators. In the first integral we have

In the second integral we have

,ta,, +a

Thus the integrals in (4) reduce to the same differential formula:

': l"' ,FW at: l"'1..wa:'- - ! .?, r+ 9u)3/ ' l '

9 l - l n

: -_: (10J10 - r) x 2.27.27 '

The Short Differential FormulaThe equations

(*) ' "= L'l'.(n'

'.(E)'f b -

L : J" Jdx? +drz.

5.38 Diagram for remembering theeguation ds = Jdx2 + dyz.

5.5 Lengths of Plain Curves 397

Of course, dx and d;t must be expressed in terms of a common variable, andappropriate limits of integration must be found before the integration in Eq. (5) isperformed.

We can shorten Eq. (5) still further Think of dr and dy as two sides of a smalltriangle whose "hypotenuse" is ds : Jdx2 -t cly2 (Frg.5.38). The differential dsis then regarded as a differential of arc length that can be integrated betweenappropdate limits to give the length of the cuNe. With /d"x, + dy2 set equal tods, the integral in Eq. (5) simply becomes the integral of ds.

DefinitionThe Arc Length Differential and the Differential Formulafor Arc Length

L : l ds

diferential formulafor arc lenglh

r.- j

Curve I Curve 2 Cur,re 3 Curve 4

5.39 The first four polygonalapproximations in the construction ofHelga von Koch's snowflake.

1 2

The length of C1e is nearly 40 and the length of C1s0 is greater than7,000,000,000,000. The lengths grow too rapidly to have a finite limit. Thereforethe snowflake curve has no lenglh. or. if you prefer. infinite length.

What went wrong? Nothing. The formulas we derived for length are for thegraphs of smooth functions, curves that are smooth enough to have a continuouslyturning tangent at every point. Helga von Koch's snowflake curve is too rough forthat, and our derivative-based formulas do not apply.

Benoit Mandelbrot's theory of fractals has proved to be a rich source of curveswith infinite length, curves that when magnified prove to be as rough and varied asthey looked before magnification. Like coastlines on an ocean, such curves cannotbe smoothed out by rnagnilication (Fig. 5.40, on the following page).

n

1/1 \ ' '

" \ 3 /

3

(i) ,(i)'3

ds -- /n4ayz

arc lengdrdifferential

:i< Curves with Infinite LengthAs you may recall from Section 2.6, Helga von Koch's snowflake curue K is thelimit curve of an infinite sequence C1, C2,...,C^,... of "triangular" polygonalcuwes. Figure 5.39 shows the first four curves in the sequence. Each time weintroduce a new vertex in the construction process, it remains as a veltex in allsubsequent curves and becomes a point on the limit curve K. This means that eachof the C's is itself a polygonal approximation of K-the endpoints of its sides allbelonging to K. The length of K should therefore be the limit of the lengths of thecurves C,. At least, that is what it should be if we apply the definition of lengthue developed for smooth curves.

What, then, is the limit of the lengths of the curves C,,? If the original equilateraltriangle C1 has sides of length 1, the total length of C1 is 3. To make C2 from C1,we replace each side of C1 by four segments, each of which is one-third as longas the original side. The total length of C2 is therefore 3(413). To get the length ofC3, we multiply by 4/3 again. We do so again to get the length of Ca. By the timewe get out to C,, we have a curve of length 3(4/3)' I.

Curve Number

Length


5-40 Repeated magnifications of afractal coastline. Like Helga Von Koch'ssnowflake curve, coasts like these are toorough to have a measurable length.

ar/rtaar/tta

Exercises 5.5

Finding Integrals for Lengths of CurvesIn Exercises 1-8:

1 0 . y : a t l z f r o m t = 0 t o , r = 4

l l . x = ( y 3 / 3 ) + l / ( 4 ) , ) f r o m ) = l t o ) : 3(Hint: 14 (dtt ld.l)2 is a perfect squarc.)

1 2 . x : O 3 / 2 p ) - y r l 2 f r o m ) : 1 t o ) : 9(Hint: | + (di( ldy)2 is a perfect square.)

13 . r : (y4 /4 ) + 1 / (8y '? ) f tom y : l tov :2(Hi t: | + (da /dy)z is a perfect square.)

l ! . a : (y3 /6 ) + l / (2y ) f rom y :2 toY=3(Hint: 1 + (dx ldy)z is a perfect square )

15 . y : (314)xa t1 - (3 /8 )12 /3 +5 , 13x =8

1 6 . y : ( x 3 1 3 ) + x 2 + x + L / ( 4 x + 4 ) , 0 = x = 2

a, Set up an inte$al for the length of tle curveSt t Crapl the curue to see whal it looks l ike.S c; Use your grapher's or computer's integml evaluator to find the

curve's length numericallY.

l . y : x 2 , - l = x = 2

2 , J : t u n " , - 1 r / 3 = a = 0

3 . ; 1 = s i n y , O : . t < 1 r

a . r : r 4 - y z , - l l 2 3 y = 1 1 2

5 . y 2 + 2 y : b c + 1 f t o m ( - 1 , - l ) t o ( 7 , 3 )

6 , ) = s i n x - . r c o s . r , 0 = t 5 t i

7 . v = I t a n t d t , 0 < x S n / 6

?| -

8 . x - I J s e d r - l d r , n l l 1 y = n 1 4

Finding Lengths of CurvesFind the lengths of the curves in Exercises 9-18. If you have a grapher'

you may want to graph these curves to see what they look like'

g . y : (1 /3 ) (x2 + 2)3 /2 f rcm x :0 to * :3

17.

18.

t t _ -a : | / s e c ' r - 1 t u . - n / 4 a ) ) 1 f t / 4

t t _y : I J 3 t ^ t d r . - 2 = x s - l

J 2

19. a) Find a curve through the point (1, l) whose length integral(Eq. 2) is

r a Ft : | , l r - ) 'a '

Jr Y 4x

b) How many such cuNes are there? Give reasons for your

answer.

a) Find a curve through the point (0, 1) whose length integral(Eq. 3) is

t ! I IL : I t l t + ^dy .

J l v ,

b) How many such curves are there? Give reasons for youranswer.

Find the length of the curvet ' -

y : I t / c o s 2 r d ,J N

f r o m i = 0 t o x = n / 4 .

The length of an astroid. The graph of the equation .r2l3 +)2/3 : 1 is one of a family of curves called astroids (not "as-teroids") because of their starlike appearance (see the accom-panying figure). Find the length of this particular astroid byfinding the length of hall the f,rst-quadrant portion, ):e , x2/\3/2, \f2/4 < r : l, and multiplying by 8.

Numerical IntegrationYou may have wondercd why so rnany of the curves we have beenworking yl{!311unusual fomulas. The reason is that the squarc.oot Jl + (dy ld.x)2 that appears in the integals for length and sur-face area almost never leads to a function whose antiderivative wecan find, In fact, the square root itself is a well-known souce ofDonelementary integrals. Most integrals for length and surface areahave to be evaluated numerically, as in Exercises 23 and 24.

23. Your metal fabdcation company is bidding for a contract to makesheets of corrugated ircn roofing like the one shown here. Thecross sectiotrs of the cofiugated sheets are to conform to the curve

ady : s i n 1 . x . 0 < r < 2 0 i n .' 2 0

If the roofing is to be stamped from flat sheets by a processthat does not stretch the material, how wide should the originalmaterial be? To find out, use numerical integration to approximatethe length of the sine cufle to 2 decimal places.

Exercises 5.5 399

Your engineering flrm is bidding for the contact to constructthe tunnel shown here. The tunnel is 300 ft long and 50 ft wideat the base. The cross section is shaped like one arch of thecurve y : 25 cos (zr/50). Upon completion, the tunnel's insidesurface (excluding the roadway) will be treated with a waterproofsealer that costs $1.75 per squaie foot to apply. How much willit cost to apply the sealer? (Hint: Vse numerical integration tofind the length of the cosine curve.)

24,

) ] : 25 cos (7tl50)

.r (fu

Theory and Examples25, Is there a smooth curve y : /(r) whose length over the interyal

0 = r : a is always eD,a'!Give reasons for your alswer.

26. Using tangent fins to detive the length formula for curves.Assume/is smooth on [a, r] and partition the interval [a,,] inthe usual way. In each subinterval [rk-r, rr] consmrct the targentl'? at the point (xr-r, "f(xr-r)), shown in the flgure.

a) Show that the length of the ,tth tangent fin over the interval[.rr-r..rr] equals v4d,'t+Tfi l lr l;;;t.

b) Show that-i- rL' -

l im ) t lengrh o f k th tangent nnr : / / l + t I t ^ ) t2dx .J"

which is the length I of the curve y : f(x) fuom a to b.

Original sheet


S CAS Explorations and ProjectsIn Exercises 27-32, use a CAS to perform the following steps for thegiven curve over the closed interval

Plot the cuNe together with the polygonal path approximationsfor n :2, 4, 8 partition points over the inteNal. (See Fig. 5.35.)Find the corresponding approximation to the length of the cuNeby summing the lengths of the line segments.Evaluate the length of the curve using an integral. Compare yourapproximations for /, : 2,4, 8 to the actual length given by the

integal. How does the actual length compare with the approxi-mations as n increases? Explain your answer

f (x) = {l-7, I : ' s I

f ( x ) : r t / 3 + t 2 / 3 , 0 . t 5 2

/ ( r ) : s in (z i '? ) , O=t =J1

f ( r ) : t 2 c o s x , o < x = 1 tY - l l

"f (xt: -::------. -; S,r S t

4 . r ' t r z

f ( x ) : 1 3 - x 2 , - l : i r 5 1

a)

b)

27.

28,

29,

c)

30.

31.

1.,

5.41 The surface generated by revolvingthe graph of a nonnegative functiony : f (x l ,aSX3b, about the x -ax is . Thesurface is a union of bands l ike the oneswept out by the arc PQ.

542 The line segment joining P and Qsweeps out a frustum of a cone.

Areas of Surfaces of RevolutionWhen you jump rope, the rope sweeps out a surface in the space around you, asurface called a surface of revolution. As you can imagine, the area of this surfacedepends on the rope's length and on how far away each segment of the rope swings.This section explores the relation between the area of a surface of revolution andthe length and reach of the curve that generates i1. The areas of more complicatedsurfaces will be treated in Chapter 14.

The Basic FormulaSuppose we want to find the area of the sufac€ swept out by revolving the graph ofa nonnegative function y : f (x), a : .x : ,, about the r-axis. We partition [a, b]in the usual way and use the points in the partition to partition the graph into shortarcs. Figure 5.41 shows a typical arc PQ and the band it sweeps out as part of thegraph of I

As the arc PQ revolves about the r-axis, the line segment joining P and Qsweeps out part of a cone whose axis lies along the;r-axis (magnified view in Fig.5.42). A piece of a cone like this is called a frustum of the cone, J?rslrrm beingLatin for "piece." The surface area of the frustum approximates the surface area ofthe band swept out by the arc P0.

The surface area of the frustum of a cone (see Fig. 5.43) is 2z times the averageof the base radii times the slant height:

Frustum surface arca : 2r 'a+

L : r (rt + r)L.

For the frustum swept out by the segment P0 (Fig. 5.aa), this works out to be

Frustum sur face area - r t f (x1 , t + " / t " , t l . / ta* , ) , + tay, l .

The area of the original surface, being the sum of the areas of the bands sweptout by arcs like arc P_Q, is approximated by the frustum area sum

D, ff 6r) + /(xr)).u{A+f + (AvrP. ( 1 )k=1

We expect the approximation to improve as the partition of [a, l] becomes fine1and we would like to show that the sums in (l) approach a calculable limit as thenorm of the Dartition soes to zero.

1

t

5.6 Areas of Surfaces of Revolution 401

The important dimensions of the frustum in Fig.5.42.

Segment lengthl

z =.,(.t4-l{i

To show this, we try to rewrite the sum in (l) as the Riemann sum of somefunction over the interval from ri to b. As in the calculation of arc length, we beginby appealing to the Mean Value Theorem for derivatives.

Ifl is smooth, then by the Mean Value Theorem, there is a point (c1, /(c1)) onthe curye between P and Q where the tangent is parallel to the segment pO (Fig.5.45). At this point,

f ' (cr) : 4 l( .

Av1 : /,(c1)A.r1.

With this substitution tbr A).r, the sums in (l) take the form

v? t)'+ f1.J^'^).

- L " , 1 r . r , 1 r - / r . r r , l u / t . , 7 ' c r r : A r , . ( 2 )l - l

At this point there is both good news and bad news.The bad news is that the sums in (2) are not the Riemann sums of any lunction

because the points r1 1, ,ti, and c1 are not the same and there is no way to makethem the same. The good news is that this does not matter. A theorem called Bliss,stheorem, fiom advanced calculus, assures us that as the norm of the partition of[zr. b] goes to zero, the sums in Eq. (2) converge to

Just the way we want them to. We therefore deline this integral to be the area ofthe sudace swepr our by the graph of/ fiom .? ro D.

DefinitionThe Surface Area Formula for the Revolution About the x-axisIf the function ./(-r) > 0 is smooth on fa, bl, the area of the surface gen_erated by revolving the curye l :.f(x) about the x axis is

1A\

I

P . \ -^ [ /r- ,/: - \ r ' = r1 r )-\i-r >1

r i r

f._.u.i_l

Dimensions associated with the drcand segment PQ.

Tangentparal le l

' r - r ' r \l'-J-rr-l

l f f is smooth, the Mean ValueTheorem guarantees the existence of apoint on arc PQ where the tangent isparallel to segment PQ.

Gi.lG;))

s: fu 2 , ,,F6 rbd:( :

J . 27t f (x)J t+( f , (x) )2dx. (3)

r =2"[;


546 Example 1 calculates the area ofthis surface,

5,47 Revolving l ine segment AB aboutthe y-axis generates a cone whose lateralsurface area we can now calculate in twodifferent ways (Example 2).

The square root in Eq. (3) is the same one that appears in the formula for the lengthof the generating curve.

EXAMPLE I Find the area of the surface generated by revolving the curvey:zJi,1 = x =2, about the x-axis (Fig. 5.a6).

Solution We evaluate the formula

with

,: 1"",,fr(fj o. ,q r,

y :2"tr,

r -- ; t-----:--7 /------i---il , t l x + L v x + r

Y r V x \/x

With these substitutions,

s= [ 'zn. rJ ig! a* :+n [ 'u iardxJr Jx J t

' t 12 Rr: +n . ' r t x+ l ) i / ' ?J r : ; (3 . , / 3 2J2 \ .

Revolution About the y-axisFor revolution about the y-axis, we interchange -r and / in Eq. (3).

dv1dx

"E'

FW:

D

Surface Area Formula for Revolution About the '-axis

If -r : g(y) > 0 is smooth on [c, d], the area of the surface generated byrevolving the curve r : g(y) about the y-axis is

z" g1y1,/1 + ({ (yD, ay. (4)fr19',:1"'EXAMPLE 2 T:he line segment x :1- y,0: l : 1, is revolved about the

:t-axis to generate the cone in Fig. 5.47. Find its lateral surface arca.

