Steward – Application of Integration

(1)

414

In this chapter we explore some of the applications of the definite integral by using it

to compute areas between curves, volumes of solids, and the work done by a varying

force. The common theme is the following general method, which is similar to the one

we used to find areas under curves: We break up a quantity

into a large number of

small parts. We next approximate each small part by a quantity of the form

and

thus approximate

by a Riemann sum. Then we take the limit and express

as an

integral. Finally we evaluate the integral using the Fundamental Theorem of Calculus

or the Midpoint Rule.

Q Q

f共xi*兲⌬x Q

The volume of a sphere is the limit of sums of volumes of approximating cylinders.

APPLICATIONS OF

INTEGRATION

(2)

0

y=© y=ƒ

S

F I G U R E 1

S=s(x, y) | a¯x¯b, ©¯y¯ƒd

x y

b a

In Chapter 5 we defined and calculated areas of regions that lie under the graphs of

functions. Here we use integrals to find areas of regions that lie between the graphs of two

functions.

Consider the region

that lies between two curves

and

be-tween the vertical lines

and

, where

and are continuous functions and

for all in

. (See Figure 1.)

Just as we did for areas under curves in Section 5.1, we divide

S

into

n

strips of equal

width and then we approximate the

i

th strip by a rectangle with base

and height

. (See Figure 2. If we like, we could take all of the sample points to be right

endpoints, in which case

.) The Riemann sum

is therefore an approximation to what we intuitively think of as the area of

S

.

This approximation appears to become better and better as

. Therefore we define

the

area

of the region as the limiting value of the sum of the areas of these

approxi-mating rectangles.

We recognize the limit in (1) as the definite integral of

. Therefore we have the

fol-lowing formula for area.

The area

A

of the region bounded by the curves

, and the

lines

,

, where and are continuous and

for all in

, is

Notice that in the special case where

,

is the region under the graph of

and our general definition of area (1) reduces to our previous definition (Definition 2 in

Section 5.1).

f

S

t

共x兲

_苷

₀

A

_苷

y

b a

关

f

共x兲

⫺

t

_共x兲兴

dx

关a, b兴

x f共x兲艌t_共_x_兲

t f x苷b

x苷a

y苷f共x兲, y苷t共x兲 2

f

⫺

t

A

_苷

lim

nl⬁

兺

n

i苷1

关

f

共x

i

*

兲

⫺

t

共x

*

i

兲兴

⌬

x

1

S

A

n

_l

⬁

(a) Typical rectangle

x y

b 0 a

f

(

xi*

)

f

(

xi*

)

-g

(

x*i

)

_g

(

xi*

)

x*i Îx

(b) Approximating rectangles x y

b 0 a

F I G U R E 2

兺

n i苷1

关

f

共x

*

i

兲

⫺

t

共x

*

i

兲兴

⌬

x

*

i

苷

x

i

f

共x

*

i

兲

⫺

t

共x

*

i

兲

⌬

_x

关a

,

b兴

x

f

共x兲

艌

t

共x兲

t

f

x

_苷

b

x

_苷

a

y

_苷

t

共x兲

y

_苷

f

共x兲

S

(3)

In the case where both and are positive, you can see from Figure 3 why (2) is true:

EXAMPLE 1

_{Find the area of the region bounded above by}

_{, bounded below by}

, and bounded on the sides by

x

_苷

0 and

x

_苷

1.

SOLUTION

_{The region is shown in Figure 4. The upper boundary curve is}

_{and the}

lower boundary curve is

. So we use the area formula (2) with

,

and :

M

In Figure 4 we drew a typical approximating rectangle with width

as a reminder of

the procedure by which the area is defined in (1). In general, when we set up an integral

for an area, it’s helpful to sketch the region to identify the top curve , the bottom curve

, and a typical approximating rectangle as in Figure 5. Then the area of a typical

rect-angle is

and the equation

summarizes the procedure of adding (in a limiting sense) the areas of all the typical

rectangles.

Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Figure 3

the right-hand boundary reduces to a point. In the next example both of the side

bound-aries reduce to a point, so the first step is to find

a

and

b

.

EXAMPLE 2

_{Find the area of the region enclosed by the parabolas}

_and

.

SOLUTION

_{We first find the points of intersection of the parabolas by solving their}

equa-tions simultaneously. This gives

, or

. Thus

,

so

or 1. The points of intersection are

and

.

We see from Figure 6 that the top and bottom boundaries are

and

The area of a typical rectangle is

and the region lies between

and

. So the total area is

(4)

Sometimes it’s difficult, or even impossible, to find the points of intersection of two

curves exactly. As shown in the following example, we can use a graphing calculator or

computer to find approximate values for the intersection points and then proceed as before.

EXAMPLE 3

_{Find the approximate area of the region bounded by the curves}

and

SOLUTION

_{If we were to try to find the exact intersection points, we would have to solve}

the equation

This looks like a very difficult equation to solve exactly (in fact, it’s impossible), so

instead we use a graphing device to draw the graphs of the two curves in Figure 7. One

intersection point is the origin. We zoom in toward the other point of intersection and

find that

. (If greater accuracy is required, we could use Newton’s method or a

rootfinder, if available on our graphing device.) Thus an approximation to the area

between the curves is

To integrate the first term we use the subsitution

. Then

, and

when .

So

M

EXAMPLE 4

_{Figure 8 shows velocity curves for two cars, A and B, that start side by side}

and move along the same road. What does the area between the curves represent? Use

the Midpoint Rule to estimate it.

SOLUTION

_{We know from Section 5.4 that the area under the velocity curve}

A

represents

the distance traveled by car A during the first 16 seconds. Similarly, the area under curve

B

is the distance traveled by car B during that time period. So the area between these

curves, which is the difference of the areas under the curves, is the distance between the

cars after 16 seconds. We read the velocities from the graph and convert them to feet per

second .

共

1 mi

兾

h

苷

52803600

ft

兾

s

兲

⬇

0.785 苷

s

2.39 ⫺

1 ⫺

共

1.18 兲

5

5 ⫹

共

1.18 兲

2

2 苷

s

u

]

1

2.39

⫺

冋

x

5

5 ⫺

x

2

2 册

0 1.18

A

⬇

1

2

y

2.39

1

du

s

u

⫺

y

1.18

0

共x

4

_⫺

_x兲

_dx

x

_苷

1.18, we have

u

⬇

2.39 du

_苷

2 x dx

u

_苷

x

2

_⫹

1 A

⬇

y

1.18

0

冋

x

s

x

2

_⫹

₁

⫺

共x

4

_⫺

_x兲

册

_dx

x

⬇

1.18 x

s

x

2

_⫹

₁

苷

x

4

_⫺

_x

y

_苷

x

4

_⫺

_x

.

y

_苷

x兾

s

x

2

_⫹

₁

1.5

_1

_1 2

y=x$-x x

œ„„„„„≈+1

F I G U R E 7 y=

t 0 2 4 6 8 10 12 14 16

0 34 54 67 76 84 89 92 95

0 21 34 44 51 56 60 63 65

0 13 20 23 25 28 29 29 30

v_A⫺v_B v_B v_A

F I G U R E 8 0 10 20 30 40 50 60

A

B

(5)

We use the Midpoint Rule with

intervals, so that

. The midpoints of the

intervals are

,

, and

. We estimate the distance between the

cars after 16 seconds as follows:

M

If we are asked to find the area between the curves

and

where

for some values of but

for other values of , then we split the

given region into several regions ,

, . . . with areas ,

, . . . as shown in Figure 9.

We then define the area of the region to be the sum of the areas of the smaller regions

,

, . . . , that is,

. Since

we have the following expression for

A

.

