chapter 16.1 - 16.3 solutions rogowski

798 CHAP TE R 14 I DIFFERENTIATION IN SEVERAL VARIABLES (LT CHAPTER 15)

11. f(x ,y, z)=3x+2y+4z, x2 +2y2 +6z2 =1

SOLUTION We find the extreme values of f(x , y , z ) = 3x + 2y + 4z under the constraint g(x , y, z) = x2 + 2y2 t

6z2 - I= 0. Step 1. Write out the Lagrange Equations. The gradient vectors are 'V f = (3, 2, 4) and 'Vg = (2x , 4y , 12z), therefore the Lagrange Condition 'V f = A 'V g is:

The Lagrange equations are, thus:

(3 , 2, 4) =A (2x, 4y, 12z)

3 = A(2x)

2 = A(4y) =>

4 = A(l2z)

3 -=AX 2

1 2 = AY

1 - = AZ 3

Step 2. Solve for A in terms of x, y , and z. The Lagrange equations imply that x i= 0, y i= 0, and z i= 0. Solving for!. we get

3 A=-

2x ' 1

A=-, 2y

1 A=-

3z

Step 3. Solve for x, y, and z using the constraint. Equating the expressions for A gives

3 2x 2y 3z

=>

Substituting x = i z and y = i z in the equation of the constraint x2 + 2 y2 + 6z2 = I and solving for z we get

G z)2

+2G z)2

+6z2

=I

_!E z2 = 1 4

Using the relations x = iz, y = iz we get

9 2 9

2 2 => Zl = .jill ' Z2 =-.jill

3 2 3 2 X! = 2 · .jill= .jill ' y,-- . ----- 2 .jill - .jill ' Z ! = .jill

9 -2 9 X2- - . -- - ---- 2 .jill - .jill '

3 -2 3 Y2-- · ------ 2 .jill - .jill '

2 Z2 =-.jill

We obtain the following critical points:

( 9 3 2 )

pI = .jill' .jill' .jill

Critical points are also the points on the constraint where 'V g = 0. However, 'V g = (2x , 4y , 12z) = (0, 0, 0) only at the origin, and this point does not lie on the constraint.

Step 4. Computing fat the critical points. We evaluate f (x, y, z) = 3x + 2y + 4z at the critical points:

27 6 8 41 {41 f (p,) = .jill+ .jill+ .jill= .jill= y 3 ~ 3.7

27 6 8 41 {41 f (P2) = -.jill - .jill - .jill = - .jill = -y 3 ~ -3.7

Since f is continuous and the constraint is closed and bounded in R3, f has global extrema under the constraint. We conclude that the minimum value off under the constraint is about -3.7 and the maximum value is about 3.7.

12. f(x , y, z) = x2 - y - z, x2 - y2 + z = 0

SOLUTION We show that the function f(x , y, z ) = x2 - y - z does not have minimum and maximum values subject

to the constraint x2 - y2 + z = 0. Notice that the curve (x , x , 0) lies on the constraint, since it satisfies the equation of the constraint. On this curve we have

f(x , y , z) = f(x , x , 0) = x2 - x- 0 = x2 - x

SEC T I 0 N 14.8 I Lagrange Multipliers: Optimizing with a Constraint (LT SECTION 15.8) 799

Since li m (x 2 - x) = oo, f does not have a maximum value subject to the constraint. Observe that the curve (0, ,JZ, z) x-->±oo

also lies on the constraint, and we have

f(x, y, z) = f (0, ,JZ, z) = 02 - ,JZ- z = - (z +viz)

Since lim - (z + ,JZ) = -oo, f does not attain a minimum value on the constraint either. z-->oo

13. f(x, y , z) = xy + 3xz + 2yz, 5x + 9y + z = lO

SOLUTION We show that f(x , y, z) = x y + 3xz + 2yz does not have minimum and maximum values subject to the constraint g (x, y, z) = 5x + 9y + z - 10 = 0. First notice that the curve c 1 : (x, x, I 0 - 14x) lies on the surface of the constraint since it satisfies the equation of the constraint. On c1 we have,

f(x, y, z) = f(x, x, 10- 14x) = x 2 + 3x(IO- 14x) + 2x(10- 14x) = -69x2 + 50x

Since lim (-69x 2 +Sox) = -oo, f does not have minimum value on the constraint. Notice that the curve c2 X-->00

(x , - x , 10 + 4x) also lies on the surface of the constraint. The values of f on c2 are

f(x, y, z) = f(x, -x, lO + 4x) = -x2 + 3x(LO + 4x)- 2x(10 + 4x) = 3x2 +lOx

The limit lim (3x2 + lOx) = oo implies that f does not have a maximum value subject to the constraint. X--+00

14. ~ Let

g(x, y) = x 3 - xy + y3

(a) Show that there is a unique point P =(a, b) on g(x , y) = I where V fp = A.Vgp for some scalar A. .

(b) Refer to Figure 12 to determine whether f(P) is a local minimum or a local maximum off subject to the constraint.

(c) Does Figure 12 suggest that f(P) is a global extremum subject to the constraint?

y

2

0

-2

- 2 0 2

FIG URE 12 Contour map of f(x, y) = x 3 + xy + y3 and graph of the constraint g(x, y) = x 3 - xy + y3 = 1.

SOLUTION

(a) The gradients off and g are V f = (3x2 + y , x + 3y2 ) and Vg = (3x2 - y, -x + 3y2 ). hence the Lagrange

Condition V f = A. V g is

or

3x2 + y = A.(3x2 - y)

x+3i =A.(-x+3i) (1)

Notice that if 3x2 - y = 0, the first equation implies that also 3x2 + y = 0, hence y = 0 and x = 0. Since the point (0, 0) does not satisfy the equation of the constraint, we may assume that 3x2 - y =1= 0. Similarly, if-x + 3 y2 = 0, the second equation implies that also x + 3y2 = 0, therefore x = y = 0. We thus may also assume that -x + 3y2 =I= 0. Using these assumptions, we have by (1):

3x2 + y A.----

- 3x2 - y' X+ 3y2

A.=---'-.,--x + 3y2

SECT I 0 N 14.8 I Lagrange Multipliers: Optimizing with a Constraint (LT SECTION 15.8) 801

Remark: Since the constraint is not bounded, we need to justify the existence of a minimum value. The values f(x, y) = xy on the constraint y =ex are f(x, ex)= h(x) = xex. Since h(x) > 0 for x > 0, the minimum value (if it exists) occurs at a point x < 0. Since

lim xex = lim X

x-+-oo x-+-oo e-x lim

X--+-00 -e-X lim -ex= 0,

x-+-oo

then for x <some negative number -R, we have lf(x)- 01 < 0.1, say. Thus, on the bounded region -R .:5 x .:50, f has a minimum value of -e- 1 ~ -0.37, and this is thus a global minimum (for all x).

16. Find the rectangular box of maximum volume if the sum of the lengths of the edges is 300 em.

SOLUTION We denote by x, y, and z the dimensions of the rectangular box.

Then the volume of the box is xyz. We must find the values of x, y and z that maximize the volume f(x, y, z) = xyz, subject to the constraint g (x, y, z) = x + y + z = 300, x 2: 0, y 2: 0, z 2: 0. (One could also argue that the sums of the lengths of the edges is 4x + 4 y + 4z = 300, but that would give a different answer, of course. Instead, we will choose to interpret the problem with the constraint x + y + z = 300).

Step 1. Write out the Lagrange Equations. The Lagrange Condition is

'Vf=).'Vg

(yz, xz, xy) =).(I, I, 1)

We obtain the following equations:

yz =).

xz =).

xy =).