Solution Here we have a calculation we can check with a formula from geometry:

^ base circumference /^Laterat surace area : -------

- X Stant nelgnl : z{z.

'.'(+)'

5.6 Areas of Surfaces of Revolution 403

To see how Eq. (4) gives the same result, we take

c : 0 , d : 1 , , d x

x . : | _ J , d y : _ t ,

and calculate

s : 1"" zo'T - , z t l:z"rtlt- i),

: "J1.

: r "J . (1- ] ' )

The results agree, as they should.

The Short Differential FormThe equations

D

s: 1"" zo',Fe,J',

'. (#)' r+ \- \ 'z : r t

': 1."'rl'.(H)'

5.48 The area of the surface swept outby revolving arc AB about the axis shown.

" b ^nele ts l2 zltpos, Ine exacr expresstondeoends on the formulas for p and ds.

are often written in terms of the arc length differential ds : Jdx\,lyz as

l b l dS:

J. ht ds and s:

J" zzx ds.

In the first of these, y is the distance from the x-axis to an element of arc lengthdr. In the second, .r is the distance from the y-axis to an element of arc length ds.Both integmls have the form

t fs:/ 2z(radiug(band width) :

Jzrods,

where p is the radius from the axis of revolution to an element of arc length ds(Fig. 5.48).

Axis ofrevolution


If you wish to remember only one fomulafor surface area, you might make it the shortdifferential form.

549 The surface generated by revolvingthecurvey=x3,6 3 x 3112, about thex-axis could be the design for achampagne glass (Example 3).

Short Differential Form

s: I zoo a,

In any particular problem, you would then express the radius function p and the arclength differential ds in terms of a ctimmon variable and supply limits of integrationfor that variable.

EXAMPLE 3 Find the area of the surface generated by revolving the curvey : .r3, 0 < x .< | 12, about the .{-axis (Fig. 5.49).

so/ution We staft with the short differential form:

s: I zooa,

: I z"t a'f _

: I

2ryJdxz + dy?.

We then decide whether to express dy in terms of dx or dx in terms of dy. Theoriginal form of the equation, y : .r3, makes it easier to express dy in terms of dr,so we continue the calculation with

! : x3 , dy :3x2 d;c, and Jdr, + dy, : I dx, + Ax, dil,

:uEaXnar .

With these substitutions, x becomes the variable of integration andrx=l /2

S : , h t y \ / d x z + d y 2Jr=o

r t / 2: I znx3^,/ t+gxadx

Jo

/ | \ / 2 \ . . . . 1 , / , s , . L . r j r | l e , , _ | e r . . , / r r r 0 _: z t t l ) l l i l t t + g t o t ' r ' | . ' , r . ' . n t c t ' . r c . r n J \ , , b \ , i r , , e

\ J6 l \ J / l n b i . k .

/ r25 \l* -' j

As with arc length calculations, even the simplest curves can provide e workout.

t t

For revolutior aboulthe r-nxi \ . lhc radius

,i.l : .,?-" +./r'

- f r o \ 3 / 2 I

; l \ ' . ;6) -1.]

n f lz5frz -l r(r L\c/

-',1: u61n17?a'

Exercises 5.6 405

a)

L

)

4.

5.

o.

7.

8.

Exercises 5.6

Finding Integrals for Surface AreaIn Exercises 1-8:

Set up an integral for the area of the surface generated by re-

volving the given cuwe about the indicated axis.

Graph the curve to see what it looks like. If you can, graph the

surface, too.

Use your grapher's or computer's integral evaluator to find the

sur-face's area numerically.

v : t a n r . 0 < - r < r 7 / 4 : r - a x i s

] - 1 2 , o 5 r : 2 ; ' r - a x i s

. r _ r , : 1 , l : l ' : 2 ; ) a x i s

_ t : s i n ) , 0 = ) r : z ; ) ax i s

r j l I - I f r o m t 4 . l r t o r l . 4 r : r ' a \ i \

r + 2 - f - r . 1 < r < 2 1 r - a x i s

r - I t c n r d t , 0 y - 2 , 3 : 1 - a x i .

I t -j : l ^ / t 2 l d t , l 5 r : V 5 ; x a x r s

1 5 . y : J z r - x t , 0 . 5 : r : 1 . 5 ; - Y a x i s

16. r:.,,6T l, I 5 r.: < 5; .r-axis

1 7 . x - . , - 1 f i , 0 : ] 1 1 ; _ \ , - a x i s

1 3 . 1 = r l , 3 r y ' : I I 1 1 3 : ' - e r i s

1 9 . r - 2 / - ' + - v . 0 a | 1 5 . 4 : f a x i .

r - tzt=1, 5/8 : t': l; ),axis

.t

21 , x :01 lq+ 1 / (8 r ' ' ? ) , 11y12 ; r - ax i s (H l r r . ' Exp ress

,ts : l7tz + d\t in terms of d), and evaluate the integral S =

/ 2zl ds with appropriate limits.)

22. -,- = Ol3)(x2 +2)tit, 0 <.r i 'D; y axis (lli,rt. Express

,lt : Jth\ ay'? in terms of dx, and evaluate the integral S :

J 2nt ds wirh appropriate limits.)

23, Testing the new definition. Show that the surface area of a

sphere of radius a is still 4zar by using Eq. (3) to find rhe area

of the surface generated by revolving the curve r : 161lll,a = x = a, about the ,r-axis.

24. Testing the new definition. The lateral (side) surface area of

a cone of height l? and base radius r should be rr"'G\ h2,

the semipeimeter of the base times the slanl height. Show that

this is still the case by linding the area of the surface generated

by revolving the l ine segment r = (r lh)x,0 = x:, t , about the

r-axis.

25. a) W.ite an integral lbr the area of the surface generated by

revolving the cuNe ) - cos r. -1t 12 : x = 1/2, about thej-axis. In Section 7..1 we will see how to evaluate such

integrals.

E r; calcuraroR Find the surface area numerically.

20.

Finding Surface Areas9. Find the lateral (side) surface atea of the cone Senerated by

revolving the line segment ) : -r/2. 0 = t : '1, about the i-axis

Check your answer with the Seometry formula

ILateral surface area : : x base circumference x slant height

10. Find the lateral surface area of the cone generated by revolving

the l ine segment y' : xl2.O: t :4 about the )]-axis. Check

your answer with the geometry formula

tLateral surface aJea -

i x base circumference x slant height.

11. Find the surface area of the cone irustum generated by revolving

rhe line se8ment ) : (x 12) + (l l2), 1 : r 5 3, about the x axis.

Check your result with the geometry formula

Frustum surface area - n(rr + /r) x slant height.

12, Fjnd the sudace area of the cone frustum generated by revolving

rhe l ine segment ) : (xl2) + (112),I : r :3, about thel '-axis.

Check your result with the geomefty formula

Frustum surface area = r(.rt + r:) x slant height.

Find the areas of the surfaces generated by revolving the curves in

Exercises 13 22 about the indicated axes. If you have a grapher, you

may want to graph these curves to see what they look like.

13. r - , x '19, 0 : .r : 2; r-axis

l . l . _ r "G .

l , q _ r ' l 5 4 : \ r x i .


26. thc tu[faG af an astroid Find the area of the surface gen-

erated by revolving about the r axis the poition of the astoidr2l3 + )2/:r : I shown here. (Hirlt: Revolve the filst-quadrantportion ) : (l ir2lr)3/2, 0 < ). = I, about the r-axis and doubleyour result.)

@ 27, in;tmelincy vroks. Your company decided to put out a deluxeversion of the successful wok you designed in Section 5.3, Ex-ercise 41. The plan is to coat it inside with white enamel and

outside wjth blue enamel. Each enamel will be sprayed on 0.5mm thick belore baking. (See diagram here.) Your manufacturingdepartment wants to know how much enamel to have on handfor a production run of 5000 woks. What do you tell them? (Ne-

glect waste and unused mateial and give your answer jn liters.Remember that I cmj : I mL, so I L - 1000 cml.)

29. The shaded band shown here is cut ftom a sphere of radius R byparallel planes ft units apafl. Show that the sudace area of theband is 2z Rft.

30, Here is a schematic drawing of the 90 ft dome used by the U.S.National Weather Service to house radar in Bozeman, Mont.

a) How much outside surface is there to paint (not countingthe bottom)?

ffi u) calcuraroR Exprcss rhe answer ro rhe nearesr squareroot,

a

?lTi

+sr t l ̂I I Lenler.. i , / Radiusi\ ----l 45ft

225 ft | \

€'r t .

\, 1-r4 r

h'I

*2 +1'2 = i62 = 256

Slt.tng bread. Did you know that if you cut a spherical loafof bread into slices of equal width, each slice will have thesame amount of crust? To see why, suppose the semicircle y =

.n42 "2

shown here is revolved about the ,r axis to generate

a sphere. Let AB be an arc of the semicircle that lies above aninterval ot' length I on the r-axjs. Show that the area swept outby AB does not depend on the location of the interval. (It doesdepend on the length of the inteNal.)

t:

Surfaces generated by curves that (rcss the axis of revolutiot. The surface area fbrmula in Eq. (3) was developed underthe assumption that the function / whose graph generated thesudace was nonnegative over the interval [a, b]. For cuNes thatcross the axis of revolution. we replace Eq. (3) with the absolutevalue formula

t l5 = | 2ttpt ls: | 2n f(x) ds. (5)

.t J

Use Eq. (5) to find the surface area of the double cone generatedby revolving the line segment 'y : x, I : * : 2, about the n-axis.

(Exercise 31, co tinue.l.) Find the area of the surface generatedby revolving the curye )' - ,'19. Ji =

":! J5, about the -r

axis. What do you think will happen if you drop the absolutevalue bars f.om Eq. (5) and attempt to find the surface area withrhe formula S : .f 2ft f (x) ds insread? Try it.

31.

) (cm)

5.7 Moments and Centers of Mass 407

Numerical IntegrationFind, to 2 decimal places, the areas of the surfaces generated byrevolving the curyes in Exercises 33-36 about the ;r-axis.3 3 . y : s i 1 y , 0 = x = n

3 4 . y = a 2 1 4 , 9 . r . 2

35. y=yasi1 2x, -2jt/3 Sr.32n/3 (the curve in Section3.4, Exercise 5)

x /-:- ^Jo . y : l2VJ6-

.xz . l ) 1x : .6 ( rhe sur face o f rhe p lumb bob inSection 5.3, Exercise 44)

37. An alternative derivation of the surfa.e area formula. As_sume/is smooth on [a, r] and parrition [d, D] in rhe usual way.In the *th subinterval [.rr t,tr] construct the tangent line to thecurye at the midpoint mk : (ak 1 + ri/2, as in the figure here.

a , Show |nar r t : I tmt , l - 1 t . r t \ - and 12 : f tnp) -

" , . Lx*J t n k )

) .

b) Show that the lengrh It oj !!9_jg!gqll4gj!g!q!-nt in theIth subinterval is Lk : lra{xi, + U,@k;6;..1:.

t k r f r k r k

' -^t

c) Show that the lateral surface area of the frustum of the coneswept out by the tangent line segment as it revolves aboutthe .r-a.r,is is 2n Itmttrl l - t f ' tmrt txt.

d) Show that the area of the surface generated by revolving) : /(.,r) abour the r-axis over [a, ]l is

f ir I(lf i l l . ' :*::-*) = fu 2,1,,t,1t ,(r,G)id,.,-- i=i \ or /(rh rrusrum / L

Moments and Centers of MassMany structures and mechanical systems behave as if their masses were concentratedat a single point, called the center of mass (Fig. 5.50, on the following page). It islmportant to know how to locate this point, and doing so is basically a mathematicalenterprise. For the moment we deal with one- and two_dimensional obiects. Ihree-dimensional objects are best done wirh rhe multiple integrals of Chapier 13.

Masses Along a LineWe develop our mathematical model in stages. The first stage is to imagrne massesm1, m2, a\d m 3 on a rigid x-axis supported by a fulcrum at the origin.

Mass vs. weight

Weight is the force that results from gravirypulling on a mass. If an object of mass m lsplaced in a location where the acceleration ofgravity is & the object's weighr there is

F : m g

(as in Newton's second law).

Fulcrumat ongrn

The resulting system might balance, or it might not. It depends on how large themasses are and how they are arranged.

Each mass mk exerts a downward force mpg equal to the magnitude of thernass times the acceleration of gravity. Each of these forces has a tendency ro tumthe axis about the origin, the way you turn a seesaw. This turning effect, called

1 torque, is measured by multiplying the force mtg by the signed distance xk

from the point of application to the origin. Masses to the left of the orisin exertnegative (counterclockwise) torque. Masses to the dght of the origin exen positive(clockwise) torque.


(4,

5-50 (a) The motion of this wrench gliding on ice seems haphazard unti l we noticethat the wrench is simply turning about its center of mass as the center glides in astraight l ine. (b) The planets, asteroids, and comets of our solar system revolveabout their collective center of mass. (lt l ies inside the sun.)

The sum of the torques measures the tendency of a system to lotate about theorigin. This sum is called the system torque.

System torque : m8xt + m28x2 + m384 (1)

The system will balance if and only if its torque is zero.If we factor out the I in Eq. (1), we see that the system torque is

g(mf i t + mzxz+ m3h) .

" r"ut{. ot tt"

Thus the torque is the product of the gravitional acceleration g, which is a fea-

ture of the environment in which the system happens to reside, and the number

(mrxt I mzxz + m3r3), which is a feature of the system itself, a constant that stays

the same no matter where the system is placed.

The number (m 1.r1 I m.zxz I mzxz) is called the moment of the system about

the origin.It is the sum of the moments mt)ct, m2x2, m3x3 of the individual masses

Mo : Moment of system about origin : l*t xr

(We shift to sigma notation here to allow for sums with more terms. For lmpxp,read "summation mt times xr.")

We usually want to know where to place the fulcrum to make the system

balance, that is, at what point t to place it to make the torque zero.

Special locatioflfor balance

The torque of each mass about the fulcrum in this special location is

rorque of z1 abour ': (lr# r;::,r:

) (*il:.-.)

: (x7 -i)mpg.

When we write the equation that says that the sum of these torques is zero, we get

the system


an equation we can solve for t:

I,to ilrxi I : o

s f ( r r i )mp :6

I t r ^ r ^ i na ' t : 0

D*r r r - l r l l 1 :g

f * r r r : " l * r

This last equation tells us to lind -r by dividing the system's moment about theodgin by the system's total mass:

system moment about origin

system mass

The point t is called the system's center of mass.

Wires and Thin RodsIn many applications, we want to know the center of mass of a rod or a thin stdpof metal. In cases like these where we can model the distribution of mass with acontinuous function, the summation signs in our fonnulas become integrals in amanner we now describe.

Imagine a long, thin stdp lying along the r-axis fiom I : a to.{ : D and cutinto small pieces of mass Lmp by a partition of the interval [a, r].

The,tth piece is Arr units long and lies approximately .r1 units from the origin.Now observe three things.

