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Chapter Six

Applications of Integration

Section 6.1

More about Areas

Goals

◦ Use integrals to find areas of more general regions,

including

regions that lie between the graphs of two functions

regions enclosed by parametric curves

Area Between Curves

Let S be the region that lies…

◦ between two curves y = f(x) and y = g(x), and

◦ between the vertical lines x = a and x = b.

Here we assume that f and g are continuous

functions and that f(x) ≥ g(x) for all x in [a, b].

We want the area of S, as illustrated on the

next slide:

Areas (cont’d)

Riemann Sum

As earlier with areas under curves, we

◦ divide S into n strips of equal width, and

◦ approximate the ith strip by a rectangle of base ∆x

and height f(xi*) – g(xi*).

As the next slide shows, the Riemann sum

is an approximation to the area of S:

* *

1

n

i ii

f x g x x

Riemann Sum (cont’d)

Definition

Thus we define the area A of S to be the

limiting value of the sum of the areas of

these approximating rectangles:

We recognize this limit as the definite

integral of f – g :

* *

1

limn

i in

i

A f x g x x

Area Formula

In the special case where g(x) = 0,

◦ S is the region under the graph of f , and so

◦ our new formula reduces to our old one.

Example

Find the area of the region enclosed by the

parabolas y = x2 and y = 2x – x2.

Solution First, using algebra we find that the

points of intersection of these two curves are

(0, 0) and (1, 1).

From the figure on the next slide we identify

the top and bottom curves

yT = 2x – x2 and yB = x2 :

Solution (cont’d)

Solution (cont’d)

The area of a typical rectangle is

and the region lies between x = 0 and x = 1.

So the total area is

2 2 22 2 2T By y x x x x x x x x

Changing the Point of View

Some regions are best treated by

regarding x as a function of y.

The next slide shows a region bounded

by the curves x = f(y), x = g(y), y = c, and y

= d, where

◦ f and g are continuous and

◦ f(y) ≥ g(y) for c ≤ y ≤ d :

Point of View (cont’d)

Point of View (cont’d)

The area of the figure is given by

If we write xR for the right boundary and xL

for the left, then as the next slide shows,

Point of View (cont’d)

Example

Find the area enclosed by the line y = x –

1 and the parabola y2 = 2x + 6.

Solution By algebra we find that the

points of intersection are (–1, –2) and (5,

4).

We solve the equation of the parabola for

x ; as the next slide illustrates,

xL = (½)y2 – 3 and xR = y – 1

Solution (cont’d)

Solution (cont’d)

Thus

Parametric Curves

The area under a curve y = F(x) from a to b is

If a curve is traced out once by the parametric

equations x = f(t) and y = g(t),

α ≤ t ≤ β, then we can calculate an area

formula using the Substitution Rule:

, where 0.b

aA F x dx F x

Example

Find the area under one arch of the cycloid

x = r(θ – sinθ) , y = r(1 – cosθ).

Solution One arch of the cycloid (shown on

the next slide) is given by 0 ≤ θ ≤ 2π.

On the subsequent slide we apply the

Substitution Rule with

◦ y = r(1 – cosθ) and

◦ dx = r(1 – cosθ)dθ :

Solution (cont’d)

Solution (cont’d)

Review

Use of Riemann sums to define areas

between curves, both regarding…

◦ y as a function of x and

◦ x as a function of y

Areas enclosed by parametric curves

Section 6.2

Volumes

Goals

◦ Define volume of a solid

◦ Use integrals to find volumes of solids of known

cross-section

◦ Find volumes of solids of revolution, using the

methods of

disks/washers

cylindrical shells

What is Volume?

What do we mean by the volume of a

solid?

How do we know that the volume of a

sphere of radius r is 4πr3/3 ?

How can we give a precise definition of

volume?

Our Starting Point: Cylinders

A cylinder is bounded by a plane region B1

(the base) and a congruent region B2 in a

parallel plane

We define the volume

of the cylinder by V =

Ah , where A is the area

of the base and h is the

height

Familiar Cylinders

Circular cylinder

◦ Base is a circle of radius r , and

so has area πr2

◦ Thus V = πr2h

Rectangular box

◦ Base is a rectangle of area lw

◦ Thus V = lwh

The base of a cylinder need

not be circular!

What About Other Solids?

What if a solid S is not a cylinder?

We can use our knowledge of cylinders:

◦ Cut S into thin parallel slices

◦ Treat each piece as though it were a cylinder,

and add the volumes of the pieces

◦ The thinner we make the slices, the closer we

will be to the actual volume of S

This leads to the method of finding

volumes by cross-sections

The Method of Cross-Sections

Intersect S with a

plane Px

perpendicular to

the x-axis

Call the cross-

sectional area A(x)

A(x) will vary as x increases from a to b

Cross-Sections (cont’d)

Divide S into “slabs” of equal width ∆x

using planes at x1, x2,…, xn

◦ Like slicing a loaf of bread!