Step 2. Solve for). in terms of x, y, and z. The Lagrange equations already_ give). in terms of x, y, and z. Equating the expressions for). we get yz = xz = xy. Step 3. Solve for x, y, and z using the constraint. We have

yz = xz

xy =xz

z(x- y) = 0

x(z- y) = 0

If x = 0, y = 0, or z = 0, the volume has the minimum value 0. We thus may assume that x f 0, y f 0, and z f 0. The first equation implies that x = y and the second equation gives z = y. We now substitute x = y and z = y in the constraint x + y + z = 300 and solve for y:

y+y+y =300

3y = 300 ::::} y = 100

Therefore, x = 100 and z = 100. The critical point is (100, 100, 100). Step 4. Conclusions. The value of f(x, y, z) = xyz at the critical point is

f(IOO, 100, 100) = 1003 = 106 cm3

The constraint x + y + z = 300, x 2: 0, y 2: 0, z 2: 0 is the part of the plane x + y + z = 300 that lies in the first octant. This is a bounded and closed set in R3. Since f is continuous on this set, f has global extreme values on this set. The minimum value is zero (obtained if one of the variables is zero), hence the value 106 is the maximum value. We conclude that the box with maximum value is a cube of edge 100 em.

y

sEcT I 0 N 14.8 I Lagrange Multipliers: Optimizing with a Constraint (LT SECTION 15.8) 811

27. Show that the minimum distance from the origin to a point on the plane ax+ by+ cz =dis

ldl

SOLUTION We want to minimize the distance P = )x2 + y2 + z2 subject to ax+ by + cz = d . Since the square

function u2 is increasing for u :::: 0, the square P2 attains its minimum at the same point where the distance P attains its minimum. Thus, we may minimize the function f(x, y , z) = x 2 + y2 + z2 subject to the constraint g(x , y, z) = ax+ by+ cz =d.

Step 1. WriteouttheLagrangeEquations. We have "\1 f = (2x , 2y, 2z ) and 'Vg = (a , b , c), hencetheLagrangeCondition "\1 f = A 'V g gives the following equations:

Assume for now that a :/= 0, b :/= 0, c :/= 0.

2x = Aa

2y = Ab

2z = AC

Step 2. Solve for A in terms of x, y, and z. The Lagrange Equations imply that

2x A=-,

a

2y A=

b ' 2z

A=c

Step 3. Solve for x, y, and z using the constraint. Equating the expressions for A give the following equations:

2x 2z a x = - z

a c c ::::}

b 2y 2z

b Y = -z

c c

( 1)

We now substitute x = ~ z and y = ~ z in the equation of the constraint ax +by + cz = d and solve for z . This gives

a ( ~ z ) + b ( ~ z) + cz = d

a2 b2 - z +-z + cz =d c c

Since a2 + b2 + c2 :/= 0, we get z = 2 ~~ 2 . We now use (1) to compute y and x: a + +c

a de ad X=- .

c a2 + b2 + c2 - a2 + b2 + c2 '

We obtain the critical point:

(2)

Step 4. Conclusions. It is clear geometrically that f has a minimum value subject to the constraint, hence the minimum value occurs at the point P. We conclude that the point P is the point on the plane closest to the origin. We now consider the case where a = 0. We consider the planes ax +by + cz = d , where a :/= 0 and a ---+ 0. A continuous change in a causes a continuous change in the closest point P. Therefore, the point P closest to the origin in case of a = 0 can be obtained by computing the limit of P in (2) as a ---+ 0, that is, by substituting a = 0. Similar considerations hold for b = 0 or c = 0. We conclude that the closest point P in (2) holds also for the planes with a = 0, b = 0, or c = 0 (but not all of them 0). The distance P of that point to the origin is

P= a2 + b2 + c2

(a2 + b2 + c2)2

ldl

28. Antonio has $5.00 to spend on a lunch consisting of hamburgers ($1.50 each) and French fries ($1.00 per order). Antonio's satisfaction from eating x 1 hamburgers and x2 orders of French fries is measured by a function U(x, , x2) = JXIX2. How much of each type of food should he purchase to maximize his satisfaction? (Assume that fractional amounts of each food can be purchased.)

SOLUTION

sEcT I 0 N 14.8 I Lagrange Multipliers: Optimizing with a Constraint (LT SECTION 15.8) 815

h / Increasing S

Level curves of S I

Critical point P = (r, h)

Level curve of V

~7reasingV

FIGURE 17

• To minimize surface areaS= 2nrh + 2nr2 for a fixed volume (subject to the constraint c(r, h)= nr2h- V) we use the Lagrange equations. Then using '\1 S = A '\1 c, we see

or

2nh +4nr = 2nArh, 2nr = Anr2

Consider the second equation, rewriting we have:

2nr- Anr2 = 0 =} nr(2- Ar) = 0 =} 2

r = O,A =r

Since this is a question about surface area, we are not interested in the point when r = 0. Using the first equation, rewriting we have:

2nh + 4nr = 2nArh 2nh + 4nr h + 2r

A= =--2nrh rh

Now to solve for r, h using the constraint to determine the critical point. Using the two derived equations for A we

have:

2 h +2r

rh =} 2rh = hr + 2r2 r(2h - h- 2r) = 0 =} h = 2r

r

Therefore, we see that the critical point is (r, h) where h = 2r.

• To maximize the volume V = nr2h for a fixed surface area (subject to the constraint c(r, h) = 2nrh + 2nr2- S)

we use the Lagrange equations. Then using '\lV =A '\lc we see

or

(2nrh, nr2 } =A (2nh + 4nr, 2nr)

A(2nh + 4nr) = 2nrh,

rh A=-

h +2r '

A(2nr) = nr2

r A=-

2

Using these two derived equations for A, we have:

r rh

h +2r =} h = 2r -

2

Therefore, we see that the critical point is (r, h) where h = 2r.

Using the contour plots in the figure, we can see that S has a minimum for a fixed value of V, but no maximum because it increases without an upper bound, whereas has V has a maximum for a fixed value of S, but no minimum because it decreases without a lower bound.

33. A plane with equation:::_ + ~ + ~ = I (a, b, c > 0) together with the positive coordinate planes forms a tetrahedron a b c

of volume V = ~abc (Figure 18). Find the minimum value of V among all planes passing through the point P = (1, 1, I).

z

J C= (0, 0, c)

8 = (0, b, 0)

~=(a, O, O) y

X

FIGURE 18

SOLUTION The plane is constrained to pass through the point P = (1, 1, I), hence this point must satisfy the equation of the plane. That is,

I I I -+-+-=1 a b c

We thus must minimize the function V(a, b, c)= ~abc subject to the constraint g(a, b, c)= ~ + t + ~ = l, a> 0, b > 0, c > 0.

Step 1. Write out the Lagrange Equations. We have V' V = (~be , ~ac, ~ab} and V' g = (- ~, - tr, -~ ), hence the

Lagrange Condition V' V = A V' g yields the following equations:

l l -be= --A 6 a2

l l -ac =--A 6 b2

l l -ab =--A 6 c2

Step 2. Solve for A in terms of a, b, and c. The Lagrange equations imply that

bca2 A=---

6 ' acb2

A=---6 '

abc2 A=---

6

Step 3. Solve for a, b, and c using the constraint. Equating the expressions for A, we obtain the following equations:

bca2 = acb2

abc2 = acb2

abc(a- b)= 0

abc(c- b)= 0

Since a, b, care positive numbers, we conclude that a = band c = b. We now substitute a = band c = bin the equation of the constraint ~ + t + ~ = l and solve for b. This gives

l l l -+-+-=1 b b b

3 -=1 ::::} b=3 b

Therefore also a = b = 3 and c = b = 3. We obtain the critical point (3, 3, 3).

Step 4. Conclusions. If V has a minimum value subject to the constraint then it occurs at the point (3, 3, 3). That is, the plane that minimizes V is

X y Z

3 + 3 + 3 = l or x + y + z = 3

Remark: Since the constraint is not bounded, we need to justify the existence of a minimum value of V = !abc under

the constraint ~ + t + ~ = 1. First notice that since a, b, care nonnegative and the sum of their reciprocals is l, none of them can tend to zero. In fact, none of a, b, c can be less than 1. Therefore, if a --+ oo, b --+ oo, or c --+ oo, then V --t oo. This means that we can find a cube that includes the point ( ~, ~, ~) such that, on the part of the constraint that is outside

the cube, it holds that V > V ( ~, ~, ~) = Tk. On the part of the constraint inside the cube, V has a minimum value m,

since it is a closed and bounded set. Clearly m is the minimum of V on the whole constraint.

866 CHAP TE R 15 I MULTIPLE INTEGRATION (LTCHAPTER16)

SOLUTION The double integral is the signed volume of the region between the graph of f(x, y) and the xy-plane over R. In (b) and (c) the function satisfies f (-x, y) = - f (x, y ), hence the region below the xy-plane, where -1 ~ x ~ 0 cancels with the region above the xy-plane, where 0 ~ x ~ l. Therefore, the double integral is zero. In (a) and (d), the function f(x, y) is always positive on the rectangle, so the double integral is greater than zero.