First, the strip's center of mass r is nearly the same as that of the system ofpolnt masses we would get by attaching each mass Am r to the point t(:

system moment

- L,r*rL ro

t r ysystem mass

Second, the moment of each piece of the strip about rhe origin is approximatelyrnAmr, so the system moment is approximately the sum of the "rrrAl?1r:

S ls tem moment r f r - Arz t .

Third, if the density of the strip at;r1 is d("ra), expressed in tems of massper unit length, and 6 is continuous, then Az1 is approximately equal to d("rp)A4(mass per unit length times length):

Lmk x 8 (xk) Lxk '

Combining these three observations gives

_ sys tem moment f x ,Am, ) -x^ar , r , )4 . r ,\ :----i----:---:

system mass -

EL/|l -

T6(t-)Ar- \z)

Density

A material's density is its mass per unitvolume. In practice, however, we tend to useLtnits we can convenientl) measure. Forwires, rods, and nairow strips we use massper unit length. For flat sheets and plates weuse mass per unit area.


To find a center of mass, divide moment bymass.

5,51 The center of mass of a straight,thin rod or strip of constant density lieshalfwav between its ends.

, / l' l 0 l

5-52 We can treat a rod of variablethickness as a rod of variable density. seeExample 2.

The sum in the last numerator in Eq. (2) is a Riemann sum for the continuousfunction ;6(x) over the closed interval la, bl. The sum in the denominator is aRiemann sum for the function 6(;) over this interval. We expect the approximationsin (2) to improve as the stdp is partitioned more finely, and we are led to the equation

,: r;:::,,)!,l " o \ x t q x

This is the formula we use to find i.

Moment, Mass, and Center of Mass of a Thin Rod or Strip Along ther-axis with Density Function 6(r)

Moment about the origin: Mo :f.b xt{i a*

I,u u<*t o,Mo

M

(3a)

(3b)

(3c)

Mass: M :

Center of mass:

EXAMPLE 1 Strips and rods of constant density

Show that the center of mass of a straight, thin strip or rod of constant density lieshalfway between its two ends.

So/ution We model the strip as a portion of the.x-axis from J : a to x: b (FiE.5.51). Our goal is to show that t: (a + b)/2, the point halfway befween d and b.

The key is the density's having a constant value. This enables us to regard thefunction 6(.r) in the integrals in Eqs. (3) as a constant (call it d), with the resultthat

: 6 ( b - a )

,'),,: urru' - o,l"u u*o':u l, ',0,:ul)

l,' u o, : u l"' o*: , ['], =

- M o

a l b

2

d . . ,, ( b -

- q - )

6(b - ,)

The ,t s cancel in rheformula fof r. tr

EXAMPLE 2 Avariable density

The 1o-m-long rod in Fig. 5.52 thickens from left to dght so that its density, insteadof being constant, is 6(-r) : t a (x/10) kg/m. Find the rod's center of mass.


solution The rod's moment about the origin (Eq. 3a) isr ' n r l 0 " t O r 1 2 I

Mo: l t x td r - l * ( r+ - - )a ' - / ( x+ ,^ l l ax- J o J o

' l u ' J n \ t u l

f x ' * ' l ' " l oo 250 .= l r - l - r r '

- l : l g . t .

L 2 301,,

The rod's mass (Eq. 3b) is

r ) u r ' n , . x , I t t - l ' oM -

Jo drx tdx :

/ . , ( ' -

i )dx -

L , * ,o lo

:

The cenler of mass tEq. 3c) is located al lhe point

_ M 2 2 5 0 1 5 0 - . . .- I 5 . )O m.' - M 3 1 5 9

I n f L f r l \ r i L n , i r r r f r t

l 0 + 5 : 1 5 k g .

j

Masses Distributed over a Plane RegionSuppose we have a finite collection of masses located in the plane' with mass flk

at the point (rr, )r) (see Fig. 5.53). The mass of the system is

System mas.: U -L^^.

Each mass n ' has a moment about each axis. Its moment about the;r-axis is nt4y1'

and its moment about the 1-axis is zrrr. The moments of the entire system about

the two axes are5.53 Each mass mk has a moment abouteach axt5.

5.54 A two-dimensional array of massesbalances on its center of mass.

Moment about Jr-axis:

Moment about y-axis:

u" : l l n1 ,v t ,

u' : l^ t r '

The r-coordinate of the system's center of mass is deflned to be

M, L^,* ,M L m o

With this choice of .x, as in the one-dimensional case, the system balances aboutthe line,r : t (Fie. 5.54).

The y-coordinate of the system's center of mass is delined to be

- _ M^ -L:r*vr .'ML^ r (s)

With this choice of t, the system balances about the line l' : t as well The torques

exerted by the masses about the line y : y cancel out. Thus, as far as balance is

concemed, the system behaves as if a1l its mass were at the single point (t' l') We

call this point the system's center of mass.

Thin, Flat PlatesIn many applications, we need to find the center of mass of a thin, flat plate: a

disk of aluminum, say, or a triangular sheet of steel ln such cases we assume the

distribution of mass to be continuous, and the formulas we use to calculate t and

t contain integrals instead of finite sums. The integrals arise in the following way,


.l srrin ^f m,.s AD

r.: i j A plate cut into thin strips parallelto the y-axis. The moment exerted by atypical strip about each axis is themoment its mass Am would exert ifconcentrated at the strip's center of mass0,,i1.

i 5j The plate in Example 3.

Imagine the plate occupying a region in the 4)-plane, cut into thin sffips parallelto one of the axes (in Fig. 5.55, the y-axis). The center of mass of a typical stripis (i, !). We tleat the strip's mass Aru as if it were concentrated at (;, i). Themoment of the strip about the /-axis is then iAn. The moment of the strip aboutthe r-axis is iA.,??. Equations (4) and (5) then become

_ M ,r - M '

Li t *LL-

As in the one-dimensional case, the. sums are Riemann sums for integrals andapproach these integrals as limiting values as the strips into which the plate is cutbecome nanower and nanower. We write these integrals symbolically as

M,, )-i Alrr' - D r \ _ ^ * '

l i d mt - --v---

.l am

(. r ,2)

and

To evaluate these integrals, we picture the plate in the coordinate plane andsketch a stdp of mass parallel to one of the coordinates axes. We then express thestrip's mass dm and the coordinates (;, i) of the stxip's center of mass in termsof x or y. Finally, we integrate j dm, i dm, and dm between limits of integrationdetermined by the plate's location in the plane.

EXAMPLE 3 The triangular plate shown in Fig. 5.56 has a constant density of6 :3 glcm2. Find (a) the plate's moment M) about the 1,-axis, (b) the plate's massM, and (c) the r-coordinate of the plate's center of mass (c.m.).

Solution

Method 1: Vertical strips (Fig. 5.57).

a) The moment Mr: The typical vertical strip has

center of mass (c.m.): (i, y) : (x, x),

length: 2x.,width: dr,

Strip c.m.is halfway.G, i) : (r, r)

Units in centimeters

5.5.2 Modeling the plate in Example 3with vertical strips.

nea'. dA :2x dx,mass: dtn : 6 dA :3 .2x dx :6x dx,

y (crn)

Mornents, Mass, and Center of Mass of a Thin Plate Covering a Regionin the xy-plane

Moment about the r-axis: U, : I y a*

Moment about the y-axis: a, : I

i a, (6)

Mass: u: I

a^ ]

center of masr - M' - M' l, : x :

M , ) = M

distance of c.m. from y-axis: i :,r.

Strip c.m.is halfway.

/ - ! , \

G.t \=f f .y )

_L1

5.58 Modeling the plate in Example 3with horizontal strips.

How to Find a Plate's Center ofMass1. Picture the plate in the,xy-plane.2. Sketch a stdp ofmass parallel to one

of the coordinate axes and find itsdimensions.

3, Find the saip's mass dm and centerof mass (i, i).

4. Inte$ate t dm, i dm, altid dm to fi\dM,, My aD:d, M.

5. Divide thg moments by the mass tocalculate t and t.


The moment of the strip about the y-axis is

i d m : x . . 6 x . d x : 6 x 2 d x .

The moment of the plate about the y-axis is therefore

f 1 1 - - 1 1M t : l i d m : I 6 x 2 d x : 2 x 3 1 : 2 g , c m .'

J Jo lo

b) The plate's mass:

c) The r-coordinate of the plate's center of mass:

_ M t 2 g ' c m 21-"

: t

tt'M

By a similar computation we could find Mx andt: Mr/M.

Method 2: Horizontal smTs (Fig. 5.58).

a) The moment M/: The y-coordinate of the center of mass of a typical horizontalstrip is y (see the figure), so

j, : v.The .x-coordinate is the .r-coordinate of the point halfway across the triangle.This makes it the average of y/2 (the strip's lefrhand .r-value) and I (the strip,srighrhand x-value):

- (Y /2) + |, : 2

) (cmJ

] F -

" !^ 2

f f t ' l l

^ t : Jdm: Jo

ox ax : ix , lo :3 g .

Y 1 v+24 ' 2 - 4

We also have

l e n s t h : \ - a : - - ! .' 2 2

width: dy,

area; dA : '---!

d\-2 '

m a s s : d m : 6 d A : 3 . 2 - ! ' - 'z

oy'

distance of c.m. to )-axis: I : +.

The moment of the strip about the y-axis isl l t ) - ' , 1

i dn : J - - - - - : . - -

4 . .

2 a y : g t + _ y - t a y .

The moment of the plate about the y-axis is

r r 2 r 3 f . y , l ' l / 1 6 \M" = I i dm: I ; t+ y , )dy : : l 4y , . l : . ( : l : 2 s .cm.

J J o d 6 L J l o 6 \ J , i

b) The plate's mass:

f f 2 l ? r , , 2 1 2 ?M: ldm- - I : t2 -y tdy : ; l zy - ' ; l = ia -2J :3s .

J J o z 2 L z ) o Z

<-r.E l--l


(b)

5.59 Modeling the plate in Example 4with (a) horizontal strips leads to aninconvenient integration, so we modelwith (b) vertical strips instead.

c) The x-coordinate of the plate's center of mass:

- M r Z g . c m 2

M 3 9 3

By a similar computation, we could find M, andt.

If the distribution of mass in a thin, flat plate has an axis of symmetly, thecenter of mass will lie on this axis. If there are two axes of symnetry, the centerof mass will lie at their intersection. These facts often help to simplify our work.

EXAMPLE 4 Find the center of mass of a thin plate of constant density 6covering the region bounded above by the parabola y : 4 - x2 and below by the;r-axis (Fig. 5.59).

so/ufion Since the plate is symmetric about the y-axis and its density is constant,the distribution of mass is synmetric about the y-axis and the center of mass lieson the y-axis. This means that .r:0. It remains to fitrd t : Mr/M.

A trial calculation with horizontal strips (Fig. 5.59a) leads to an inconvenientintegration

fM, :

Jo 23 yr/a - y dy.

We therefore model the distribution of mass with veftical strips instead (Fig. 5.59b).The typical vertical strip has

r L - . 2 tcenter of mass {c .m): { i . i ) - { r . - ; : }

\ z /

length: 4 - x2,

width: dx,

area: dA: (4 x2) dx,

mass: dm : 6 dA : 6(4 - x2) dx,

distance from c.m to r-axis: t:+.

The moment of the strip about the x-axis is

4 - x 2 3i a ^ =

i . 6 t 4 x ' t dx = -14 - s ' 1 ' i x .

The moment of the plate about lhe x-a\is is

f

ru": lra*: l .uuro-",t0":XI' l .u-2"+"\a':2fa'

The mass of the plate is

u : I a* : l,';r+ -.\a* :3] a.

(7)

(8)

5.7 Moments and Centers of Mass 4'15

Therefore,_ M, (256/1s) 6t - M - ( 3 2 / 3 ) 6

-

The plate's center of mass is the point

/ - 8 \( t . t ) :10 ' ; l '\ J . /

EXAMPLE 5 Variable density

Find the center of mass of the plate in Exarnple 4 if the density at the point (.r, y)i.s 6 : 2x2, twice the square of the distance from the point to the y-a,\rs.

so/ut on The mass dishibution is still symmetric about the y-axis, so "r = 0. With6 : 2x2, Eqs. (7) and (8) become

I t 2 8 fM, : l i dm : I i t + - * ' t ' d * : I x2 t4 - x2 \2dx

J J - t 2 l - z

85

tr

12 ro48: / 1 t 6 x ' : - 8 * n + x 6 t d , = ; .

I , l U J

r 1 2 7 )u - - l d m = | 6 t 4 - x 2 r d x : | 2 x 2 r 4 - x 2 t d x

J J t J - z

1 2 ) \ A: I Gx2-zxa ld - - ( s )

J - 2 ^ -

1 5 '

Therefore,

_ M, 2048 15 8' M 1 0 5 2 5 6 7

The plate's new center of mass is

G,t : (0 , ; ) .

(7't

A typical small segment ofw iehasdm=6ds=6ad0 .y = $ -

( , , :(d cos d, , s ind)

(b)

5,60 The semicircular wire in Example 6.(a) lhe dimensions and variables used infinding the center of mass. (b) The centerof mass does not l ie on the wire.

E

EXAMPLE 5 Find the center of mass of a wire of constant density d shapedlike a semicircle of radius c.

solution We model the wire with the semicircle y: "@=7

Gig. 5.60). Thedistribution of mass is symmetric about the y-axis, so i : 0. To find y, we imaginethe wire divided into short segments. The typical segment (Fig. 5.60a) has

length: ds : a d0,

mass: dtn : 6 ds : 6 a d0, Mass per unit length times length

distance of c.m. to x-axis: i : a sir 0.


Hence,ffr

| \ qm t ^t

-- !;". : ""!-I a m

6a2f- cos9lf, 26att 7(

The center of mass lies on the axis of symmetry at the point (O,2a/T), abo''Jt

two-thirds of the way up from the origin (Fig. 5.60b). tr

CentroidsWhen the density function is constant, it cancels out of the numelator and denom-

inator of the formulas for 7 and y. This happened in nearly every example in this

section. As far as -r and y were concerned, 6 might as well have been l. Thus,

when the density is consta.nt, the location of the center of mass is a feature of the

geometry of the object and not of the material from which it is made. In such cases

engineers may call the center of mass the centroid of the shape, as in "Find the

centroid of a triangle or a solid cone." To do so, just set d equal to 1 and proceed

to flnd - and y as before, by dividing moments by masses.

Exercises 5.7

Thin Rods1. An 80lb child and a l00lb child are balancing on a seesaw. The

80lb child is 5 ft from the fulcrum. How far from the fulcrumis the l00lb child?

2. The ends of a log are placed on two scales. One scale reads 100kg and the other 200 kg. Where is the log's center of mass?

3. The ends of two thin steel rods of equal length are welded togetherto make a dght-angled frame. Locate the frarne's center of mass.(Ilizt Wherc is the center of mass of each rod?)

4. You weld the ends of two steel rods into a right-angled frame.One rod is twice the length of the other. Where is the frame'scenter of mass? (llirt Where is the center of mass of each rod?)

Exercises 5-12 give density functions of thin rods lying along variousintervals of the .r-aris. Use Eqs. (3a--c) to find each rcd's momentabout the odgin, mass, and centet of mass.