Choose a sample point in each *ix 1 ,i ix x

Cross-Sections (cont’d)

The ith slab (or, slice of bread) is roughly a

cylinder with volume

◦ Here is the base of the “cylinder”

and is its height

Adding the volumes of the individual slabs

gives an approximation to the volume of S:

*

1

n

ii

V A x x

*iA x x

*iA x

x

Precise Definition of Volume

Now we use more and more slabs

◦ This corresponds to letting n ∞

The larger the value of n, the closer each slab

becomes to an actual cylinder

◦ In other words, our approximation becomes better and better as n ∞

So we define the volume of S by

*

1

limn b

i ani

V A x x A x dx

Example: Volume of a Sphere

Here

So

(as the usual formula predicts!)

2 2 2A x y r x

2 2 34

3

r

rV r x dx r

A Special Case

A solid of revolution is formed by rotating a

plane region about an axis

About Solids of Revolution

The slabs are always in one of two shapes:

◦ Either the shape of a disk…

…as was the case in the sphere problem above

◦ or the shape of a washer

A washer is the region between two concentric circles

Let’s look at two examples of each type

◦ First, an example of disks, using the solid in the

preceding slide

First Example of Disks

Problem Find the volume of the solid

obtained by rotating about the x-axis the

region under the curve from 0

to 1.

◦ See the diagram two slides back

Solution Each cross-section is a disk, so

Here the radius is , so

y x

2radius of cross-sectionA x

x

2

A x x x

First Example of Disks (cont’d)

So the volume of the solid of revolution is

In general: Whenever we use the disk

method, we will start with the formula

121

002 2

xV xdx

2radius of cross-sectionA x

Second Example of Disks

Problem Find the volume of the solid obtained by rotating the region bounded by y = x3 , y = 8 , and x = 0 about the y-axis

Here the slices are perpendicular to the y-axis, rather than the x-axis

Second Example of Disks (cont’d)

Solution The cross –

section is given by

So the volume is

2 2 / 33A y y y

8 8 2 / 3

0 0

85 / 3

0

3 96

5 5

V A y dy y dy

y

First Example of Washers

Problem Find the volume of the solid

obtained by rotating about the x-axis the

region enclosed by the curves y = x and

y = x2

On the next slide are shown the region and

the solid

First Example of Washers (cont’d)

First Example of Washers (cont’d)

Solution To find A(x) we

subtract the area of the

inner disk from that of the

outer disk:

22 2

2 4

A x x x

x x

First Example of Washers (cont’d)

Therefore

In general: Whenever we use the washer

method, we will start with the formula

13 51 8 2 4

0 00

2

3 5 15

x xV A x dx x x dx

2 2outer radius inner radiusA

A Bigger Picture

The formula

can be applied to any solid for which the

cross-sectional area A(x) can be found

This includes solids of revolution, as

shown above…

…but includes many other solids as well

b

aV A x dx

Example Problem A solid has a circular base of radius

1 . Parallel cross-sections perpendicular to the

base are equilateral triangles. Find the volume

of the solid.

Example (cont’d)

Solution Here the cross-sections are

equilateral triangles with

2 2

2

12 3

21

2 1 3 12

3 1

A x y y

x x

x

Example (cont’d)

So

At right is shown

the completed solid.

1

14 3 / 3V A x dx

Review

Use of Riemann sums to define volume

◦ Cylinders

Volumes of solids with known cross-

section

Solids of revolution

◦ Method of disks/washers

6.3 Volume by Cylindrical Shells

Sometimes the disk/washer method is difficult for a solid of revolution

An alternative is to divide the solid into concentric circular cylinders

This leads to the method of cylindrical shells

Volumes by Cylindrical Shells

Let’s consider the problem of finding the volume of the

solid obtained by rotating about the y-axis the region bo

unded by y = 2x2 – x3 and y = 0. (See Figure 1.)

If we slice perpendicular to the y-axis, we get a washer.

Figure 1

Volumes by Cylindrical Shells

But to compute the inner radius and the outer radius of the washer

, we’d have to solve the cubic equation y = 2x2 – x3 for x in terms of

y; that’s not easy.

Fortunately, there is a method,

called the method of cylindrical

shells, that is easier to use in

such a case. Figure 2 shows a

cylindrical shell with inner radius r1,

outer radius r2, and height h.

Figure 2

Volumes by Cylindrical Shells

Its volume V is calculated by subtracting the volume V1 of the inner

cylinder from the volume V2 of the outer cylinder:

V = V2 – V1

= r22h – r1

2h = (r22 – r1

2)h

= (r2 + r1) (r2 – r1)h

= 2 h(r2 – r1)

Volumes by Cylindrical Shells

If we let r = r2 – r1 (the thickness of the shell) and r = (r2 + r1)

(the average radius of the shell), then this formula for the volume of

a cylindrical shell becomes

and it can be remembered as

V = [circumference] [height] [thickness]

Volumes by Cylindrical Shells

Now let S be the solid obtained by rotating about the y-axis the region bounded by

y = f (x) [where f (x) 0], y = 0, x = a and x = b, where b > a 0. (See Figure 3.)