Exercises 1. Compute the Riemann sum S4,3 to estimate the double integral off (x, y) = xy over R = [ l, 3] x [I, 2.5] . Use the

regular partition and upper-right vertices of the subrectangles as sample points.

SOLUTION The rectangle Rand the subrectangles are shown in the following figure:

y

2.5

2 Pn p22 Pn p42

1.5 p" p21 PJI p41

L----+--~~-+--~ X 0 1.5 2 2.5 3

The subrectangles have sides of length

3- l 2.5- I /::;.x = -

4- = 0.5, L:;.y = -

3- = 0.5 =} t:;.A = 0.5 · 0.5 = 0.25

The upper right vertices are the following points:

pll = (1.5, 1.5) p12 = (1.5, 2) pl3 = (1.5, 2.5)

p21 = (2, 1.5) p22 = (2, 2) p23 = (2, 2.5)

p31 = (2.5, 1.5) p32 = (2.5, 2) p33 = (2.5, 2.5)

We compute f(x , y) = xy at these points:

f(Pll) = 1.5 · 1.5 = 2.25 f(P2J)=2 · 1.5= 3 j(P31) = 2.5 · 1.5 = 3.75 j(P41) = 3 · 1.5 = 4.5

Hence, S4,3 is the following sum:

4 3

j(P12) = 1.5 · 2 = 3 j(P22) = 2 · 2 = 4 j(P32) = 2.5 · 2 = 5 f(P42) = 3 · 2 = 6

p4l = (3, 1.5) p42 = (3, 2) p43 = (3, 2.5)

j(P13) = 3.75 j(P23) = 5 j(P33) = 6.25 j(P43) = 7.5

s4,3 = "L "L f(P;j)t:;.A = o.25(2.25 + 3 + 3.75 + 4.5 + 3 + 4 + 5 + 6 + 3.75 + 5 + 6.25 + 7.5) = 13.5 i=lj=l

2. Compute the Riemann sum with N = M = 2 to estimate the integral of .jx + y over R = [0, I] x [0, 1]. Use the regular partition and midpoints of the subrectangles as sample points.

SOLUTION The rectangle Rand the subintervals are shown in the following figure:

y

0.75 pl2 p22 --~-- ---· I

I

0.5 I

I I I I I

0.25 --+ --Pill

--- -+ 1P21 I

X

0 0.25 0.5 0.75

The subrectangles have sides of length /::;.x = i = 0.5 and L:;.y = i = 0.5 and area t:;.A = 0.5 · 0.5 = 0.25. The midpoints of the subrectangles are:

pll = (0.25, 0.25) , p21 = (0.75, 0.25),

p\2 = (0.25, 0.75), p22 = (0.75, 0.75)

SECTION 15.1 I lntegrationinTwoVariables (LTSECTION 16.1) 867

We compute the values of f(x, y) = -Jx + y at the sample points:

f(PII) = ./0.25 + 0.25 = 0.707

f(P21) = ./0.75 + 0.25 = 1

f(PI2) = ./0.25 + 0.75 = 1

f(P22) = ./0.75 + 0.75 = 1.225

Hence, Sn is the following sum:

2 2

s22 = L L f (Pij) ~A= 0.25(0.707 + 1 + 1 + 1.225) = 0.983 i=l )=I

In Exercises 3-6, compute the Riemann sums for the double integral J fn f(x, y) dA, where R = [1, 4] x [1 , 3], for

the grid and two choices of sample points shown in Figure /6.

y y

3LQ· : . : : : X

:Lg_ .. ·· .. · .. . . I

X 0 I 2 3 4 0 I 2 3 4

(A) (B)

FIGURE 16

3. f(x,y)=2x+y

SOLUTION The subrectangles have sides of length ~x = 431 = 1 and ~y = y = 1, and area ~A = ~x ~y = I. We find the sample points in (A) and (B):

(A)

(B)

Pu = (1.5, 1.5) P21 = (2.5, 1.5) P31 = (3.5, 1.5)

pl2 = (1.5, 2.5) p22 = (2.5, 2.5) p32 = (3.5, 2.5)

y

0 2

(A)

4

pll = (1.5, 1.5) p21 = (2, 1) p31 = (3.5, 1.5)

p21 = (2, 3) p22 = (2.5, 2.5) p23 = (4, 3)

y

4

2 ~31

p 21

X

0 2 4

(B)

The Riemann Sum S32 is the following estimation of the double integral:

3 2 3 2

~~ fCx,y)dA~s32=LLf(P;J)~A=LLf(P;J) R i=l )=I i=l )=I

872 CHAPTER 15 I MULTIPLEINTEGRATION (LTCHAPTER16)

12. Use the following table to compute a Riemann sum S3,3 for f(x, y) on the squareR= [0, 1.5] x [0.5, 2] . Use the regular partition and sample points of your choosing.

Values of f(x, y)

2 2.6 2.17 1.86 1.62 1.44 1.5 2.2 1.83 1.57 1.37 1.22 1 1.8 1.5 1.29 1.12 1 0.5 1.4 1.17 I 0.87 0.78 0 I 0.83 0.71 0.62 0.56

0 0.5 1.5 2

SOLUTION The subrectangles and our choice of sample points are shown in the figure:

y

2 PJJ Pn Pn

1.5 pl2 p22

p32 p31

pll

0.5 p21

X

0 0.5 1.5 2

Each subrectangle is a square of side 0.5, hence the area of each subrectangle is L'.A = 0.52 = 0.25. By the given data, the sample points are:

f (Pll) = /(0.5, I) f (P21) = f(l, 0.5) j(P3J) = /(1.5,1)

f (P12) = f(O, 1.5) f (P22) = /(1, 1.5) f (P32) = /(1, 1)

f (P13) = /(0.5, 2) f (P23) = f(l, 2) f (P33) = /(1.5, 2)

The Riemann sum S33 that corresponds to these sample points is the following sum:

3 3

S33 = L L t ( PiJ) L'.A = 0.25(1.5 + 1 + 1.12 + 2.2 + 1.57 + 1.29 + 2.11 + 1.86 + 1.62) ~ 3.58 i=lj=l

{I t 3 3 13. CR 5 Let SN ,N be the Riemann sum for Jo Jo ex -y dy dx using the regular partition and the lower left-hand

vertex of each subrectangle as sample points. Use a computer algebra system to calculate SN ,N for N = 25, 50, 100.

SOLUTION Using a computer algebra system, we compute SN ,N to be 1.0731 , 1.0783 , and 1.0809.

14. CR 5 Let SN ,M be the Riemann sum for

using the regular partition and the upper right-hand vertex of each subrectangle as sample points. Use a computer algebra system to calculate S2N,N for N = 25 , 50, 100.

SOLUTION Using a computer algebra system, we compute S2N ,N to be 14.632, 14.486, and 14.413.

In Exercises 15-18, use symmetry to evaluate the double integral.

15. j Jn x3 dA , R = [-4, 4] x [0, 5]

SOLUTION The double integral is the signed volume of the region between the graph of f(x, y) = x3 and the xy-plane. However, f(x, y) takes opposite values at (x, y) and ( -x, y):

f(-x, y) = (-x)3 = -x3 =- f(x, y)

874 CHAPTER 15 I MULTIPLE INTEGRATION (LTCHAPTER 16)

20. fo213

x3ydxdy

SOLUTION

21. [918

I dx dy 14 -3

SOLUTION

1-118 22. ( -5) dx dy

-4 4

SOLUTION

= fo2 x3 (y22 0 dy

= 4 fo2

x3 dx

= 4. x441: = 16

= 19

11dx

= llxl:

= 99-44 =55

[~1

18

(-5)dxdy = L~ (18(-5)dx) dy

23. 11 frr x 2 sin y dy dx

-do

= L~l ( -5xl:) dy

= 1-l (-40 + 20)dy -4

= 1-l (-20)dy -4

1

-1

= -20y -4

= 20 - (80) = -60

SOLUTION We first evaluate the inner integral, treating x as a constant, then integrate the result with respect to x. This gives