5 . 6 ( a ) : 4 , O = x = 2

6 . 6 ( : t ) = 4 , l s " { : 3

6 ( . r ) = l + ( r / 3 ) , 0 : - t = 3

6 ( x , ) : 2 - ( t / 4 ) , O = r < 4

6 ( x ) : t + ( t / J i ) , r s x 3 4

d ( r l : 3 ( r - l l 2 l r 5 / 2 ) . 0 , 2 5 : . r < I

2 - x , 0 : r < lx , | . x = 2

) t + 1 , 0 : . r < l2 , l = r = 2

I

I

10.

11.

8.

9.

12. 6 (.r.) :

Thin Plates with Constant DensityIn Exercises 13-24, find the center of mass of a thin plate of constantdensity 6 covering the given region.

13. The region bounded by the parabola y: x2 and the line y = 4

14. The region bounded by the parabota ), : 25 -.r2 and the x-axis

15. The rcgion bounded by the parabola ): t - tc2 and the line

16. The region enclosed by the parabolas y : x2 - 3 and y = -2x2

17. The region bounded by the J-axis and the curve x : 1, - )3,o<Ys t

18. The region bounded by the parabola , = y' - y and the line

19. The rcgion bounded by the r-axis and the curve y:cosr,- t / 2 3 x = n / 2

7 1

24,

20. The region between the x axis and the curve ) : sec2 't, -z/4 <

x = f t 1 4

21. The region bounded by the parabolas ) = 1r2 4x and )' :

22, a) The region cut from the lirst quadrant by the circle x2 1

b) The region bounded by the i-axis and the semicircle ):

J'-vCompare your answer with the answer in (a).

The "triangular" region in the first quadrant between the circlejr2 + )2 = 9 and the lines :r = 3 and y = 3. (llinl. Use geometry

to find the area.)

The region bounded above by the curye ): l/x3, below by the

curve ): -1/n3, and on the left and r ight by the l ines r: I

and x : zz > 1 Also, lind lim"--;r.

Thin Plates with Varying DensityFind the center of mass of a thin plate covering the region be-

tween the r-axis and the curve 1 : 27,12, I : i 5 2, if the plate's

density at the point (r, ),) is 6(.r) - 1'?.

Find the center ofmass ofa thin plate covering the region bounded

below by the parabola ) : t2 and above by the line y : x if the

plate's density at the point (x, r) is d(x) : 12r'

The region bounded by the cufr'es ): +4/vA and the lines

x : I and r : 4 is revolved about the y-axis to generate a solid

Find the volume ol the solid.

Find the center of mass of a thin plate covering the region

if the plate's density at the point (r, y) is 6(,r) = l/x.

c) Sketch the plate and show the center of mass in your sketch

28. The region between the curve ) : 2/.t and the r-axis from "r - I

to.r - 4 is revolved about the r-axis to generate a solid.

a) Find the volume of the solid.

b) Find the center of mass of a thin plate coveling the region

il the plate's density at the point (.r, y) is 6(r) : r/x.c) Sketch the plate and show the center of mass in your sketch.

toward the opposite vertex is the point where the triangle's three

medians intersect. Show that the centroid lies at the inteNection

of the medians by showing thai it too lies one-third of the way

from each side toward the opposite vertex. To do so, take the

following steps.

1. Stand one side ofthe triangle on the t-axis as in Fig. 5 61(b)

Express dm in terms of I and d).2, Use similar triangles to show that L : (blh)(h - y) Sub-

stitute tlis expression for l, in your formula for dm.

(b)

' :: The triangle in Exercise 29. (a) The centroid.(b) The dimensions and variables to use in locatingthe center o{ mass.

3. Show that t : l /3.4. Extend the argument lo the other sides.

Use the result in Exercise 29 to nnd the cenfroids of the triangles

whose vertices appear in Exercises 30-34. (Hint: Draw each triangle

first.)

30. ( -1 ,0 ) , (1 ,0 ) , (0 , 3 )

32 . (0 ,0 ) , (a ,0 ) , (0 , a )

34. (0,0), (a,0). (a /2, b)

Thin Wires35. Find the moment about the x-axis of a wire of constant density

that lies along the curvs ,v: .v& from jr = 0 to:t :2.

36. Find the moment about the i-axis of a wire of constant density

that lies along the cuNe ) : xr from,t = 0 to r : 1.

37. Suppose the density of the wire in Example 6 is d:f tsind(l constant). Find the center of mass.

38. Suppose the density of the wire in Example 6 is , - 1 + k cos rl(f constant). Find the center of mass.

(a)

Exercises 5.7 417

31. (0 ,0 ) , (1 .0 ) , (0 , l )

33. (0,0), (.r, 0), (0, r)

26.

',1

a)b)

_.L

4'18 Chapter 5: Applications of lntegrals

Engineering FormulasVerify the statements and formulas in Exercises 39-42'

39. The coordinates of the centroid of a differentiable plane cuNe

are

Whatever the value of p > 0 in the equation )-: xz l(4p), the

)-coordinate ol the cenffoid of the parabolic segment shown here

is 1 : (3/5)a.

41. For wires and thin rods of constant density shaped like circular

arcs centered at the origin and symrnetric about the ) axis' the

) -coordincle ol lhe center of mas' i"

a s i na ac

d s

z i;\N Y,'

42, (ContinLratioll of Etercise 4l)

a) Show that when d is small, the distance I lrom the centroid

to chord AB is about 2i/3 (in the notation of the figure herc)

by taking the following steps.

l. Show that

slnd - av cos d

J x 4 slength

J ) a slength

(e)

sina - cr cos cv

c t - c t c o s ( l

and use TRACE to show that Ilm"-a- f @) x 213

(You will be able to confiIm the suggested equality

in Section 6.6, Exercise 74 )

ffi l) CnfcufnfoR The enor (difference between d atd2hl3)

is small even for angles greater than 45' See {or yourself

by evaluating the right-hand side of Eq (9) for a :0 2'

0 .4 ,0 .6 ,0 .8 , and 1 .0 rad .

h

ZZ z. cnneHeR craptr

: : : : E""'*:*.",;:--:".",:,::;*:*;; "*".InSc ience. the te lmle fe6spec i f i ca l l y toa forceac t ingonabodyandthebody 'ssubsequent displacament This section shows how to calculate work The appltca-

tions run frorJcompressing railroad car springs and emptying subteranean tanks

to forcing electrons together and lifting satellites into orbit'

Work Done bY a Constant ForceWhen a body moves a distanca d along a straight line as a result of being acted

on by a force of constant magnitude F in the direction of motion, we calculate the

Joules

The joule, abbreviated J and pronounced' jeuel. i \ named after rhe En!l i5h phy.icislJames Prescott Joule (1818-1889). Thedefining equation is

I joule - ( l newton)(l meter).

t n s y m b o l s , 1 J - 1 N . m .It takes a force of about I N to lift an

apple from a table. If you lift it I ln youhave done about 1 J of work on the apple. Ifyou then eat the apple you will haveconsumed about 80 food calories, the heatequivalent of nearly 335,000 joules. If thisenergy were dircctly useful for mechanicalwo[k, it would enable you to lift 335,000more apples up I m.

5.8 Work 419

work W done by the force on the body with the formula

W : Fd (Constant-forca formula for work). (1)

Right away we can see a considerable difference between what we are used tocalling work and what this formula says work is. If you push a car down the sbeet,you will be doing work on the car, both by your own reckoning and by Eq. (l).But if you push against the car and the car does not moveJ Eq. (l) says you willdo no work on the car, even if you push for an hour.

From Eq. (1) we see that the unit of work in any system is the unit of forcemultiplied by the unit of distance. In SI units (SI stands for Systime International,or lntemational System), the unit of force is a newton, the unit of distance is ameter, and the unit of work is a newton-meter (N , m). This combination appearsso often it has a special name, the joule. In the British sysrem, the unit of work isthe foot-pound, a unit frequently used by engineers.

EXAMPLE 1 If you jack up the side of a 2000-lb car 1.25 ft to change a nre(you have to apply a constant vetical force of about 1000 lb) you will perform1000 x 1.25 : 1250 ft-lb of work on the car. In SI units, you have applied a forceof 4448 N through a distance of 0.381 m to do 4448 x 0.381 :: 1695 J of work.

Work Done by a Variable ForceIf the force you apply varies along the way, as it will if you are lifting a leakingbucket or compressing a sprirg, the fomula lll: Fd has to be replaced by anintegral formula that takes the variation in F into account.

Suppose that the force performing the work acts along a line that we can modelwith the r-axis and that its magnitude F is a continuous function of the position. Wewant to find the work done over the interval ftom _r : a to r : b. We partition La, D]in the usual way and choose an arbitrary point cr in each subinteryal [.:rp 1, r1]. Ifthe subinteryal is short enough, f being continuous, will not vary much from;r1 1to .rt. The amount of work done across the interval will be about F(ca) times thedistance A;1, the same as it would be if F were constant and we could apply Eq.(l). The total work done from a to b is therefore approximated bv the Rienannsum

! F(ci)Ar1. (2)t : l

We expect the approximation to improve as the norm of the pafiition goes ro zero,so we define the work done by the force from 4 to , to be the integral of f. fromato b.

DefinitionThe work done by a variable force F(,r) directed along the.r-axis fromx : q t o x : b i s

D

l"'F (x) dx. (3)

4ZO Chapter 5: Applications of Integrals

5.62 The leaky bucket in ExamPle 3

The units of the integral are joules if F is in newtons and x is in meters' and

foot-pounds if F is in pounds and t in feet'

E{AMPLE 2 The work done by a force of F(r): I /';r2 N along the x-axis

f r o m . t : l m t o x : 1 0 m i s

= -l -u 1 :0.9 J.10 !

i" 10,: J,. r r L ol - r l

" d x = - - l I l r

EXAMPLE 3 A leaky 5lb bucket is lifted from the ground into the air by

pulling in 20 ft of rope at a constant speed (Fig 5'62)' The rope weighs 0'08 lbift'

ih" ui"t"t ,tu.tt wiih 2 gal of water (16 lb) and leaks at a constant rate' It finishes

&aining just as it rcaches the top' How much work was spent

a) lifting the water alone;b) lifting the water and bucket together;c) lifting the watel bucket, and rope?

Solution

^) The water elone. The force required to lift the water is equal.to the water's

weight, which varies steadily from 16 to 0 lb over the 20-ft lift When the

bucket is r ft off the ground, the water weighs

/ 2 0 - x \ / ' { \ " 4 x "6 1 ' 1 : 1 6 { = - ) = 1 6 ( l - . ) A ) : 1 6 - T r b .

/ ' ' " \ '

. / \original'weight ProPirrtionleft

The work done is

w: I F (x ) dx

f z o t 4 r \ f " r z - l 2 0: I ( 'o -T) a" : l to ' -? l :320-160=160rr ' lb 'J o \ t / L r l o

b) The water and bucket ngether Accotding to Eq (l), it takes 5 x 20: 100

ft.lb to lift a 5Jb weight 20 ft' Therefore

160 + i00 : 260 ft ' lb

of work were spent lifting the water and bucket together'

c) The water, bucket, and rope. Now the total weight at level r is

/ 4 Y \F(x )= l16 - ; l

\ ) , /-v-

variablewelgntof watel

at elevation .r

Use L .q . ( l ) fo f \ . t f i : tb lc lo l . ces

lb/ft ft

l l(0.08)(20- L 5 +

conslanrwergntof bucket

\_'--v-'

weight of ropepaid out atelevation x

5.8 Work 421

The work lifting the rope is

f20Work on rope :

;l

[ "120: I l . 6 r - 0 . 0 4 x ' | - 1 2 - l o - 1 6 f t . I b .L J o

The total work for the water, bucket, and rope combined is

160+ lob+ f t -276 f t . t b . D

(0.08)(20 x)d,r - 1,,'o lr.u-o.orr)0,

Amount compressed

(b)

5.63 The force F needed to hold a springunder compression increases l inearly asthe spring is compressed.

Hooke's Law for Springs: F = kxHooke's law says that the force it takes to stretch or compress a spring I length

units from its natural (unstressed) length is proportional to .r. In symbols,

(4)

The constant i, measured in force units per unit length, is a characteristic of the

spring, called the force constant (or spring constant) of the spring. Hooke's law(Eq. 4) gives good results as long as the force doesn't distort the metal in the spdng.

We assume that the forces in this section are too small to do that.

EXAMPLE 4 Find the work required to compress a spring from its natuml

length of I ft to a length of 0.75 ft if the force constant is ft: 16 lb/ft.

Sorution We picture the uncompressed spring laid out along the r-axis with its

movable end at the origin and its fixed end at x : I ft (Fig. 5.63). This enables us

to describe the force required to compress the spring from 0 to,r with the formula

F: 16,r. To compress the spring from 0 to 0.25 ft, the force must increase from

F ( 0 ) : 1 6 . 0 : 0 l b t o F ( 0 . 2 5 ) : 1 6 . 0 . 2 5 : 4 l b .

The work done by F over this interual is

f0.25 10.25 Eq. (3) wirh a : 0,w : l l . 6 x d . x : 8 x 2 | : 0 5 f t . l b . b : 0 . 2 5 , r ( j r ) :

JO _10 lo. I

tr

EXAMPLE 5 A spring has a natural length of I m. A force of 24 N stretches

the spring to a length of 1.8 m.

a) Find the force constant /t.b) How much work will it take to stretch the spring 2 m beyond its natural length?

c) How far will a 45-N force stretch the spring?

Solution

a) The force constant. We find the force constant from Eq. (a). A force of 24 N

shetches the spring 0.8 m, so

24 : k(0.8)

k : 24/0.8 : 30 N/m.

Work done by Ff romr=0 to - I=0 .25

Eq . ( 4 ) w i t h F :24 . r : 0 .8


-r (m)

5.64 A 24-N weight stretches this spring0.8 m beyond its unstressed length.

b) The work to stretcll the spring 2 m. We imagine the unstressed spdng hangingalong the r-axis with its free end at -{ - 0 (Fig. 5.64). The force required tostretch the spdng.r m beyond its natural length is the force required to pullthe free end of the spring -{ units from the origin. Hooke's law with I :30says that this force is

F(n) : 30x '

The work done by F on the spring from r :0 m to r :2 m is

l 2w: I 30x dx

J o

c) How far will a 15-N Jbrce stretchequation F: 30.r to find

" l t: ls;r'ln = 60 J.

the spring? We substitute F:45 in the

-t----:-:\,*l' ',,#

IF

45 : 30:r. or r : 1.5 m.

A 45-N force will stretch the spdng 1.5 m. No calculus is required to find this.

l

Pumping Liquids f rom ContainersHow much work does it take to pump all or part of the liquid from a container?To find out, we imagine lifting the liquid out one thin hodzontal slab at a timeand applying the equation W : F d to each slab. We then evaluate the integral thisleads to as the slabs become thinner and more numerous. The integral we get eachtime depends on the weight of the liquid and the dimensions of the cortainer, butthe way we nnd the integral is always the same. The next examples show what todo.

EXAMPLE 6 How much work does it take to pump the water from a fullupdght circuiar cylindrical tank of radius 5 m and height l0 m to a level of 4 mabove the top of the tank?