We divide the interval [a, b] into n subintervals [xi – 1, xi] of equal

width x and let be the midpoint of the i th subinterval.

Figure 3

Volumes by Cylindrical Shells

If the rectangle with base [x i – 1, xi] and height is rotated

about the y-axis, then the result is a cylindrical shell with average

radius , height , and thickness x (see Figure 4), so by

Formula 1 its volume is

Figure 4

Volumes by Cylindrical Shells

Therefore an approximation to the volume V of S is give

n by the sum of the volumes of these shells:

This approximation appears to become better as n

. But, from the definition of an integral, we know that

Volumes by Cylindrical Shells

Thus the following appears plausible:

The best way to remember Formula 2 is to think of a typical shell, c

ut and flattened as in Figure 5, with radius x, circumference 2x, hei

ght f (x), and thickness x or dx:

This type of reasoning will be helpful in other situations, such as wh

en we rotate about lines other than the y-axis.

Volumes by Cylindrical Shells

Figure 5

Example 1 – Using the Shell Method

Find the volume of the solid obtained by rotating

about the y-axis the region bounded by y = 2x2 – x3 and

y = 0.

Solution:

From the sketch in Figure 6 we see that a typical shell h

as radius x, circumference 2x, and height f (x) = 2x2 – x3

Figure 6

Example 1 – Solution

So, by the shell method, the volume is

It can be verified that the shell method gives the same a

nswer as slicing

cont’d

Review

Solids of revolution

◦ Method of cylindrical shells

Section 6.4

Arc Length

Goals

◦ Define arc length of a smooth curve given

parametrically

◦ Give an integral formula for arc length

Introduction

What do we mean by the length of a

curve?

If the curve is a polygon, then the length is

simply the sum of the lengths of the line

segments that form the polygon.

Our approach to the length of a general

curve C is shown on the next slide:

Introduction (cont’d)

Introduction (cont’d)

We define the length by…

◦ first approximating it by a polygon, and then

◦ taking a limit as the number of segments of

the polygon is increased.

We will assume that C is described by the

parametric equations

x = f(t), y = g(t), a ≤ t ≤ b

Introduction (cont’d)

We also assume that C is smooth, that is, that the derivatives f (t) and g (t) are

◦ continuous, and

◦ not simultaneously zero, a < t < b.

This ensures that C has no sudden change

in direction.

Inscribed Polygon

We divide the parameter interval [a, b] into n

subintervals of equal width ∆t.

The preceding figure shows the points

P0, P1, …, Pn corresponding to the endpoints of

the above subintervals.

The polygon with these vertices approximates

C, and so the length of this polygon approaches that of C as n ∞

Inscribed Polygon (cont’d)

Thus we define the length of C to be the

limit of the lengths of these inscribed

polygons:

Now we work toward a more convenient

expression for L :

11

limn

i in

i

L P P

Riemann Sum

If we let ∆xi = xi – xi-1 and ∆yi = yi – yi-1,

then the length of the ith line segment of

the polygon is

But from the definition of a derivative we know that f (ti) ≈ ∆xi/∆t if ∆t is small.

Therefore ∆xi ≈ f (ti)∆t and ∆yi ≈ g (ti)∆t.

22

1i i i iP P x y

Riemann Sum (cont’d)

Therefore

so that

Riemann Sum (cont’d)

This is a Riemann sum for the function

suggests that

With the restriction that no portion of the

curve is traced out more than once, this is the

correct arc length formula:

2 2

, and so our argumentf t g t

2 2b

aL f t g t dt

Arc Length Formula

As an example we find the length of the arc of

the curve x = t2, y = t3 that lies between the

points (1, 1) and (4, 8), shown on the next

slide:

Example (cont’d)

Solution

We note that the given portion of the

curve corresponds to the parameter

interval 1 ≤ t ≤ 2, so the formula gives

Solution (cont’d)

Substituting u = 4 + 9t2 leads to

This number is just slightly larger than the

distance joining the endpoints (1, 1) and (4,

8)—as the figure would suggest.

A Special Case

If we are given a curve y = f(x), a ≤ t ≤ b,

then we can regard x as a parameter.

With the parametric equations x = x,

y = f(x), our formula becomes

Example

Find the length of one arch of the cycloid

x = r(θ – sinθ) , y = r(1 – cosθ).

Solution We know from earlier work that

one arch is described by the parameter

interval 0 ≤ θ ≤ 2π. Since

Solution (cont’d)

we have

A computer algebra system gives the value of

8r for this integral.

◦ Thus the length of one arch of a cycloid is eight

times the radius of the generating circle.

Review

Use of Riemann sums to define arc length

using polygonal approximation

Integral formula for arc length