11

frr x 2 sinydydx = 11 x 2 (-cosy)lrr dx = 11

x 2 (-cosn +cosO)dx -I lo -I y=O -1

32.12 ~

4

e3x-y dydx

SOLUTION

lo4 1o5

dydx 33. --0 o .Jx + Y

SOLUTION

lo812

xdxdy 34. 0 I Jx2 + y

SOLUTION

35• [ 2 [ 3 ln(xy) dy dx

11 11 y

SOLUTIO


1 2 ~4 e3x-y dydx = 12 ~4 e3x. e-y dydx

= 12 e3x dx. ~4 e-y dy

{4 {5 dydx {4({5 dy ) 1o 1o .Jx + y = 1o 1o ,Jx + Y dx

= fo4 (2,Jx + YI:=O) dx

= 2 fo\.Jx + 5- ../X) dx

= 2 G<x + 5)3/2- ~x3f2) 1:

= 2 G . 27 _ ~ . 8) _ 2 G . 53/2 _ o) 32 20 76 20 = 36- - - -./5 = - - -./5 ~ 10.426 3 3 3 3

{8 [ 2 xdxdy {8 [ 2 ( xdx ) 1o 1t Jx2+y = 1o Y 1t Jx2+y dy

= fo\J4 + y- Jl+Y)dy

= -(4 + y)3/2- -(1 + y)3/2 2 2 18 3 3 0

= ~(12312 - 27)- ~(8- 1) 3 3

68 = -3 + 16.J3 ~ 5.047


= - [ln(3x)]2 - [ln(x)]2 dx 112 2 I

= ~ [2(1n6)2 - (ln3)2]- f2ln(3x)dx - ~ [2(1n2)2 - o] + f2

lnxdx

=(In 6)2- ~(In 3)2

- [x ln(3x)- xl~] -(In 2)2

+ [x ln x - xl~] l

= (In 6)2 - 2 (In 3)2 - (In 2)2 - (21n 6 - 2 - In 3 + I) + (2ln 2 - 2 - 0 + 1)

1 = (In 6)2 - 2 (In 3)2 - (In 2)2 - 21n 6 + In 3 + 1 + 2ln 2 - 2 + 1

1 = (In 6)2 - 2 (In 3)2 - (In 2)2 - 21n 6 + In 3 + 21n 2 ~ 1.028

loll3 1 36. dxdy

o 2 (x + 4y)3

SOLUTION We calculate the inner integral with respect to x, then we compute the resulting integral with respect to y. This gives

f'f3 1 3 dxdy=f'(f

3 (x+4y)-3 dx)dy=f'_~(x+4y)-2 1 3 dy 1o 12 (x + 4y) 1o 12 1o 2 x=2

=-~fo' (<3+4y)-2 -(2+4y)-2) dy=-/4 (-(3+4y)-l +(2+4y)-')l~

= -~ (2;4y- 3 ;4J I~=-~ ( G- D- G- D) = ~ G -~ + ~ - D = s

16

In Exercises 37-42, use Eq. ( 1) to evaluate the integral.

37. JJn ~dA, R= [-2,4] x [1,3]

SOLUTION We compute the double integral as the product of two single integrals:

!"{ ~dA=14

{ 3 ~dydx= 14

xdx· {3 ~dy

1n Y -2 1, Y -2 1, y

= Ox2

[2) (In yO= ~(16-4)-(ln3-ln1)

= 61n3

38. J fn x 2 ydA, n = [-1, I] X [0, 2]

SOLUTION We compute the double integral as the product of two single integrals:


39. JJR cosxsin2ydA, R= [0, If] x [0, If]

SOLUTION Since the integrand has the form f(x, y) = g(x)h(y), we may compute the double integral as the product of two single integrals. That is,

f·r r 12 r 12 ( r 12

) ( r 12 ) Jn cosxsin2ydA = Jo Jo cosxsin2ydxdy = Jo cosxdx Jo sin2ydy

( lrr/2) ( 1 lrr/2) n ( 1 1 ) = sinx0

- 2cos2y0

=(sin 2 -sino) - 2cosrr+ 2cos0

= (I - 0) G + D = I

40. 1· r _Y_ dA. n = [o. 21 x [o. 4] ln x + 1

SOLUTION We evaluate the integral as the product of two single integrals. This can be done since the function has the form f(x, y ) = g(x)h(y).

fh y !o4 !o2 y (!o2 dx ) (!o4 ) --dA- --dxdy- -- ydy nx+l o ox+l ox+I o

= (ln(x + 1)0 ( y2

2 1:) = (ln3 -In I) ( ~- O:) = 8ln3 ~ 8.79

41. ffn exsinydA, R=[0, 2]x[O.i]

SOLUTION We compute the double integral as the product of two single integrals. This can be done since the integrand has the form f(x, y) = g(x)h(y). We get

ffn ex sinydA = forr/4

fo2

ex sinydxdy = (fo2

ex dx) (forr /\inydy)

42. j fn e3x+4Y dA, R = [0, I] x [1, 2]

SOLUTION We can compute this double integral as the product of two single integrals:

f fn e3x+4y dA = fo I 12 e3x+4y dy dx = fo 1 12 e3x e4y dy dx

= fo 1

e3x dx ·12

e4y dy = ( ~e3x 0 ( ~e4Y 0 !3 lg I 13 8 = 3ce -I)· 4(e - e ) =

12 (e - 1)(e -e)

43. Let f(x,y) = mxy2 , where m is a constant. Find a value of m such that ffn f(x,y)dA = 1, where R

(Q, 1] X (Q, 2].

SOLUTION Since f(x, y) = mxy2 is a product of a function of x and a function of y, we may compute the double integral as the product of two single integrals. That is,

We compute each integral:

f 1 xdx=~x2 11

=~(I 2 -o2)=~ lo 2 0 2 2

!o2 I 12 I 8 i dy = - Y3 = - (23 - o3) = -

0 3 0 3 3

sEcT I 0 N 15.2 I Double Integrals over More General Regions (LT SECTION 16.2) 887

Exercises 1. Calculate the Riemann sum for f (x, y) = x - y and the shaded domain D in Figure 19 with two choices of sample

points, • and o. Which do you think is a better approximation to the integral off over D? Why?

SOLUTION The subrectangles in Figure 17 have sides of length D.x = D.y = I and area D. A = 1 · 1 = 1.

(a) Sample points • · There are six sample points that lie in the domain D . We compute the values off (x, y) = x - y at these points:

The Riemann sum is

f(J, 1) = 0, /(2, I)= I ,

/(1,2)=-1, /(2, 2) = 0,

/(1, 3) = -2, /(2,3) = - 1

S3,4 = (0- I -2 +I+ 0- 1) · 1 = -3

(b) Sample points o. We compute the values of f(x , y) = x- y at the eight sample points that lie in D :

/(1.5 , 0.5) = 1, /(1.5, 3.5) = -2,

The corresponding Riemann sum is thus

/(0.5, l.5) = -1, /(1.5 , 1.5) = 0 , /(2.5 , 1.5) = 1,

/(0.5, 2.5) = -2, /(1.5 , 2.5) = -1 , /(2.5, 2.5) = 0.

S34 = (1- 1-2 + 0- I- 2 + 1 + 0) · I = -4.

2. Approximate values of f(x, y) at sample points on a grid are given in Figure 20. Estimate j fv f(x, y) dx dy for

the shaded domain by computing the Riemann sum with the given sample points.

y

FIGURE 20

SOLUTION The subrectangles have sides of length D.x = 0.5 and D.y = 0.25, so the area is D. A = 0.5 · 0.25 = 0.125. Only nine of the sample points lie in D , hence the corresponding Riemann sum is

s5,4 = (2.5 + 3.3 + 2 + 2.3 + 3 + 3 + 2.9 + 3.5 + 3.5) . 0.125 = 3.25

3. Express the domain Din Figure 21 as both a vertically simple region and a horizontally simple region, and evaluate the integral of f(x , y) = xy over D as an iterated integral in two ways.

y

FIGURE 21


SOLUTION The domain V can be described as a vertically simple region as follows:

0 ~ x ~ I , 0 ~ y ~ 1 - x 2

y

x=O x=l

Vertically simple region

(I)

The domain V can also be described as a horizontally simple region. To do this, we must express x in terms of y, for nonnegative values of x. This gives

y = I - x 2 ::::} x 2 = 1 - y ::::} x = jl"=Y

y

y=l

OsxsvT=Y

Horizontally simple region

Therefore, we can describe V by the following inequalities:

O~y~1, O~x~jl"=Y (2)

We now compute the integral off (x, y) = xy over V first using definition ( 1) and then using definition (2). We obtain

!~ lallal-x

2 lal xy2~ l -x2 Ia! x ( 2 ) Ia! x(1 x2)

2 xydA = xydydx = - dx = - (l-x2) -02 dx = - dx

V 0 0 0 2 y=O 0 2 0 2

=~ {I(x-2x3+x5)dx=~ (x2 _x4 +x6)11 =~(~-~+~)=_!_ 2 }0 2 2 2 6 0 2 2 2 6 12

Using definition (2) gives

!'{ {I {~ {I yx2~~ {I y ( 2 ) Jv xydA = Jo Jo xydxdy = Jo 2 x=O dy = Jo 2 ( !l"=Y) -02 dy

= {I~(I- y)dy= t(~-y2 )dy=y2 _Y311=~-~=_!_ }0 2 }0 2 2 4 6 0 4 6 12

The answers agree as expected.