So/ution We draw the tank (Fig. 5.65), add coordinate axes, and imagine the waterdivided into thin horizontal slabs by planes perpendicular to the }-axis at the pointsof a partit ion P of the inteNal 10, 101.

The typical slab between the planes at ) and l + A,v has a volume of

Ay : r (radius):(rhickness) : r(5)':Ay : 25r Ly m3.

The folce F required to liti the slab is equal to its weighr,

1{'rr

5.65 To find the work it takes to pumpthe water from a tank, think of Iiftingthe water one thin slab at a time.

How to Find Work Done DuringPumping

1. Draw a figure with a coordinatesystem.

2. Find the weight F of a thinhorizontal slab ot' liquid.

3. Find the wolk AW it takes to lift theslab to its destination.

4. Integrate the work expression fron]the base to the surface of the liquid.

F : 9 8 0 0 A y

= 9800(25zAy) = 245,0002Ar N.

The distance through which F must act is about (14 - ))lifiing the slab is about

AW : force x distance : 245,000r (14 r)A) J.

The work it takes to lift all the water is approximatelyl 0 i t )

w ''LLw : I z+s,ooo" r.t+ _ y) a) J.

( )S00 i \ , r r

m, so the work done

y = z x o r ' = ) t

5.8 Work 42i

This is a Riemann sum for the functio n 245,00M (14 _ y) over the interval 0 <t :; 10. The work of pumping the tank dry is th. fi",it,f ,;;;;,

,; ji;l--' O:l ro

w : Jo

24s,0oor(14 _ ytdy= 245.0002 ['u l+ _ r) atJo

= 245,00M1uy r,2t'0L "-71"=245'ooon[eol

x 69,272,118 ^, 69.3 x 106 J.A l-horsep_ower output motor rated, at 746 J/sec could empty the tank in a littleIess than 26 h

- -"-'- v'rPLJ ure

o

EXAMPLE 7 The conical tank in Fig. 5.66 is filled to within 2 ft of the top;:l*; fi : i""",ff

ng 57 tblrt3 . How m"uch ;; ;;,; ;.T,pump the ol to

Solution We imagine the oil di*. r;I':1 *: points of a ,*;i':r'il:,t#;'fo;.0d.0,"n.'

perpendrcurar rorne ryplcat stab between the planes ut y ana y i Ay nas a volume of about

Alz = n(radiust2( th ic l / l \2iness, = tr \r, )

o, __ it,

tt n,.

The force F(y) required to lift this slab is equal to its weight,

F(y) : 57 Av : 5!r, o, tt. w;-jshr : w€1cnr per unrl

,'- .T:lt:l-": thyuch__1hich F()) must act ro tift rhis slab ro the level of thenm ot rhe cone is about (10 _ y) ft, so the work a""" fif,irg ,fr" lji"b ,. uOout

5inAW :

--::119 - y)y2Ay fr . tb.

The work done lifting all the slabs from y : 0 to ),-: g to the rim is approximately

w - f f r ro- y t f ay f t . tb .

This is a Riemann sum for the funcnon g7r /4)(10 _ y)y2 onthe interval trom) = 0 to l, _- 8. The work of pumoing the oil a G" ,irn,rr,,rr""iirniio, *"r. ,urn,as the norm of the partition go", to ,".o.

f8 s't -W: I

' - i - \ l o_ v t y l d t

J o + - - '

a7 - rE= ;

| : ' o y 2 _ y 3 ; d y

57T f loyr va'l8= 4 | . -

-7 ' - 30 .561 f r ' l b ._ J 0

T- v(5, 10)

1, = , y

5.66' The olive oil in Example 7.

tr


Exercises 5.8

Work Done by a Variable Forcel The workers in Example 3 changed to a larger bucket that held

5 gal (40 lb) of watet but the new bucket had an even largerleak so that it, too, was empty by the time it reached the top.Assuming that the water leaked out at a steady rate, how muchwork was done lifting rhe water? (Do not include the rope andbucket.)

2. The bucket in Example 3 is hauled up twice as fast so that thereis still I gal (8 lb) of water left when the bucket reaches thetop. How much work is done lifring the water this time? (Do notinclude the rope and bucket.)

3. A mountain climber is about to haul up a 50-m length of hangingrope. How much work will it take if the rope weighs 0.624 N/m?

4. A bag of sand originally weighing 144 lb was lifted at a constantrate. As it rose, sand also leaked out at a constant late. The sandwas half gone by the rime rhe bag had been lifted 18 fr. Howmuch work was done lifting the sand this far? (Neglect the weightof the bag and lilling equipment.)

5. An electric elevator with a motor at the top has a multishandcable weighing 4.5 lb/ft. When the car is at the lirst floor, 180 ftof cable are paid out, and effectively 0 ft are out when the car isat the top floor. How much work does the motor do just liftingthe cable when it takes the car from the first floor to the top?

6. When a paticle of mass zr is at (_r, 0), it is attracted toward theorigin with a force whose magnitude is &/r2. If the particle startsfrom rest at r - D and is acted on by no other forces, find thework done on it by the time it reaches r - a,0 < a < b.

7. Suppose that the gas in a circular cylinder ofcross-section areaAis being compressed by a piston. Ifp is the pressure ofthe gas inpounds per square inch and yis the volume in cubic inches, showthat the work done in compressing the gas from state (pr, %) tostate (p2, V2) is given by the equation

f ( P ' v ' )Work : / pdv.

J( t1,.vn

(I1irr. In the coordinates suggested in the figvehete,dV : A dt.The force against the piston is pA.)

(Continuation oJ Exercise Z) Use the integral in Exercise 7 tonnd the work done in compressing the gas from Vt:243 in3

to V2 : 32 in3 if pr : 50 1b/in3 and p and y obey the gas lawpV'a : constant (for adiabatic processes).

Springs9, It took 1800 J ofwork to stretch a spring ftom its natural length

of 2 m to a lenBth of 5 m. Find the spring's force constant.

10. A spring has a natural lengrh of 10 in. An 8001b force stretchesthe spdng to 14 in. (a) Find the force constant. (b) How muchwork is done in stretching the spring from 10 in. to 12 in.?(c) How far beyond its natural length will a 1600-lb force stretchthe spring?

11. A force of2 N will stetch a rubber band 2 cm (0.02 m). AssumingHooke's law applies, how far will a 4-N force shetch the rubberband? How much work does it take to stretch the rubber bandthis far?

12. If a force of 90 N stetches a spring I m beyond its natural length,how much work does it take to stretch the spdng 5 m beyond itsnatuml length?

13. Subway car springs. It rakes a force of 21,'114Ib to compressa coil spring assembly on a New York City Transit Authoritysubway car from irs free height of 8 in. to its fully compressedheight of 5 in.

a) What is the assembly's force constant?b) How much work does it take to compress the assembly the

first half inch? the second half inch? Answer to the nearestin . lb.

(Data courtesy of Bombardier, Inc., Mass Transit Division, forspring assemblies in subway cars delivered to the New york CityTransit Authority from 1985 to 1987.)

14. A bathrcom scale is compressed 1/16 in. when a l50lb personstands on it. Assuming the scale behaves like a spring that obeysHooke's law, how much does someone who compresses the scale1/8 in. weigh? How much work is done compressing the scale1/8 in.?

Pumping Liquids from Containers

The Weight of WaterBecause of variations in the earth's gavitational field, theweight of a cubic foot of water at sea level can vary fromabout 62.26 lb at the equator to as much as 62.59 lb near thepoles, a variation of about 0.57o. A cubic foot rhat weighsabout 62.4 lb in Melboume and New York Ciry will weigh62.5 lb in Juneau and Stockholm. While 62.4 is a typicalligure and a common textbook value. there is considerablevanat1on.

15. The rectangular tank shown here, with its top at ground level, isused to catch mnoff water. Assume that the water weighs 62.41b/ft3.

a) How much work does it take to empty the tank by pumpingthe water back to ground level once the tank is full?

b) If the water is pumped to ground level with a (5/11)-hpmotor (work output 250 ft . lb/sec), how long will it taketo empty the full tank (to the nearest minute)?

c) Show that the pump in part (b) will lower the water level10 ft (halfway) during the first 25 min of pumping.

d) What are the answen ro parts (a) and (b) in a location wherewarer weighs 62.26lblft32 62.59 lbfif.l

Exercises 5.8 425

19. A vertical right circular cylindrical tank measures 30 ft high and20 fr in diameter It is full of kerosene weighing 51.2 lb/ft3. Howmuch work does it take to pump the kerosene to the level of thetop of the tank?

20. The cylindrical rank shown here can be filled by pumping waterfrom a lake 15 ft below the bottom of the tank. There are twoways to go about it. One is to pump the water through a hoseattached to a valve in the bottom of the tank. The other is toattach the hese to the rim of the tank and let the water pour in.Which way will be faster? Give reasons for your answer

@ 21. CALCULATOR The truncated conical container shown here isfull of strawberry milkshake that weighs 4/9 ozlin3. As you cansee, the container is 7 in. deep, 2.5 in. across at the base. and3.5 in. across at the top (a standard size at Brigham,s in Boston).The straw sticks up an inch above the top. About how muchwork does it take to suck up the milkshal<e through the straw(neglecting friction)? Answer in inch-ounces.

- 17.5

(r.15,',7t

' : . 1 4

22, a)

b)

Dimensions in inches

Suppose the conical container in Example 7 contains milk(weighing 64.5 lb/ft3) instead of olive oil. How much workwill it take to pump the contents to the rim?How much work will it take to pump the oil in Example 7to a level 3 ft above the cone's rim?

16. The rectangular cistem (storage tank for rainwater) shown herehas its top 10 ft below ground level. The cistem, cunently full, isto be emptied for inspection by pumping its contents to groundlevel.

a) How much work will it take to empty the cistem?b) How long will it take a ( 1/2)-hp pump, rared at 275 ft . lb/sec,

to pump the tank dry?c) How long will it rake the pump in part (b) ro empry rhe

tank halfway? (It will be less rhan half rhe time required toempty the rank completely.)

d) What are the answers to parts (a) (c) in a location wherewater weighs 62.261b1ft31 62.59 tbfit3'l

Cround level

",w

10 t r

How much work would it take to pump tie water from the tankin Example 6 ro rhe level of the rop of the rank (instead of 4 mhigher)?

Suppose that, instead of being full, the tank in Example 6 is onlyhalf full. How much work does it take to pump the remainingwater to a level 4 m above the top of the tank?

t7,

18.

24.


To design the interior surface of a huge stainless steel tank, you

rcvolve the curve ) : t2, 0 s t : 4, about the y-axis. The con-tainer, with dimensions in meters, is to be filled with seawater,which weighs 10,000 N/m3. How much work will it take to emptythe tant by pumping the water to the tank's top?

We model pumping from spherical containers the way we do fromother containe$, with the axis of integration along the verticalaxis of the sphere. Use the f,gure here to find how much work

it takes to empty a full hemispherical water reservoir of radius5 m by pumping the water to a height of 4 m above the top of

the reservoir. Water weighs 9800 N/m3.

25. You are in charge of the evacuation and repair of the storage tankshowtr her€. The tank is a hemisphere of radius 10 ft and is full

of benzene weighing 56 lb/ft3. A firm you co[tacted says it canempty the tank for 1/20 per foot-pound of work. Find the workrequircd to empty the tank by pumping the benzene to an outlet2 ft above the top of the tank. If you have $5000 budgeted forthe job, can you afford to hirc the fiIm?

26. Your town has decided to drill a well to inqease its water supply.As the town engineer, you have determined that a water towerwill be necessary to provide the pressure needed for distribu-tion, and you have designed the system shown here. The wateris to be pumped from a 300-ft well through a vertical 4-in. pipe

into the base of a cylirdrical tart 20 ft in diameter atrd 25 fthigh. The base of the tank will be 60 ft aboveground. The pumpis a 3-hp pump, mted at 1650 ft . lb/sec. To the nearest hour, how

long will it take to fill the tank the fiIst time? (Include the timeit rakes to fill the pipe., Assume water weiShs 62 4 lblfi3.

Subme$ible pump

Other ApplicationsR Zl, erttirg a satellite in orbit. T\e stretrgth of the earth's grav-

itational field varies v/ith the distance r from the earth's center,

and the magnitude of the $avitational folce experienced by a

satellite of mass rrt dudng and after launch is

, , . , _ AMG, v t _

1 2

Here, M :5,975 x 102 kg is the eatth's mass, G :6.6'720 x10 lr N . m2kg-2 is the universal gravitational constant, and r ismeaswed in meters. The work it takes to lift a 1000-kg satelliteftom the earth's surface to a circular orbit 35,780 km above theearth's center is therefore given by the integral

f 15 ?80 000 l000MG _Work :

/ ... ̂ ^^ I ,r Jou,.t.

J6.3?0.000

Evaluate the integral. The lower limit of integration is the earth'smdius in meters at the launch site. Cfhis calculation does not takeinto account energy spent lifting the launch vehicle or energyspent bringing the satellite to o$it velocity.)

E 28. Forcing electrons together Two electrons r meters apart rcpeleach other with a force of

2 3 x 1 O 2 eF : -------;- newtons'

Suppose one electron is held fixed at the point (1, 0) onthe r-axis (units in meters). How much work does it taketo move a second electrol along the t-axis ftom the point(-1,0) to the origin?Suppose an electron is held fixed at each of the points(-1,0) and (1, 0). How much work does it take to move athird electron along the r-axis from (5, 0) to (3, 0)?

Work and Kinetic Energy29. If a variable force of magnitude F(t) moves a body of mass

a)

b)

m along the ir-axis frcm -rr to 12, the body's velocity u can bewritten as di/d/ (where / represents time). Use Newton's SecondLaw of Motion F : m(dD/dt) and the Chain Rule

du du t lx du

dt dx dt -dt

to show that the net work done by the force in moving the bodyfrom ,r1 to 12 is

l ' . l . t .W : I F ( r ) d x = . n u i -

i n u i .J,

where u1 and u2 are the body's velocities at xr and 12. In physicsthe expression (1/2)mv2 is called tte kinetic energy of the bodymoving with velocity v. Therefore, the work done by the forceequals the change in the body's kinetic energ),, and we can findthe wo* by calculating this change.

In Exercises 30-36, use the result of Exercise 29.

30. Tennis. A 2-oz tennis ball was served at 160 frlsec (about 109mph). How much work was done on the ball to make it go thisfast? (To find the ball's mass from its weight, express the weightin pounds and divide by 32 ft/sec2, the acceleration of gravity.)

31, Baseball. How many foot-pounds of work does it take to throwa baseball 90 mph? A baseball weighs 5 oz :0.3125 lb.

32. Golf. A 1.6-02 golf ball is driven off the tee at a speed of 280ft/sec (about 191 mph). How many foot-pounds of work are donegetting the ball into the air?

E 33. lennis. Dudng the match in which Pete Sampras won the 1990U.S. Open men's tennis championship, Sampras hit a serve that

5.9 Fluid Pressures and Forces 427

Weight Ys. MassWeight is the force that results from gravity pulling on a mass.The two are related by the equation in Newton's second law,

Weight : mass x acceleration.