4. Sketch the domain

V : 0 ~ x ~ 1, x 2 ~ y ~ 4 - x 2

and evaluate J fv y dA as an iterated integral.

SOLUTION The domain V is shown in the following figure:

Y x= I

890 CHAPTER 15 I MULTIPLE INTEGRATION (LTCH APTER 16)

y

2 4

We use Theorem 2 to evaluate the double integral as follows:

x2ydA = x 2ydx dy = - dy= - (2y) 3 -o3 dy= -·8y3 dy !~ 1

2

1

2y

1

2 x3y 12y 1 2 y ( ) 1 2 y

'D 0 0 0 3 x=O 0 3 0 3

= { 2 8y4 d = ~ 512 = 256 ~ 17.07 ]0 3 Y 15 Y 0 15

7. (C)

SOLUTION The domain in (C) is a horizontally simple region, described by the inequalities

0 _:::: y _:::: 2, y _:::: X _:::: 4

y

-7., I i- ----------

-!"-+---+--+---+-X I 2 3 4

Using Theorem 2 we obtain the following integral:

!'{ x2y dA = { 2 { 4 x2ydxdy = { 2 x3y lx=4 dy = { 2 ~ (43- Y3 ) dy = { 2 (64y- Y4 ) dy lv lo ] y lo 3 x=y lo 3 lo 3 3

= 32 Y2 - Ysj2 = 32. 22 - 2s = 608 ~ 40.53 3 15 0 3 15 15

8. Sketch the domain V defined by x + y _:::: 12, x ~ 4, y ~ 4 and compute j fv ex+y dA.

SOLUTION The domain V = [(x, y) : x + y _:::: 12, x ~ 4, y ~ 4} is shown in the following figure:

y

x+y=l2

To compute the integral we described Vas a vertically simple region by the following inequalities (see figure):

y

4SySI2-x

~----+---~--~-X 4 8

4 _:::: X _:::: 8, 4 _:::: y _:::: 12 - X

sECT 1 0 N 15.2 I Double Integrals over More General Regions (LT SECTION 16.2) 891

Using Theorem 2, we obtain the following integral:

9. Integrate f(x, y) = x over the region bounded by y = x 2 andy= x + 2.

SOLUTION The domain of integration is shown in the following figure:

y

-2 2

To find the inequalities defining the domain as a vertically simple region we first must find the x-coordinates of the two points where the line y = x + 2 and the parabola y = x2 intersect. That is,

x + 2 = x 2 ::::} x 2 - x- 2 = (x- 2)(x +I}= 0

::::} X 1 = -1, xz = 2

We describe the domain by the following inequalities:

-] ~X ~ 2, x 2 ~ y ~X+ 2

y

-2 2

We now evaluate the integral of f(x, y) = x over the vertically simple region D:

fl 121x+2 12 lx+2 12 xdA= x dydx= xy dx= x(x+2-x2)dx

V - 1 x2 -1 y=x2 -1

12 x3 x412 (8 ) ( 1 1) I = (x 2 +2x-x3)dx=-+x2 -- = -+4-4- --+1-- =2--1 3 4 -1 3 3 4 4

10. Sketch the region D between y = x 2 andy = x(l - x). Express D as a simple region and calculate the integral of f(x, y) = 2y over D.

SOLUTION The region D between y = x 2 andy= x( l - x) is shown in the following figure:

y

sEcT I 0 N 15.2 I Double Integrals over More General Regions (LT SECTION 16.2) 893

12. Calculate the double integral of f(x, y) = y2 over the rhombus R in Figure 24.

y

4

-4

FIGURE 24 lxl + 11YI :S l

SOLUTION Since f (x, - y) = f (x, y) and since the rhombus is symmetric with respect to the x-axis, the double integral equals twice the integral over the upper half of the rhombus. Moreover, since f (-x, y) = f (x, y) and R is symmetric with respect to they-axis, the double integral over R equals twice the integral over the right half of the rhombus. Therefore, denoting by D the part of the rhombus in the first quadrant, we have

(l)

We now compute the double integral over D . We express D as a vertically simple region. The line connecting the point (0, 4) and (2, 0) has the equation

4-0 Y- 4 =

0 _

2 (x - 0) =} y - 4 = -2x =} y = 4- 2x

Thus, D is defined by the following inequalities:

0 :::: x :::: 2, 0 :::: y :::: 4 - 2x

y

0:Sy:S4- 2x

We now compute the integral over Dusing Theorem 2:

!~ 2 !o2 !o4-2x 2 !o2 y3~4-2x !o2 (4- 2x)3 (4- 2x)412 44 32 y dA= y dydx= - dx= dx= =0+-=-

V o o o 3 y=O o 3 12·(-2) 0 24 3

Combining with (1) we get

J fn i d A = 4 . 3

3

2 ""' 42.67

13. Calculate the double integral of f(x, y) = x + y over the domain D = {(x , y): x 2 + y2 :::: 4, y :=: 0} .

SOLUTION

y

2 Osysv'4-x2

-2 2

The domain V

The semicircle can be described as a vertically simple region, by the following inequalities:

-2:::: x :::: 2, 0:::: y :::: J 4- x2

898 CHAPTE R 15 I MULTIPLE INTEGRATION (LTCHAPTER 16)

19. f(x, y) = x; 0::;: x ::;: I,

SOLUTION We compute the double integral of f(x, y) = x over the vertically simple region D, as the following iterated integral :

= xex dx - x dx = xex dx - - = xex dx - -11 2 11 11 2 x211 11 2 1 o o o 2o o 2

y

OL---------~----x

The resulting integral can be computed using the substitution u = x 2 . The value of this integral is

Combining with (1) we get

[ 1 2 e- I Jo xex dx = -2-

f·r e- 1 1 e-2 lv xdA = -2-- 2 = -

2- ~ 0.359

20. f(x, y) = cos(2x + y); ! S x S If, 1 S y S 2x

SOLUTION The vertically simple region D defined by the given inequalities is shown in the figure:

y

(I)

We compute the double integral of f(x , y) = cos(2x + y) over D as an iterated integral, as stated in Theorem 2. This gives

!~ f n /2 f 2x frr /2 12x cos(2x + y) dA = cos(2x + y) dy dx = sin(2x + y) dx

V 1/2 I 1/2 y= l

= r /2 (sin(2x + 2x)- sin(2x + 1)) dx = {n/

2 (sin(4x)- sin(2x + I)) dx

11 /2 11 /2

= _ cos4x + cos(2x +I) lrr/2 =_cos~ +cos(~+ I) _ (- cos2 + cos2)

4 2 x=l/2 4 2 4 2

= -~ + cos(rr +I) _ cos2 = -0.416 4 2 4

21. f(x, y) = 2xy; bounded by x = y, x = y2

SOLUTION The intersection points of the graphs x = y and x = y2 are (0, 0) (I, 1). The horizontally simple region 'D is shown in the figure:


The domain 1) is a horizontally simple region, as shown in the figure.

y

0 2 4

From the sketch of TJ, we see that 1) can be expressed as a vertically simple region. Rewriting the equation of the curve x = ,JY as y = x 2 , we obtain the following inequalities for TJ:

4

2

0

0 :S x :S 2, 0 :::: y :::: x 2

y

2 4

The integral in reversed order of integration is thus

!o2 !ox2 !o2 lx2 !o2 !o2 JxJ+ldydx= Jx3+1y dx= Jx3+1(x2 -o) dx= Jx3+l-x2dx

0 0 0 y=O 0 0

This integral is easy to compute using the substitution u = x 3 + 1, du = 3x2 dx. This gives

!o2 !ox2 !o2 19 du 2 19 2 52 Jx3+l dydx = JxJ+l. x 2 dx = ../U ·- = -u312 = -(9312 - I)=-~ 5.78 00 0 l 3 9 19 9