Thus,

Newtons : kilograms x m/sec2,

Pounds : slugs x ft/sec2 .

To convert mass to weight, multiply by the acceleration ofgravity. To convert weight to mass, divide by the accelerationof gravity.

was clocked at a phenomenal 124 mph. How much work didSampras have to do on the 2-oz ball to get it to that speed?

R 34. football. A quarrerback threw a 14.5-oz ioorbatl 88 lvsec (60mph). How many foot-pounds of work were done on the ball toger ir to this speed?

Ei 3S. SoftO"ti. How much work has to be perlbrmed on a 6.5-02 soft-ball to pitch it 132 fr./sec (90 mph)?

36. A ball bearing. A 2-oz steel ball bearing is placed on a ver-tical spring whose force constant is ,t: l8 lb/ft. The spring iscompressed 3 inches and released. About how high does the ballbearing go?

Fluid Pressures and ForcesWe make dams thicker at the bottom than at the top (Fig. 5.67) because the pressureagainst them increases with depth. It is a remarkable fact that the pressure at anypoint on a dam depends only on how far below the surface the point is and not onhow much the surface of the dam happens to be tilted at that point. The pressure,in pounds per square foot at a point & feet below the surface, is always 62.4h. Thenumber 62.4 is the weight-density of water in pounds per cubic foot.

The formula, pressure : 62.4h, makes sense when you think of the unitsinvolved:

x ft.

As you can see, this equation depends only on units and not on the fluid involved. Thepressure & feet below the surface of any fluid is the fluid's weight-density times lz.

lb lb

ff ft35.67 To withstand the increasingpressure, dams are built thicker as theygo down.


Weight-density

A fluid's weighfdensity is its weight per unit

volume. Typical values (lb/ft3) are

4284964.5

10057b4

62.4

The Pressure-Depth Equation

In a fluid that is standing still, the pressure p at depth ft is theweight-density l, times lr.'

fluid's

( 1 )GasolineMercuryMilkMolassesOlive oilSeawaterWater

Fluid Force on a Constant-Depth Surface

p A : w h A

EXAMPLE 1 rl're Great i'/lcl.::t:.. :':. '.. :

At 1:00 pM. on January 15, 1919, an unusually warm day, a 9O-ft-high, 90-ft-diameter cylindrical metal tank in which the Pudtan Distilling Company was stodng

molasses at the comer of Foster and Commercial streets in Boston's North End

exploded. The molasses flooded into the streets, 30 ft deep, tapping pedesftians

and horses, knocking down buildings, and oozing into homes. It was eventually

tracked all over town and even made its way into the suburbs (on trolley cars andpeople's shoes). It took weeks to clean up.

Given that the molasses weighed 100 lb/ft3, what was the total force exerted

by the molasses against the bottom of the tank at the time it blew? Assuming the

tank was full, we can find out from Eq. (2):

Total force : whA - (100)(S0)(n (45)'?) \ 57.255.526 lb.

How about the force against the walls of the tank? For example, what was the

total force against the bottom foot-wide band of tank wall (Fig. 5.69)? The area ofthe band was

(2)

F\F{\*T

5.59 These containers are fi l led withwater to the same depth and have thesame base area. The total force istherefore the same on the bottom ofeach container. The containers'shapes donot matter here.

SHADED BAND NOT'IO SCALE

5.1a.,! schematic drawing of the molassestank in Example 1 . How much fo rce d idthe lowest foot of the vertical wall haveto withstand when the tank was full? lttakes an integral to find out. Notice thatthe proportions of the tank were ideal.

In this section we use the equation P : u,h to derive a formula for the total

force exerted by a fluid against all or part of a vertical or horizontal containing

wall.

The Constant-Depth Formula for Fluid ForceIn a container of fluid with a flat horizontal base, the total force exerted by the

fluid against the base can be calculated by multiplying the area of the base by thepressure at the base. We can do this because total force equals force per unit area(pressure) times area. (See Fig. 5.68.) If E p, and A are the total fbrce, pressure,

and area. then

F : total force : force per unit area x area

: preSSUIe x arca : PA

A : 2Trh - 2r (45)(\) : gott fe .

The force exerted by a fluid againstone side of a thin horizontal strip isabout AF: p ressure x a rea = wx (s t r iPdepth) x r0)Ay. The plate here is f lat,but it might have been curved instead,l i ke the ver t i ca l wa l l o f a cy l indr ica l tank .Whatever the case, the strip length ismeasured along the surface of the plate.

5.9 Fluid Pressures and Forces 429

The tank was 90 ft deep, so the pressure near the bottom was approximately

p - u)h : (100)(90) = 9000 lb/ftr.

Therefore the total force against the band was approximately

F - uhA: (9000)(901r ) ^, 2, 544. 690 lb.

But this is not exactly right. The top of the band was 89 ft below the surface, not

90. and the pressure there was less. To find out exactly what tbe force on the bandwas, we need to take into account the variation of the pressure across the band.

The Variable-Depth FormulaSuppose we want to know the force exerted by a fluid against one side of a verticalplate submerged in a fluid of weight-density }y. To llnd it, we model the plate asa region extending from y : a to | = b in the .r-v-plane (Fig. 5.70). We pa ition

ftr, Dl in the usual way and imagine the region to be cut into thin horizontal st ps

by planes perpendicular to the )-axis at the partition points. The typical stdp tiom _\'to ) + A} is A_"- units wide by a()') units long. We assume L(1) to be a continuousfunction of 1

The pressure varies across the strip from top to bottom, just as it did in the

molasses tank. But if the strip is narrow enough, the pressure will remain close toits bottorn edge value of u x (strip depth). The fbrce exerted by the fluid againstone side of the strip will be about

AF = (pressure along bottom edge) x (area)

= rrl x (strip depth) x Z(1)A_r.

The force against the entire plate will be aboutb h

)- lr : \-ru x (strip depth) x L(),)4,-). (3)/ 2

- ' 1 - '

The sum in (3) is a Riemann sum for a continuous function on [4, b]. and we erpectthe approximations to improve as the nolm of the partition goes to zero. We definethe force against the plate to be the limit of these sums.

Definit ionThe lntegral for Fluid ForceSuppose that a plate submerged vertically in fluid of weight-density w runsfrom r, : a to ) : b on the )-axis. Let Z()) be the length of the horizontalstrip measured from left to right along the surface of the plate at level -r,.Then the force exefied by the fluid against one side of the plate is

f hF= I w . (strip depth) . L(y) dy. (4)

J.

EXAMPLE 2 A flat isosceles right triangular plate with base 6 1t and height3 ft is submerged vertically, base up. 2 ft below the surface of a swimming pool.

Find the fbrce exerted by the water against one side of the plate.

Strip length ai level l,


.r (f0

To find the force on one side of thesubmerged plate in ExamPle 2, we canuse a coordinate system like the one here.

\OTTO SCALE Bottom level

I The molasses tank with thecoordinate origin at the bottom(Example 3) .

Solution We establish a coordinate system to work in by placing the origin at

the plate's bottom vertex and running the ]-a-\is upward along the plate's axis of

symmetry (Fig.5.71). (We will look at other coordinate systems in Exercises 3 and

4.) The surface of the pool lies along the line 1 : 5 and the plate's top edge along

the line y : 3. The plate's right-hand edge lies along the line 1' : x, with the upper

right vertex at (3, 3). The length of a thin sftip at level ) is

L(1') = )'a : 2Y '

The depth of the strip beneath the surtace is (5 - ;v). The force exerted by the water

against one side of the plate is therefore

r= [0 , ' f ) ' j l ) L r y t c t yJ . \ dep rh /

f 3: I 62.4(s Y)2! dY

Jn

(5) - )') d)f 3

124.8 I

124.8 I ;L Ly2 +],: 1684.8 lb.

How to Find Fluid Force

Whatever coordinate system you use, you can find the fluid force against

one side of a submerged vertical plate or wall by taking these steps.

l. Find expressions for the length and depth of a typical thin horizontal

strip.2. Multiply their product by the fluid's weight-density w and integrate over

the interval of depths occupied by the plate or wall

EXAMPLE 3 We can now calculate exactly the force exerted by the molasses

against the bottom l-ft band of the Puritan Distilling Company's storage tank when

the tank was full.The tank was a right circular cylindrical tank 90 ft high and 90 ft in diameter'

Using a coordinate system with the origin at the bottom of the tank and the )-axispointing up (Fig. 5.72), we find that the typical hodzontal strip at level ) has

Strip depth: 90 - f,

Strip length: zx tank diameter:902.

The force against lhe band is therelore

t t IForce: / ru(depth)(length) dy = , 100(90 r)(902)dr

Jo Jo

f l: 90002 I (90 - r) dr - 2's30,553 lb

Jt)

Surface level of fluid

5.73 The force against one side of thep la te i s w. h . p la te a rea .

5.9 Fluid Pressures and Forces 43'l

As expected, the force is slightly less than the constant-depth estimate followingExample 1.

Fluid Forces and CentroidsIf we know the location of the centroid of a submerged flat vertical plate (Fig.5.73), we can take a shortcut to find the force against one side of the plate. FromEq. (4) ,

7 bF : I w > t s r r i p d e p t h ) , 4 L t y t ( l y

J,,

: , 1."

r"oiv depth) x z(y) r/l

: u x (moment about surface level line of region occupied by plate)

: u x (depth of plate's centroid) x (area of plate).

Fluid Forces and Centroids

The force of a fluid of weighrdensity w against one side of a submergedflat vertical plate is the product of 4 rhe distance t from the plare's centroidto the fluid sudace. and the Dlate's area:

F = w h A . (s)

EXAMPLE 4 Use Eq. (5) to find the force in Example 2.

Sorution The centroid of the triangle (Fig. 5.71) lies on the y-axis, one-third ofthe way from the base to the verler,. .o h -- 3. The rriangle's area is

4 : 1(base)(height)

I

1(6)(3) : e

Hence,

F : wh .q : 62.4)(3)(9)

: 1684.8 Ib. D

Equation (5) says that the fluid force on one side of a submerged flat vefticalplate is the same as it would be if rhe plate's entire area lay ir units beneath thesurface. For many shapes, the location of the centuoid can be found in a table, andEq. (5) gives a practical way to find F: Of course, the centroid's location was foundby someone who performed an integration equivalent to evaluating the integral inEq. (4). We recommend for now that you practice your mathematical modeling bydrawing pictures and thinking things through the way we did when we developedEq. (4). Then check your results, when you conveniently can, with Eq. (5).


Exercises 5.9

The weight-densities of the fluids in the following exercises can belbund in the table on page 428.

1. What was the total fluid force against the cylindrical inside wallof the molasses tank in Example I when the tank was full? halffu11?

2, What was the total fluid lbrce against the bottom l-ft band ofthe irside wall of the molasses tank in Example I when the tankwas half full?

3. Calculate the fluid force on one side of the plate in Example 2using the coordinate system shown here.

Calculate the fluid force on one side of the plate in Example 2usinB the coordi late sy' lem sho\ n here.

The plate in Exercise 7 is revolved 180' about line AB so lhatpart of the plate sticks out of the lake, as shown here. What forcedoes the water exert on one face of the plate now?

9, The vertical ends of a watering trough are isosceles triangles likethe one shown here (dimensions in feet).

a) Find the flujd lorce against the ends when the hough is full.g b) CALCULATOR How many inche. do you ha\e lo louer rhc

water level in the trough to cut the fluid force on the endsin half? (Answer to the nearest half inch.)

c) Does it matter how long the trough is? Give reasons foryout answer.

Surface level

1 f r,il

4 f r

IJ

8.

4.

r (fi)

5.

7.

The plate in Example 2 is lowered another 2 ft into the water.What is the fluid force on one side of the plate now?

The plate in Example 2 is raised to put its top edge at the surfaceof the pool. What is the fluid force on one side of the plate now?

The isosceles ffiangular plate shown here is submerged verticallyI ft below the surface ol a freshwater lake.

a) Find the fluid force against one face of the plate.b) What would be the fluid force on one side of the plate if

the water werc seawater instead of freshwater?

The vertical ends of a watering trough are squares 3 ft on a side.

a) Find the fluid force against the ends when the trough is full.

b) CALCULATOR How many inches do you have to lower thewater level in the trough to reduce the fluid lbrce by 25%?

c) Does it matter how long the trough is? Give reasons foryour answer,

' (fr)

10.

rG0

J (IT]

) (fi)

12.

The viewing poftion of the rectangular glass wiDdow in a typicalfish tank at the New England Aquarium in Boston is 63 in. wideand runs from 0.5 in. below the water's surface to 33.5 in. belowthe surface. Find the fluid force against this portion ofthe window.The weighFdensity of seawater is 64 lb/ft3. {b case you werewondering, the glass is 3/4 in. thick and the tank walls extend 4in. above the water to keep the ish from jumping out.)

A horizontal rectangular fteshwater f,sh tank with base 2 x 4 ftand height 2 ft (interior dimensions) is filled to within 2 in. ofthe top.

a) Find the fluid force against each side and end of the tank.b) ffthe tank is sealed and stood on end (without spilling), so

that one of the square ends is the base, what does that doto the fluid forces on the rectangular sides?

CALCULATOR A rectangular milk carton measures 3.75 x 3.75in. at the base and is 7.75 in. tall. Find the force of the milk onone side when the carton is full.

CALCULATOR A standard olive oil can measures 5.75 by 3.5in. at the base and is 10 in. tall. Find the fluid force against thebase and each side when the can is full

Exercises 5.9 433

E 19. CALCULATOR The end plates of the tough shown herc weredesigned to withstand a fluid force of 6667 lb. How many cubicfeet of water can the tank hold without exceeding this limitation?Round down to the nearest cubic foot.

(-4, l0)

t'r (ft)

20. Water is running into the rectangular swimming pool shown hereat the Iate of 1000 ft3lh.

a) Find the fluid force against the triangular drain plate aftert h of filling.

b) The dmin plate is designed to withstand a fluid force of 520lb. How high can you fill the pool without exceeding thislimitatiol?

I rr.

Iu.

15. A semicircular plate 2 ft in diameter sticks straight down intofresh water with the diameter along the surface. Find the folceexerted by the waier on one side of the plate.

16. A tank truck hauls milk in a 6-ft-diameter horizontal right circularcylin&ical tant. How much force does the milk exert on eachend of the tank when the tank is half fuU?

17. The cubical metal tank shown here has a parabolic gate, heldin place by bolts and designed to withstand a fluid force of 160lb without rupturilg. The liquid you plan to store has a weight-density of 50 1b/ff.

a) What is the fluid force on the gate when the liquid is 2 ftdeep?

b) What is the maximum height to which the container can befilled without exceeding its desi$ limitation?

10 fr

, (f0

Parabolic sate Enlarged view of- Parabolic gate

The rectangular tank shown here has a I ft x I ft square window1 ft above the base. The window is desimed to withstand a fluidforce of 312 lb without cracking.

a) What fluid force witl the window have to withstand if thetank is filled with water to a depth of 3 ft?

b) To what level can the tank be filled wilh water withoutexceeding the window's design limitation?

Triangular dlain plate

) (ft)

J (ft)

Parabolic gate] (ft)

Enlarged view of drain plate


A vertical rcctangular plate 4 units long by , units wide is sub-

merged in a fluid of weight density n with its long edges parallel

to the fluid's sudace. Find the average value of the pressure along

the vertical dimension of the plate. Explain your answer.

(Continuation of Exercis? 21.) Show that the force exerted by the

fluid on one side of the plate is the average value of the pressure

(found in Exercise 21) times the area of the plate.

warer pours inlo rhe tan-k here at lhe rale of 4 ftl/min. The lank's

cross sections are 4-ft-diameter semicircles. One end of the tank

is movable, but moving it to increase the volume compresses a

spring. The spdng constant is ,t - 100lb/ft. If the end ofthe tank

moves 5 ft against the spdng, the water will drain out ol a safeiy

hole in the bottom at the rate of 5 ft3/min. Will the movable end

reach the hole before the tank overflows?

Movable end

The Basic Pattern and OtherModeling ApplicationsThere is a pattern to what we did in the preceding sections. In each section we

wanted to measure something that was modeled or described by one or more

continuous functions. In Section 5.1 it was the area between the graphs of two

continuous functions. In Section 5.2 it was the volume of a solid. In Section 5.8 it

was the work done by a force whose magnitude was a continuous function, and so

on. In each case we responded by partitioning the interyal on which the function

or functions were defined and approximating what we wanted to measure with

Riemann sums over the interval. We used the integral defined by the limit of the

Riemann sums to define and calculate what we wanted to measure. Table 5.2 showsthe pattem.

Literally thousands of things in biology, chemistry, economics, engineering.

finance, geology, medicine, and other fields (the list would fill pages) are modeled

and calculated by exactly this process.This section reviews the process and looks at a few more of the integrals it

leads to.

Displacement vs. Distance TraveledIf a body with position function s(t) moves along a coordinate line without changing

direction, we can calculate the total distance it travels from t :.t to t - b byinte$ating its velocity functior u() from r : a to t - b, as we did in Chapter 4. If

the body changes direction one or more times during the trip, we need to integrate

the body's speed lu(t)l to find the total distance traveled. Integrating the velocity

will give only the body's displacement, s(D)-s(a), the difference between its

initial and final positions.To see why, partition the time interval a A t < b into subintervals in the usual

way and let Atr denote the length of the lrh interval. If At1 is small enough,the body's velocity u(t) will not change much from /1 I to ft and the right-hand

Water in

Side view

5.10 The Basic Pattern and Other Modeling Applications 435

Table 5.2 The phases of developing an integral to calculate something

Phase 1 Phase 2 Phase 3

We describe or model something we wantto measure in tems of one or more corr-tinuous functions def,ned on a closedinterval [a, r].

We partition [a, ,] into subinterr'als oflength A.rr and choose a point ct in eachsubinterval.

We approximate what we want to m€asurewith a finite sum.

We identify the sum as a Riemann sumof a cortinuous function over [d, ,].

The approximations improve as thenorm of thg partition goes to zero.

The Riemann sums apprcach a limit-ing integral.

We use the integral to define and cal-culate what we originally wanted tomeasule.

The area between the curves ) = f(-I).y = g(r) on [a, ,] when "f(x) Z S(r)

I [/(cr) - g(cr)] Arr , , l ip" I [ " f t . r l B(cr ) l A. r&

J. tfG\ - eG)) dx

The volume of the solid def,ned by revolv-ing the curve ): R(-x), a 5 r 5 r, aboutthe r'axis.

I z[R(c*)] '?A-rrlpot

zlR(cr)]'?A'r

J, ttlR(x))'dx

The \ orl done by a conl inuous lariableforce of magnitude F("I) directed alongthe r-axis from a to b

t a(c*) a.rk ]1poI r(cr) l"r

l."1ro'


(F (a) + C)

= [ ' u f t \ a t .

endpoint value u(t1) will give a good approximation of the velocity throughout the

interval. Accordingly, the change in the body's position coordinate during the tth

time interval will be about

u(tk) Ltk.

The change will be positive if rr(4) is positive and negative if u(te) is negative.

In either case, the distance traveled during the ftth interval will be about

,u(r4) la/r .

The tolal trip distance wil l be approximatell

i- . lu l t t ) L t t . (1 )t= l

The sum in Eq. (1) is a Riemann sum for the speed lu(t)l on the interval lri, b].We expect the approximations to improve as the nom of the partition of [4, b] goesto zero. It therefore looks as if we should be able to calculate the total distancenaveled by the body by integrating the body's speed from d to b. In practice, thistums out to be the right thing to do. The mathematical model predicts the distancecorrectly every time.

If we wish to predict how far up or down the line from its initial position a

body will end up when a trip is over, we integrate u instead of its absolute value.

To see why, let s (/) be the body's position at time t and let F be an antiderivative

of r. Then

s ( t ) :F ( r )+c

for some constant C The displacement caused by the trip from t : q to t : b ls

s(D) s(a) : (F(b) { c)

: F (b) - F (.a)

EXAMPLE 1hr f2 sec was

The velocity of a body moving along a line from / : 0 to / =

u(t) = 5cost r/sec.

Find the total distance fiaveled and the body's displacement.

: t."r ( t ) l d t

Displacement : l.'

,<l a,

r ( r ) : 5 c o s t

5.'10 The Basic Pattern and Other Modeling Applications 437

Solution

f3n /2Distance haveled :

/ 5costld/ Distance i \ the in legfal o i spccLl .

E

6

f l l

F o

t n / 2 r 3 n / 2: l 5cos rd r / t - 5co " r rd r

J o J r / 2

T/2 13a /2: 5 s i n r l - 5 s i n / l

JO )nl2

: 5 ( l - 0 ) - 5 ( -1 - l ) : s+ 10 : 15 m

Displacement : lo'"

'' s

"o", a,

Di\ fhccmc t is ihc intcgrLr l

s(0) +5: r r*, ] ' "" : s(-1) - 5(o) : -s m

During the trip, the body traveled 5 m forward and 10 m backward fordistance of 15 m. This displaced the body 5 m to the left (Fig.5.74).

Delesse's RuleAs you may knoq the sugar in an apple starts tuming into starch as soon as theapple is picked, and the longer the apple sits around, the starchier it becomes. Youcan tell fresh apples from stale by both flavor and consistency.

To find out how much starch is in a given apple, we can look at a thin sliceunder a microscope. The cross sections of the starch granules will show up clearly,and it is easy to estimate the proportion of the viewing area they occupy. Thistwo-dimensional propofiion will be the same as the three-dimensional proportionof uncut starch granules in the apple itself. The apparently magical equality of thesepropofiions was f,rst discovered by a French geologist, Achille Emest Delesse, inthe 1840s. Its explanation lies in the notion of average value.

Suppose we want to find the proportion of some granular material in a solidand that the sample we haye chosen to analyze is a cube whose edges have lengthl,. We picture the cube with an r-axis along one edge and imagine slicing thecube with planes perpendicular to points of the interval [0, f] (Fig. 5.75). Call theproportion of the area of the slice at x occupied by the granular matedal of interest(starch, in our apple example) r(x) and assume r is a continuous function of x.

a totaltr

5.74 The velocity and displacement ofthe body in Example 1 .

575 The steps leading to Delesse's rule:(a) a slice through a sample cube; (b) thegranular material in the slice; (c) the slabbetween consecutive slices determined bva partit ion of [0, L].

s ( t ) : 5 s i n l + r ( 0 )


Delesse's rule

Achille Ernest Delesse was a mid nineteenth

century mining engineer interested in

determining the composition of rocks. To

find out how much of a particular mineral a

rock contained, he cut it through, polished an

exposed iace, and covered the tace with

trrnsparent waxed paper. trimmed to size. He

then traced on the paper the exposed poftions

of the rnineral that interested him. Aftef

weighing the paper, he cut out the mineral

traces and weighed them. The ratio of the

weights gave not only the proportion of thc

surface occupied by the mineral but, more

importanl, the proportion of the entire tock

occupied by the mineral. This rule is st i l l

used by petroleum geologists today. A

two dimensional analogue of it is used to

determine the porosities of the ceramic lllters

l h , r l e \ t r r c l o rgJn i c mo lecu le \ i n chem i . t r y

laboratoies rnd screen out micrcbes in water

puri l iers.

Now pafiition the interval 10, tl into subinteNals in the usual way lmagine

the cube sliced into thin slices by planes at the subdivision points. The length At1

of the frth subinteNal is the distance between the planes at "rt r and nk lf the

planes are close enough together, the sections cut from the grains by the planes

will resemble cylinders with bases in the plane at xi. The proportion of granular

material between the planes will be about the same as the proportion of cylinder

base area in the plane at -rr, which in tum will be about r(.:r1). Thus the amount ol

granular material in the slab between the two planes will be about

lPropor l ion) Y ls l rb vo lume) = r l r / )L2 Ar t .

The amount of granular material in the entire sample cube will be about

I r (x i )L 'z AT 1 .

This sum is a Riemann sum for the function r1r)Lr o\'er the interval [0,4]. We

expect the approximations by sums like these to improve as the norm of the sub-

division of [0, L] goes to zero and thercfore expect the integral

r (x) L2 dx

to give the amount of glanular material in the sample cube

We can obtain the proportion of granular material in the sample by dividing

this amount by the cube's volume, L3. If we have chosen our sample well, this will

also be the proportion of granular material in the solid from which the sample was

taken. Putting it all together, we get

Proportion of granular Propoftion of granular

material in solid material in the sample cube

L - | r \ x ) a x 1 r L-- i J" 'ato*

t:

I rQ)L2 dx

L3 L3

: average value of r(r) over [0, Z]

: propo ion of area occupied by granular

material in a typical cross section.

This is Delesse's rule. Once we have found -, the average of r(,v) over [0, U' we

have found the proportions of granular material in the solid

In practice, i is found by averaging over a number of cross sections- There

are several things to watch out for in the procass. In addition to the possibility

that the granules cluster in ways that make representative samples difficult to find,

there is the possibility that we might not recognize a granule's tace for what it

is. Some cross sections of normal red blood cells look like disks and ovals, while

others look suryrisingly like durnbbells. We do not want to dismiss the dumbbells

as experimental error the way one research group did a few years ago

Useless Integrals - Bad ModelsSome of the integrals we get from forming Riemann sums do what we want, but

othen do not. It all depends on how we choose to model the problems we want to

solve. Some choices are good: others are not. Here is an example.

5.10 The Bas ic Pat te rn and Other Mode l ing App l ica t ions 439

Modeled by revoh,inglrn rpproprialc curvcr . : J ( _ r l , d < r < r .

Cone frustlrln rpproximatioll

'lhc suface afea ol Lhe

gcometdc model is

. , f , , i. s = 2 7 r l r . \ r l t + i ; ] . i , :

. !

We use the sudace area fbrmula

s = l.'' z, 1r,t

because it has predictive value and always gives results consistent with inlormalion

tiom other sources. In other words, the model we used to derive the tbmrula (Fig.

5.76) was a good one.Why not find the surfhce area by approximating with cylindrical bands instead

of conical bands, as suggestcd in Fig. 5.77? The Riemann sums we get this way

converge just as nicely as the ones based on conical bands, and the resulting integml

is simpler Instead of Eq. (2), we get

2tr .l (x) dx.

Atier all, we migllt argue, we used cylinders to derive good volume tbrmulas. so

why not use them again to derive sudace area fonnulas?The answer is that the lbrmula in Eq. (3) has no predictive value and alnost

never gives results consistent with oLher calculations. The comparison step in the

modeling process fails tbr this formula.There is a molal here: Jusl because we end up with a nice looking integral

does not mean it will do what we want. Constructing an integral is not enough-we

have to test it too (Exercises 15 and 16).

The Theorems of PappusIn the third century, an Alexandrian Greek named Pappus discovered trvo formulas

that relate centroids to surfaces and solids of revolutiitn. The fomrulas provide

shortcuts to a number of otherwise lengthy calculations.

Theorem 'l

Pappus's Theorem for VolumesIf a plane region is revolved once about a line in the plane that does nol

cut through the rcgion's interior, then the volume of the solid it generates is

equal to the region's area times the distance tmveled by the regiot]'s centroid

during the revolution. If p is the distance liom the axis of revolution to the

centrcid, then

.A sol id of revolut ionwhose surface area

we waDl to know

Compare( l t i r J )

The rrea of thesolid's surface

should be S.

5.76 The modeling cycle for surface area.

(b)

5 r: Why not use (a) cylindrical bandsinstead of (b) conical bands toapproximate surface area?

F(!J,' \2)

(3)s: I"'

l / : 2 r p A . (4)


578 The region R is to be revolved(once) about the x-axis to generate asolid. A 1700-year-old theorem says thatthe solid's volume can be calculated bymultiplying the region's area by thedistance traveled by its centroid duringthe revolution.

5.79 With Pappusl f irst theorem, we canfind the volume of a torus withouthaving to integrate (Example 2).

5.80 With Pappus's first theorem, we canlocate the centroid of a semicircularregion without having to integrate(Example 3).

Proof We draw the axis of revolution as the r-axis with the region R in the fir:rquadrant (Fig. 5.78). We let ,()) denote the length of the cross section of Rperpendicular to the }-axis at )'. We assume l(y) to be continuous.

By the method of cylindrical shells, the volume of the solid generated brrevolving the region about the x-a\is is

l d l d\t : | 2r (shell radius)(shell height) dy :2r I y L(y) dy. (5)

J , J ,

The y-coordinare of R s centroid is ,

f d PI i d A I y L ( y ) d y

- J " - J , '' A A

so that

y L ( y ) d y : A y .

Substituting At for the last integral in Eq. (5) gives y:2ryA.With p equal tot, we have V :2trpA. tr

EXAMPLE 2 The volume of the torus (doughnut) generated by revolving acircular disk of radius a about an axis in its plane at a distance D Zt,? from itscenter (Fig. 5.79) is

v : 2ir (b)(r q2) :21r2ba2.

EXAMPLE 3 Locate the centroid of a semicircular region.

So/ution We model the region as the region between the sernicircle y : !/A=7(Fig. 5.80) and the x-axis and imagine revolving the region about the x-axis rogenerate a solid sphere. By symmety, the x-coordinate of the centroid is i:0.With t : p in Eq. (4), we have

V (4/3)ra3 4! : -'

2n A 2n(1 /21na2 3n- "

t,"

-.t

LI

Distance from axis ofrevolution to centroid

Exercises 5.10 441

Theorem 2Pappus's Theorem for Surfac€ AreasIf an arc of a smooth plane curve is revolved once about a line in the planethat does not cut through the arc's interior, then the area of the sur{acegenerated by the arc equals the length of the arc times the distance traveledby the arc's centoid during the revolution. If p is the distance from the axrsof revolution to the centroid, then

S : 2 t t p L .

l ds .

The )-coordinate of the arc's cenboid is

I x=h

I ia 'r _

r,=bl d t

J F .