Trying to compute the double integral in the original order we find that the inner integral is impossible to compute:

!ol1l 3 35. xeY dydx

0 y=x

SOLUTION The limits of integration define a vertically simple region 1) by the following inequalities:

0 :S X :S 1, X :S y :S 1

This region can also be described as a horizontally simple region by the following inequalities (see figure):

y

y=x

O:Sy:SI, O:sx:sy

SECT I 0 N 15.2 I Double Integrals over More General Regions (LT SECTION 16.2) 907

We thus can rewrite the given integral in reversed order of integration as follows:

y

We compute this integral using the substitution u = y3, du = 3y2 dy. This gives

lo ll y 3 !o'l 3 !o' 1 euil e-1 xeY dx dy = - eY y2 dy = eu · - du = - = -- ~ 0.286 o o 0 2 o 6 6 0 6

Trying to evaluate the double integral in the original order of integration, we find that the inner integral is impossible to compute:

lo'j' 4 36. xeY dydx

0 y=x2f3

SOLUTION The limits of integration define a vertically simple region D by the following inequalities :

0 ::; x ::; 1, x 213 ::; y ::; 1

The region D shown in the figure can also be described as a horizontally simple region.

y

We rewrite the equation of the curve y = x 213 as x = y312 and express the domain D as a horizontally simple region by the following inequalities:

0 ::: y ::: I , 0 ::: X ::: y312

y

0


Finally we substitute the results above to obtain the following solution:

Jfv (x + 1)dA = J!v, (x + 1)dA + Jfv2

(x + 1)dA = 10+ 14 = 24

. ~ny . . . 43. Calculate the double mtegral of f(x, y) =--over the regwn V m F1gure 26.

y

FIGURE 26

SOLUTION To describe V as a horizontally simple region, we rewrite the equations of the lines with x as a function of y, that is , x = y and x = 2y. The inequalities for V are

I .:::: y .:::: 2, y .:::: X .:::: 2y

y

x=y

We now compute the double integral of f(x, y) = o/ over V by the following iterated integral:

!~ sin y i 2 j2Y sin y i 2

sin y 12Y f 2 sin y -dA = -dxdy = - x dy = -(2y- y)dy

'D Y I y Y I Y x=y I Y

f2 sin y f2 12 = -- · y dy = sin y dy = -cosy =cos I -cos 2 ~ 0.956

I Y I I

44. Evaluate J fv x dA for V in Figure 27.

y

FIGURE 27

SOLUTION We compute the integral using decomposition of a domain into smaller domains. We denote by 'D2 and 'D1 the right semicircles of radius 2 and 1, respectively.

y y y

SECT I 0 N 15.2 I Double Integrals over More General Regions (LT SECTION 16.2) 913

Then V2 is the union of Vt and V, which do not overlap except on a boundary curve. Therefore,

Jfv2 xdA = Jfvl xdA + Jfv xdA

or

Jfv xdA = Jfv2 xdA- Jfvl xdA (1)

To express Vt and V2 as horizontally simple regions, we write the equations of the circles in the form x = JI=Y2 and

x = J 4 - y2, respectively. The equations describing the regions V1 and V2 are

y

2

-+--+--x

-2

-2::: y::: 2,

0 :'::X:':: j4- y2

v2 We compute the double integrals over V2:

-1 ::: y::: 1,

O::=x::=Jl=Y2 VI

fh 12 !oy4-Yi 12 x21v4-? 12 ( J4="?)2

- o2

12 ( y2 ) X dA = x dx dy = - dy = dy = 2- - dy

'D2 -2 0 -2 2 x=O -2 2 -2 2

[2 2 y312 8 16 = lo (4- Y ) dy = 4y - 3 o = 8- 3 = 3

y

The integral over Vt is

fh 11 loy'~=? 11 x21v'l=? 11 ( JI=Y2)2

- o2

11 ( 1 Y2) X dA = X dx dy = - dy = dy = - - - dy

'D1 -1 0 -1 2 x=O -1 2 -1 2 2

r I 2 y311 1 2 = lo (l - Y ) dy = Y - 3

0 = 1 - 3 = 3

We now combine (1), (2), and (3) to obtain the following solution:

!'{ xdA=~-~= 14 ::::::4.67 lv 3 3 3

45. Find the volume of the region bounded by z = 40- lOy, z = 0, y = 0, y = 4- x2.

(2)

(3)

SOLUTION The volume of the region is the double integral off (x, y) = 40 - I Oy over the domain V in the xy-plane

between the curves y = 0 andy = 4 - x2. This is a vertically simple region described by the inequalities:

926 CHAPTER 15 I MULTIPLE INTEGRATION (LT CHAPTER 16)

X 4. f(x,y,z) = 2 ; [0,2] x [2,4] x [-1,1]

(y + z)

SOLUTION We write the triple integral as an iterated integral in any order we choose, and then evaluate the resulting integrals successively. We get:

ff.f f(x,y,z)dx=11

f4

f2

x 2

dxdydz=1' f4(f

2 x

2dx)dydz

la -1l2 lo (y+z) -Ih lo (y+z)

11 14 ( 1 !a2 ) 11 14 1 x212 = xdx dydz = - dydz -I 2 (y+z)2 0 -1 2 (y+z)2 2 x=O

11 14 2 1' (14 2 ) = dydz = dy dz -1 2 (y+z)2 -1 2 (y+z)2

11 2 14 11 ( 2 2 ) I' = -- dz= --+-- dz=-21n(4+z)+21n(2+z)

-I y + z y=2 -1 4 + z 2 + z z=-1

34 = -21n5 +21n3- (-21n3 + 2ln 1) = -21n5 +41n3 =In 52 ~ 1.176

5. f(x, y, z) = (x- y)(y- z); [0, I] x [0, 3] x [0, 3]

SOLUTION We write the triple integral as an iterated integral and evaluate the inner, middle, and outer integrals successively. This gives

Jjfa (x- y)(y- z)dV = fo' fo3

fo3 (x- y)(y- z)dzdydx = fo' fo

3 (fo3

(x- y)(y- z)dz ) dydx

=fa' fo\x- y) (yz- ~z2) C0dydx = fo' fo\x- y) (3y- D dydx

= f1 f3

((3x+~) y-~x-3i)dydx= f'(~x+~)i-~xy- y3 13

dx lo lo 2 2 lo 2 4 2 y=O

= f 1 ((~x + ~) · 9- ~x · 3- 27) dx = f

1 -

27 dx =-

27 = -6.75 lo 2 4 2 }0 4 4

z 6.f(x,y ,z) =-; l:Sx:s3, O:s y:s2, 0:Sz:S4

X

SOLUTION We write the triple integral as an iterated integral and evaluate it using iterated integral of a product function. We get

Jjfa f(x, y, z) dV = i3

fo2

fo4 ~ dzdydx = (fo\dz) (fo2

1 dy) (i3

~ dx)

= U z2 1:) ( yl:) (In xi:) =8·2·(1n3-lnl)= J6ln3

7. f(x, y, z) = (x + z)3 ; [0, a] x [0, b] x [0, c)

SOLUTION We write the triple integral as an iterated integral and evaluate it to obtain

f(x,y,z)dV= (x+z) 3 dzdydx= dydx ffi loa lab lac loa lab (x + z)41c

B 0 0 0 0 0 4 z=O

= f a fb (<x + c)4- x4) dydx = fa (x + c)4- x4 ylb dx lo lo 4 4 lo 4 y=O

= fa ~ [ (x + c)4- x4] dx = ~ [ (x + c)S - xs] Ia Jo 4 4 5 5 x=O

b (a + c)5

- a5 - c5 b [ s s s] - = - (a +c) -a - c

4 5 20

SEC T I 0 N 15.3 I Triple Integrals (LT SECTION 16.3) 927

8. f(x, y, z) = (x + y- d; [0, a] x [0, b] x [0, c]

SOLUTION We evaluate the triple integral using Theorem I. This gives

(x + y - c)3 (x + y)3

3 +

3 dydx

= {a ( (x + y- c)4 + (x + y)4lb ) dx ]0 12 12 y=O

= {a_ (x + b- c)4 (x + b)4 (x- c)4 _ x 4 dx ]0 12 + 12 + 12 12

(x+b-c)5 (x+b)5 (x-c)5 x 5 la ..:._ __ __:_ + + - -60 60 60 60 x=O

(a+ b- c)5 (a+ b)5 (a- c)5 a5 (b- c)5 b5 (-c)5

60 + 60 + 60 -60+ 60 -60+~

In Exercises 9-14, evaluate J J fw f(x, y, z) dV for the function f and region W specified.