Hence

f ' -I Y d s : Y L .J, --

Substituting tl, for the last inte$al in Eq. (7) gives S : 2rjL. With p equal to1', we have S : 2tt p L.

(6)

The proof we give assumes that we can model the anis of revolution as the r-axisand the arc as the graph of a smooth function of r.

Proof V'/e dmw the axis of revolution as the r-axis with the arc extending fromx = a to x = b in the fint quadrant (Fig. 5.81). The area of the surface generatedby the arc is

2try d s =r" l.':'':1.',: (7)

J ,:,, , o'L

Figure for Pappus's area theorem.

EXAMPLE 4 The surface area of the torus in Example 2

S : 2 r ( b ) ( 2 r a ) : 4 t r z b a .

Exercises 5.10

Distance and Displacementln Exercises 1 8, the iunction u(1) is the velocity in meters per second.,i a body moving along a coordinate line. (a) Graph lr to see where it:. positive and negative. Then find (b) the total distance traveled by therdy during the given time interval and (c) the body's displacement.

1.

4.

u ( r ) : 5 c o s r ,

1_'la) = srn T t,

u ( r ) : 6 s i n 3 r ,

, ( / ) = 4cos2t .

0 = t = 2 n

o : r : 2O = t = n / 2

! = t = 1 t

5-

6.

1

8.

9,

M2 Chapter 5: Applications of Integrals

v ( t ) - 4 9 9 . 8 t , 0 : l : 1 0

u ( r ) : 8 - 1 . 6 / , 0 : 1 5 1 0

v(t) : 6t2 - 18t + 12: 6(t - l(t - 2), 0 s t 3 2

u(t) : 6t2 - 18t + 12 : 6(t - I)(t - 2), 0 = t 33

The function .r : (l /3)t3 - 3t2 + 8t gives the position of a bodymoving on the horizontal .r-axis at time r > 0 (s in meters, r inseconds).

a) Show that the body is moving to the dght at time I = 0.b) When does the body move to the left?c) What is the body's position at time I = 3?d) When t : 3, what is the total distance the body has traveled?

e) GRAPHER Graph .r as a function of t and comment on therelationship of the graph to the body's motion.

The function s : -t3 + 6t2 - 9t gives the position of a bodymoving on the horizontal .r-axis at time , > 0 (.t in meters, t inseconds).

a) Show that the body is moving to the left at r = 0.b) When does the body move to the right?c) Does the body ever move to the right of the origin? Give

reasons fot your answer.d) What is the body's position at time r :3?e) What is the total distance the particle has taveled by the

time t : 3?

f) GRAPHER Gruph .r as a function of t and comment on therelationship of the graph to the body's motion.

Here are the velocity graphs of two bodies moving on a coordinateline. Find the total distance haveled and the body's displacementfor the given time interval.

Delesset Rule13. The photograph here shows a grid superimposed on the polished

face of a piece of granite. Use the grid and Delesse's rule toestimate the proportion of shrimp-colored gnnular material inthe rcck.

The photograph here shows a grid superimposed on a micrcscopicview of a stained section of human lung tissue. The clear spacesbetween the cells are cross sections ofthe lung's air sacks (calledah)eoli, accett on the second syllable)- Use the grid and Delesse'srule to estimate the proportion of air space in the lung.

Modeling Surface Area15. Modeling sLirface area. The lateral surface area of the cone

swepr out by revolving lhe l ine segment ) : xlJl.O 1r = J3.about the 1-axis should be ( U2xbase circumferencexslant height): (112)(2tr)(2): 22. What do you get if you use Eq. (3) withf (*) : , /J i l

Time(sec,

Velocity(ir./sec)

Time(sec)

Velocity(in-/sec)

0I2345

6'7

89

10

0t222l0

, 1 3

- t l- 6

260

ttna

10.

aara

14,

t (sec)

CALCULATOR The table at the top of the next column shows thevelociry of a model train engine moving back and fofth on a hackfor 10 sec. Use Simpson's rule to find the resulting displacementand total distance traveled.

v (m/sec)

v (m/sec)

w 12.

r ' : * , 0 < r < r E1: l

16.

(18, I )

The onlY surface for which Eq (3)

gives the area we want is a cylinder. Show that Eq (3) gives S :

2rrh fot the cylinder swept out by revolving the line segment

y : r,0 1 ,r < i, about the .ir-axis.

To find the volume of water dis-

placed by a sailboat, the common practice is to partjtion the

waterline into l0 subinteNals of equal length' neasurc the cross

section arca A(r) of the submerged poftion of the hull at each

parlition point, and then use Simpson's rule to estimate the inte-

gral of A(,r) from one end of the waterljne to the other' The table

here lists the area measurements at "stations" 0 through 10' as

the partition points are called, for the cruisrng sloop Pipedream'

shown here. The common subinterval length (distance between

consecutive stations) is i :2 54 ft (about 2' 6 1/2", chosen for

the convenience of the builder).

Exercises 5.10 M3

64 lb/Itr . How many pounds of water does P ipedream displace'!

(Displacement is given in pounds for small craft, and long tons

u long ton : 2240 lbl for larger vessels..i

(Data from Sftene's Elements of yacht DesiStr, Francis S- Kinney'

Dodd, Mead & ComPanY, Inc., 1962)

18. (Cotltinuatio'1 of Exercise /7/ A boat's

pdsmatic coefficient is the ratio of the displacement volume to

the volume of a prism whose height equals the boat's waterline

length and whose base equals the area of the boat's largest sub

merged cross section. The best sailboats have prismatic coefli-

cients between 0.51 and 0 54.Find Pipedredn's prismatic coefll-

cient, given a waterline length of 25.4 ft and a largest submerged

cross section area of 16.14 ft2 (at Station 6).

The Theorems of PaPPus19. The square region with vertices (0, 2), (2' 0)' (1' 2). and (2,4)

is revolved about the,r axis to generate a solid Find the volume

and sutface area of the solid

20, Use a theorem of Pappus to lind the volume generated by re-

volving about the line .I :5 the tdangulal region bounded by

the coordinate axes and the line 2ir + ) :6 (As you saw in

Exercise 31 of Section 5.7, the centroid of a ffiangle lies at the

intenection of the medians, one-third of the way from the mid-

point of each side toward the opposite vertex .)

21. Find the volume of the torus generated by revolving the circle

1 x - 2 . t ' - 1 : : I c b o u t l h e r a \ i :

22. Use the theorems of Pappus to find the lateral surface area and

the volume of c r iSht circular cone

23. Use the second fheorem of Pappus and the fact that the suface

area of a sphere of radius a is 4t tr2 to find the centroid of the

semicircle 1' - './eF =.

24. As found in Exercise 23, the centroid ol dre semicircle y -

^,62 ;ri lies at the point (0,2d/1). Find the alea of the surface

swept out by revolving the semicircle about the line I - a'

25. The area of the region R enclosed by the semiellipse -v:(b I ")

n@-'V anl ttte ;;-axis is ( I /2) z a ll and the volume of the

eliipsoid generated by revolving R about the.lr-axis is ('1/3)z aDl'

Find the cenffoid ofR. Notice the remarkable fact tbat Lhe location

is independent of d

26. As found in Example 3, the centroid ot' the region enclosed

by the i-axis and the semicircle .,- - 'JE = lies at the point

(0,4al3r). Find the volume of the solid generated by revolving

this region aboul the lrne Y = ,-1

27. The region of Exercise 26 is revolved abouf the line ) : i d

to generate a solid. Find the volume of the solid'

28. As found in Exercise 23. the centroid of the semicircle I -

,[2 - x'1 lies at the point (0, 2a/t). Find the area of the surface

generated by revolving the semicircle about the line t =:t - a'

29. Find the moment about the i-axis of the semicilcular region 1n

Example 3. If you use resuks already known' you will not need

to integrate.

4 , "

a) Estinate Pipedreon's displacement volume to the nearest

cubic foot.

Station Submerged area (ft2)

I

2345678

l0

01.0?3.84'7.82

t2.201 5 . 1 816.1 ,114.009.213.210

in the table are fbr seawater,b) The figures which weighs

4M Chapter 5: Applications of Integrals

Quesrroxs ro Guroe YouR Rpvrew

1. How do you deflne and calculate the area of the region betweenthe graphs of two continuous functions? Give an example.

2. How do you define and calculate the volumes of solids by themethod of slicing? Give an example

3, How are the disk and washer methods for calculating volumesderived from the method of slicing? Give examples of volumecalculations by these methods.

4. Describe the method ol cylindrical shells. Give an example.

5. How do you define and calculate the length of the graph of asmooth function over a closed interval? Give an €xample. Whatabout functions that do not have continuous first derivatives?

6. How do you define and calculate the area of the surface sweptout by revolving the graph of a smooth function ) : /(.r), d :"r : b, about the jr-axis? Give an example.

7. What is a center of mass?

8. How do you locate the center of mass of a straight, narow rod orstrip of material? Give an example. If the density of the matedalis constant, you can tell right away where the center of mass is.Where is it?

9, How do you locate the center of mass of a thin llat plate ofmaterial? Give an example.

10. How do you define and calculate th€ work done by a variableforce directed along a portion ofthe r-axis? How do you calculatethe work it takes to pump a liquid from a tank? Give examples.

11. How do you calculate the force exerted by a liquid against apofiion of a vertical wall? Give an example.

12. Suppose you know the velocity function u(r) of a body that willbe moving back and forth along a coordinate line from time , : ato time /: r. How can you predict how much the motion willshift the body's position? How can you predict the total distancethe body will travel?

13. What does Delesse's rule say? Give an example.

14. What do Pappus's two theorems say? Give examples of how theyare used to calculate surface areas and volumes and to locatecentroids.

15. There is a basic pattern to the way we constucted integrals inthis chapter. What is it? Give examples.

CneprBR PRecrrcB ExERCTSES

AreasFind the areas of the regions enclosed by the curves and lines inExercises 1-12.

1 , ) : x , ) : l / r 2 , x : 2

2 . y : x , y : r / " t r . x=z3 . J i+ . , / r y - 1 , r : 0 , y :Q

]

4 . x 3 + u $ = l , : r : 0 , ) : 0 , f o r 0 S . t : 1

)

5 . j r - l ) _ , , r : u . ) : J

6 . x : 4 - y 2 , t : 0

7 , y 2 : 4 x . y = 4 x - 2

8 . , y 2 = 4 . r * 4 , y : 4 x - 1 6

9 . ) : s i n r , y : x , 0 = a : n 1 4

1 0 . l : s i n - r l , r : 1 . - n 1 2 : x = f t / 2

1 1 . ) : 2 s i n t c , ) : s i n 2 r , 0 = i J z

1 2 . y : 8 c o s , r , ) : s e c 2 x . I t 1 3 < x = 7 t / 3

13, Find the area of the "triangular" region bounded on the letl by

r + y - 2.on the right by ) = rr, and above by ) - 2.

14. Find the area of the "triangulai' region bounded on the left by

y - " f , o n t h c r i g h t b y ) : 6 - n , a n d b e l o w b y 1 : I

15. Find the exheme values of /(r) : .rl 3lv2 and nnd the area oI

the region enclosed by the graph of/ and the ,t-axis

16, Find the area of the region cut from the first quadrant by the

ctyre xt /2 + ))t /2 - at /2 .

17. Find the total area of the region enclosed by the curve r : 12/la n d t h e l i n e s t = ] a n d ) : l

18, Find the total area of the region between the curves ): sinr

a n d J : c o s t f o r 0 = t < 3 r 1 2 .

VolumesFind the volumes of the solids in Exercises 19-24.

19. The solid lies between planes perpendicular to the jr-axis at t : 0

and.r : l. The cross sections perpendicular to the jl-axis between

these planes are circular disks whose diameters run ftom the

parabola y = r2 to the parabola ) : .r'/;

20, The base of the solid is the region in the first quadrant between

the line 1 : x and the parabola ) - 2./4. The cross sections

of the solid perpendicular to the ).-axis are equilateral t angles

whose bases stretch from the line to the curve.

21, The solid lies between planes perpendicular to the t-axis at

a : r 14 and x: 5ir/4. The cross sections between these phnes

are circular disks whose diametels run lrom the curve ) : 2cosr

to t he cu rve ) : 2s inn

22. The solid lies between planes perpendicular to the,r axis at ,t : 0

and r :6. The cross sections between these planes :rre squares

whose bases run from the i-axis up to the cu Ne xt /2 + !1/2 : J6.

The solid lies between planes perpendicular to the ir-axis at r. - 0

and,r:4. The cross sections ot' the solid perpendicular to the

ir-a,\is between these planes arc circular disks whose diameters

run from the curve x2 :4) to the curve -12 :'h.

The base of the solid is the region bounded by the parabola

Practice Exercises 445

y2 :4x and the line ,r = I in thc ,l-)-plane Each cross section

perpendicular to the-r-axis is an equilateral triangle with one edge

in the plane. (The triangles all lie on the same side of the plane.)

25, Find the volume of the solid generated by revolving the region

bounded by the r-axis, the cutve r':3:t4, and the lines i : I

and r - - I about (a) the 'I-axis; (b) the )-axis: (c) the line i - 1;

(d) the l ine ) = 3.

26. Find the volume of the solid generated by revolving the "tri-

angular" region bounded by the cuNe -! : '1/x3 and the lines

r - 1 and ) - 1/2 about (a) the:r-axis; (b) the ] axis; (c) the

line 'I : 2; (d) the line 1 - 4.

27, Fjnd the volume of the solid generated by revolving the region

bounded on the left by the parabola r : )2 + I and on the right

by the line r : 5 about (a) the i axis; (b) the )'-axis: (c) the line

28. Find the volume of the solid generated by revolving the region

bounded by the parabola )2 :4x and the line lr - t about (a)

the r-axis; (b) the ]-axis; (c) the line li :4: (d) the line -1 :'1.

29, Find the volume of the solid generated by revolving the "trian-

gular" region bounded by the,r-axis, the line jr :1/3, and the

curve ): tan ir in the first quadrant about tle -t-axis

30. Find the volume of the solid generated by revolving the region

boLrnded by the curve ) - sin:r and the lines -r - 0, 'r : rr, and

I : 2 abou t t he l i ne l : 2 .

31. Find tbe volume of the solid generated by revolving the region

between the x-axis and the curve \i :,!2 2l about (a) the r-

axis; (b) the l ine ] - - l ; (c) the l ine jr - 2; (d) the l ine ) ' = 2

32, Find the volume of the solid generated by revolving about the

.r-axis the region bounded by r-. = 2tanr, -r ' :0,x: r /4, ancl

x : n 14. (The region lies in the first and third quadnnts and

resembles a skewed bow tie.)

33. A round hole of radius .'6 ft is bored through the center of a

solid sphere of radius 2 ft. Find the volume of material removed

trom the sphere.

ffi:a. carcufnfOn The profile of a football resembles thc ellipse

shown here. Find the football's volume to the nearest cubic inch.

Lengths of CurvesFind the lengths of the curves in Exercises 35 38.

3 5 , y : 1 r / : - ( l l 3 ) x 3 / 2 , l = x = 1

3 6 , r : ) ' : / r , 1 : ) : 824,

chapter 5 thomas 9 ed f 2008

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