9. f(x, y, z) = x + y; W: y ::5 z ::5 x, 0 ::5 y ::5 x, 0 ::5 x ::5 I

SOLUTION W is the region between the planes z = y and z = x lying over the triangle D in the xy-plane defined by the inequalities 0 ::5 y ::5 x, 0 ::5 x ::5 1.

y

We compute the integral, using Theorem 2, by evaluating the following iterated integral:

10. f(x, y, z) = ex+y+z; W: 0::5 z ::5 l , 0::5 y ::5 x, 0::5 x ::5 I

SOLUTION W is the region between the planes z = 0 and z = l lying over the triangle D in the xy-plane described in Exercise 9.

y

0

928 CHAPTE R 15 I MULTIPLE INTEGRATION (LTCHAPTER 16)

We compute the triple integral as the following iterated integral:

=fa' (e2x+l -e2x -ex+! +ex) dx = ~e2x+l _ ~e2x -ex+! +ex~~

= ~e3 - ~e2 - e2 + e- (~e- ~e0 - e + e0 ) 2 2 2 2

1 3 3 2 3 I I 3 2 = 2e - 2e + 2e- 2 = 2ce - 3e + 3e- I)

11. f(x, y, z) = xyz; W: 0 ~ z ~ 1, 0 ~ y ~ ~. 0 ~ x ~ 1

SOLUTION W is the region between the planes z = 0 and z = 1, lying over the part D of the disk in the first quadrant.

y

y=~

1J

OL--------4-----x

Using Theorem 2, we compute the triple integral as the following iterated integral:

Jffw xyzdV = Jfv (fa' xyzdz) dA = JJv x~z2 C/A = JJv x; dA

11 (!o~ xy ) !o' xy2,~ !o' x(1-x2) = - dy dx = - dx = dx 0 0 2 0 4 y=O 0 4

= fa I x ~ x3 dx = x82 - ~: ~~ = ~ - 116 = /6

12. f(x, y, z) = x; W: x 2 + y2 ~ z ~ 4

SOLUTION Here, W is the upper half of the solid cone underneath the plane z = 4. First we must determine the projection D of W onto the xy-plane. First considering the intersection of the cone and the plane z = 4 we have:

x2 + y2 = 4

which is a circle centered at the origin with radius 2. The projection into the xy-plane is still x 2 + y2 = 4. The triple integral can be written as the following iterated integral:

ff·r xdV = j'f (14 xdz) dA = j'f xzj

4

dA lw lv x2+y2 lv z=x2+y2

= ffv x(4- x 2 - y2)dA = fa2rr fa2

rcos0(4- r 2 cos20- r2 sin2 0)rdrd0

= -r3 cos I} - -r5 cos 0 dO !o2rr 4 1 12

0 3 5 0

!o2rr 64 64 12rr = - cos 0 dO = - sin 0 = 0 0 15 15 0

930 CHAPTER 15 I MULTIPLE INTEGRATION (LTCHAPTER16)

IS. Calculate the integral of f(x, y, z) = z over the region Win Figure 10 below the hemisphere of radius 3 and lying over the triangle Din the xy-plane bounded by x = 1, y = 0, and x = y.

FIGURE 10

SOLUTION

X

The upper surface is the hemisphere z = j9- x2 - y2 and the lower surface is the xy-plane z = 0. The projection ofV onto the xy-plane is the triangle D shown in the figure.

0

We compute the triple integral as the following iterated integral:

Iff. fh (lo./9-xLy2 ) fh z21./9-xLy2 fh 9 _ x2 _ y2

zdV= zdz dA= - dA= dA v v o v2o v 2

= r'([X9-X2 -y2dy)dx= f'9y-X2y- fiX dx= r'(9x_2X3)dX lo lo 2 lo 2 y=O lo 2 3

= 9:2 - x64~~ = 2 /2 16. Calculate the integral of f(x, y, z) = e2

Z over the tetrahedron Win Figure I l.

X

FIGURE 11

SECT I 0 N 15.3 I Triple Integrals (LT SECTION 16.3) 935

21. Find the volume of the solid in the octant x ~ 0, y ~ 0, z ~ 0 bounded by x + y + z = 1 and x + y + 2z = 1.

SOLUTION The solid W is shown in the figure:

y

X

~ The upper and lower surfaces are the planes x + y + z = 1 (or z = I - x- y) and x + y + 2z = 1 (or z = ------z- ), respectively. The projection of W onto the xy-plane is the triangle enclosed by the line A B : y = 1 - x and the positive x and y-axes.

y

Using the volume of a solid as a triple integral, we have

Volume(W) = !!'{ l dV = j"f (ll -x-y I dz) dA = j"f zll-x-y dA lw Jv (l-x-y)/2 Jv z=( l-x-y)/2

f·r ( 1 -x-y) j"f 1-x-y = Jv (1- x- y)- 2 dA = Jv 2 dA

= r'(rl -x l- x-ydy)dx= r' y-xy-fll-xdx lo lo 2 lo 2 y=O

Ia! 1-x -x(l-x)- (17)

2

I !a' ( x2 1) = dx = - - - x + - dx 0 2 2 0 2 2

= ~ ( x:- x: +~X) I~=~(~-~+ D = /2

22. Calculate J J fw y dV, where W is the region above z = x 2 + y2 and below z = 5, and bounded by y = 0 and

y = l.

SOLUTION The region W is shown in the figure:

X


The upper surface is the plane z = 5 and the lower surface is the paraboloid z = x2 + y2. The projection of W onto the xy-plane is the part of the disk x 2 + y2 ~ 5 between the lines y = 0 and y = I.

y

The triple integral of j(x, y, z) = y over W is equal to the following iterated integral:

ff·r ydv =j"f ({5

ydz)dA=j"f yzl5

dA=J·r y (5-x2 -i)dA Jw Jv 1x2+y2 Jv z=x2+y2 Jv

= f y (5- x 2 - y 2) dx dy = 2 y 5x- ::.._- ix dy l(l~ ) !ol ( 3 )I~ Jo -~ o 3 x=O

[1

( x3

) ~~ t ( 3/2 1 3/2) = 2 lo Y (5- i) X- 3 x=O dy = 2 lo y (5- i) - 3 (5- y2

) dy

{l 4( )3/2 = lo 3 5- i y dy (I)

We compute the integral using the substitution u = 5- y2, du = -2y dy:

!!!. !o I 4 3/2 14 2 15 2 4 15

ydV= -(5-y2) y dy = --u312du= -u312du=-u512 w 0 3 5 3 4 3 15 4

= ~~ (55/2- 45!2) ~ 6.37

23. Evaluate Jffw xzdV, where W is the domain bounded by the elliptic cylinder x4

2 + Y

9

2

x2 + y2 + z2 = 16 in the first octant x 2: 0, y 2: 0, z 2: 0 (Figure 13).

1 and the sphere

X

FIGURE 13

SOLUTION

y

X

The upper surface is the sphere x2 + y2 + z2 = 16, or z = J 16 - x2 - y2, and the lower surface is the xy-plane, z = 0. The projection of W onto the xy-plane is the region in the first quadrant bounded by the ellipse (x /2)2 + (y /3)2 = 1,11

y=~J4-x2.

SECT I 0 N 15.3 I Triple Integrals (LT SECTION 16.3) 937

y

2

v

L---+---~--~--- x

0 2

Therefore, the triple integral over W is equal to the following iterated integral :

Jf·r j'f ( [./16-xLyz ) j'f 1

1

./16-xLyz lw xzdV = lv lo xzdz dA = lv 2xz

2 z=O dA

=~j'f x( 16- x2 -i)dA=~ {2 [i~ l6x-x3 - xidydx

2 lv 2 lo lo

1lo2 1 li~ = - 16xy- x 3y- -xy3 2 0 3 y=O

1 lo2 ~ 3 3 ~ 9 2 3/2 = - 24xv4-x~--xv4-x~--x(4-x) dx 2 0 2 8

The first and third integrals are simple u-substitution problems. For the second integral, let us use u = 4 - x 2 and thus du = -2x dx and x 2 = 4- u. Thus, we can write

-- x 3)4-x2dx=-- x · x 2)4-x2dx 3lo2 3 lo2 4 0 4 0

310 = - (4- u).,fii du 8 u=4

Therefore we have:

!!.[ xzdV= 12 f2 xJ4- x2dx-~ f 2x 3)4-x2dx-.2_ f

2x(4-x2)312

lw lo 4 lo 16 lo

lo2 ~ 16 9 lo2 2 3/2 =12 xv 4 -x~dx---- x(4 -x) dx

0 5 16 0

= -6 · - (4- X ) - - + - · - (4- X ) 2 ( 2 3/212) 16 9 2 ( 2 5/21

2)

3 0 5 32 5 0

16 9 126 = -4 (0 - 8) - 5 + 80 (0 - 32) = 5

24. Describe the domain of integration and evaluate:

SOLUTION The domain of integration W is defined by the fo llowing inequalities:

428 C HAPTER 12 I VECTOR GEOMETRY (LTCHAPTER 13)

Exercises In Exercises 1-4, convert from cylindrical to rectangular coordinates.

1. (4,rr,4)

SOLUTION By the given data r = 4, () = rr and z = 4. Hence,

X = r COS () = 4 COS 11' = 4 · (- 1) = -4

y = r sine = 4 sin 11' = 4 ° 0 => (x, y, z) = (-4, 0, 4)

z =4

2. (2, ~ · -8) SOLUTION We are given that (r, (), z) = (2, 2f, -8) . Hence,

11' 1 X = r COS() = 2 COS - = 2 · - = 1

3 2

y = r sine = 2 sin ::_ = 2 .J3 = .J3 => (x' Y' z) = ( 1, .J3, - 8) 3 2

z = -8

3. (o. ~· D SOLUTION We haver= 0, () = }-, Z = i· Thus,

x =rcose =O · cos::_ =0 5

11' ( 1) y = r sine = 0 · sin S = 0 => (x, y, z) = 0, 0, 2

1 z = 2

SOLUTION Conversion to rectangular coordinates gives

x = r cos() = 1 ·cos ::_ = 1 · 0 = 0 2

y = rsin e = I · sin ::_ = 1 . 1 = 1 => (x , Y, z) = (0, 1, - 2) 2

z = -2

In Exercises 5-10, convert from rectangular to cylindrical coordinates.

5. (1, -1, 1)

SOLUTION We are given that x = 1, y = -1, z = 1. We find r:

r = J x2 + y2 = J 12 + ( -1 )2 = Jz

Next we find () . The point (x, y) = ( 1, -1) lies in the fourth quadrant, hence,

y -I tan e = - = - = -1,

X 1

3rr - <() <2rr 2 - -

7rr B=-

4

sECT I 0 N 12.7 I Cylindrical and Sphertcal Coordinates (LT SECTION 13.7) 429

We conclude that the cylindrical coordinates of the point are

(r, e, z) = (h. 7:, 1). 6. (2, 2, I)

SOLUTION We are given that (x, y, z) = (2, 2, 1). We first find r:

r = J x2 + y2 = )22 + 22 = .J8 = 2v'2

Next we find e. The point (x, y) = (2, 2) lies in the first quadrant hence 0 ~ e ~ !f. Therefore,

y 2 tanB=-=-=1,

X 2 0<8 < ::_ - - 2

The cylindrical coordinates of the point are

(r,B,z) = (2v'2, ~ · 1).

7. (1, v'3, 7)

SOLUTION We have X = 1, y = ../3, Z = 7. We first find r :

r = J x2 + y2 = J 12 + ( J3) 2 = 2

Since the point (x, y) = ( 1, ../3) lies in the first quadrant, 0 ~ e ~ !f. Hence,

tanB=~= ../3 =v'3 X 1 '

0 <8< ::_ - - 2

The cylindrical coordinates are thus

(r, e, z) = ( 2, ~ , 7) .

8. 0· 3:.9) SOLUTION We are given that (x, y, z) = G, ¥, 9). Hence,

7r 8=-

3

Since the point (x, y) = G· ¥) is in the first quadrant, 0 ~ e ~ !f. Therefore,

y 3../3/2 tane =- = -- = v'3 X 3/2 '

The cylindrical coordinates are thus

9. (~ . ~ . 2) SOLUTION We have x = h' y = h' z = 2. We find r:

7r o < e <- - 2 7r

8=-3

Since the point (x, y) = ( h, h) is in the first quadrant, 0 ~ e ~ !f, therefore,

y 5/../2 tane =- = -- = 1 X 5j../2 '

The corresponding cylindrical coordinates are

7r o < e < - - 2

(r, e, z) = (5, ~· 2) .

7r 8=-

4

430 CHAPTER 12 I VECTOR GEOMETRY (LT CHAPTER 13)

10. (3, 3.J3, 2)

SOLUTION We have x = 3, y = 3.J3, and Z = 2, hence,

The point (x, y) = (3, 3.J3) is in the first quadrant hence 0 ~ 8 ~ ~.Therefore,

tane = ~ = 3.J3 = .J3,

X 3

The cylindrical coordinates of the given point are

7f o < e <- - 2

(r, e, z) = ( 6, ~ , 2) .

In Exercises 11-16, describe the set in cylindrical coordinates.

11. x2 + y2 ~ l

7f e =-3

SOLUTION The inequality describes a solid cylinder of radius I centered on the z-axis. Since x2 + y2 = r 2, this

inequality can be written as r 2 ~ I .

12. x2 + y2 + z2 ~ I

SOLUTION Since x2 + y2 = r2 , this inequality can be written as

r 2 + z2 ~ I or r 2 ~ I - z2

13. y2 + z2 ~ 4, x = 0

SOLUTION The projection of the points in this set onto the xy-plane are points on they axis, thus 8 = ~ or 8 = ~

Therefore, y = r sin ~ = r · 1 = r or y = r sin ( *) = -r. In both cases, y2 = r 2, thus the inequality y2 + z2 ~ 4

becomes r 2 + z2 ~ 4. In cylindrical coordinates, we obtain the following inequality

7f 8=-

2

37r or e = T

14. x2 + y2 + z2 = 4, x :::: 0, y :::: 0, z :::: 0

SOLUTION We express z in terms of x and y. Since z :::: 0 we get

x2 + i + z2 = 4 => z2 = 4 - (x2 + il => z = J 4 - (x2 + y2)

The cylindrical coordinates are (r, e, z) where x2 + y2 = r 2. Substituting into (I) gives

z = )4- r 2

We find the interval for 8. The given set is the part of the sphere x2 + y2 + z2 = 4 in the first octant.

7f Hence, the angle e is changing between 0 and i .

We combine (2) and (3) to obtain the following representation:

z = )4- r 2 ,

15. x2 + y2 ~ 9, x :::: y

o < e < ~ - - 2

(I)

(2)

(3)

SOLUTION The equation x2 + y2 ~ 9 in cylindrical coordinates becomes r 2 ~ 9, which becomes r ~ 3. However, we also have the restriction that x :::: y. This means that the projection of our set onto the xy plane is below and to the right of the line y = x. In other words, our e is restricted to - 3rr I 4 ~ e ~ rr I 4. In conclusion, the answer is:

r ~ 3, -3rrl4 ~ e ~ rrl4

SECT I 0 N 12.7 I Cylindrical and Spherical Coordinates (LT SECTION 13.7) 431

16. y 2 + z2 s 9, x 2: y

SOLUTION The region X 2: y in the xy-plane is determined by the inequalities 54 :5 8 :5 2rr, 0 :5 8 < i (and the origin). Since y = r sin e, the region y 2 + z2 :5 9 can be written as

We obtain the following description in cylindrical coordinates:

o < e < :!'._ - - 4 or

In Exercises 17-24, sketch the set (described in cylindrical coordinates).

17. r = 4

5rr - <B <2rr. 4 - -

SOLUTION The surface r = 4 consists of all points located at a distance 4 from the z-axis. It is a cylinder of radius 4 whose axis is the z-axis. The cylinder is shown in the following figure:

y

71' 1s. e = 3

SOLUTION The equation 8 = ~ defines the half plane of all points that project onto the ray 8 = ~ in the xy-plane. This half plane is shown in the following figure:

4

19. z = -2

SOLUTION z = -2 is the horizontal plane at height -2, shown in the following figure:

y

X

chapter 16.1 - 16.3 solutions rogowski

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