The area between the curves

and

and between

and

is

When evaluating the integral in (3), however, we must still split it into integrals

corre-sponding to

,

, . . . .

EXAMPLE 5

_{Find the area of the region bounded by the curves}

_,

, and .

SOLUTION

_{The points of intersection occur when}

_{, that is, when}

(since

). The region is sketched in Figure 10. Observe that

when

but

when

. Therefore the required

area is

In this particular example we could have saved some work by noticing that the region

is symmetric about

and so

M

A

_苷

2 A

1

_苷

2 y

␲兾4 0

共

cos

x

⫺

_sin

_x兲

_dx

x

_苷

␲

_兾

₄

苷

2 s

2 ⫺

2 苷

冉

_s

1

2 ⫹

1 s

2 ⫺

0 ⫺

1 冊

⫹

冉

⫺

0 ⫺

1 ⫹

1 s

2 ⫹

1 s

2 冊

苷

[

sin

x

⫹

cos

x

]

0

␲兾4

⫹

[

⫺

_cos

x

⫺

_sin

x

]

␲兾4 ␲兾2

苷

y

␲兾4 0

共

cos

x

⫺

_sin

_x兲

_dx

⫹

y

␲兾2 ␲兾4

共

sin

x

⫺

_cos

_x兲

_dx

A

_苷

y

␲兾2

0

ⱍ

cos

x

⫺

_sin

_x

ⱍ

_dx

_苷

_A

₁

⫹

_A

₂

␲

_兾

₄

艋

x

艋

␲

_兾

₂

sin

x

艌

_cos

x

0 艋

x

艋

␲

_兾

₄

cos

x

艌

_sin

x

0 艋

x

艋

␲

_兾

₂

x

_苷

␲

_兾

₄

sin

x

_苷

cos

x

_苷

␲

_兾

₂

x

_苷

0 y

_苷

cos

x

y

_苷

sin

x

V

A

2

A

1

A

_苷

y

b a

ⱍ

f

共x兲

⫺

t

_共x兲

ⱍ

_dx

x

_苷

b

x

_苷

a

y

_苷

t

共x兲

y

_苷

f

共x兲

3

ⱍ

f

共x兲

⫺

t

_共x兲

ⱍ

_苷

再

f

共x兲

⫺

t

共x兲

t

共x兲

⫺

f

共x兲

when

f

共x兲

艌

t

_共x兲

when

t

共x兲

艌

f

共x兲

A

_苷

A

1

⫹

A

2

⫹ ⭈ ⭈ ⭈

S

2

S

1

S

A

2

A

1

S

2

S

1

S

x

t

共x兲

艌

f

共x兲

x

f

共x兲

艌

t

共x兲

y

_苷

t

共x兲

y

_苷

f

共x兲

苷

4 共

93 兲

苷

372 ft

y

16 0

共

v_A

⫺

v_B

兲

dt

⬇

⌬

t

关

13 ⫹

23 ⫹

28 ⫹

29 兴

t

4

苷

14 t

3

苷

10 t

2

苷

6 t

1

苷

2 ⌬

t

_苷

4 n

_苷

4

0 x

y

a b

y=ƒ y=© S¡

S™ S£

F I G U R E 9

F I G U R E 1 0

0 x

y

x=0 A¡

y =cos x _y=_sin_x

A™

π

4 π2

(6)

Some regions are best treated by regarding

x

as a function of

y

. If a region is bounded

by curves with equations

,

, and

, where and are

contin-uous and

for

(see Figure 11), then its area is

If we write

for the right boundary and

for the left boundary, then, as Figure 12

illustrates, we have

Here a typical approximating rectangle has dimensions

and

.

EXAMPLE 6

_{Find the area enclosed by the line}

_{and the parabola}

.

SOLUTION

_{By solving the two equations we find that the points of intersection are}

and

. We solve the equation of the parabola for

x

and notice from

Figure 13 that the left and right boundary curves are

We must integrate between the appropriate -values,

and

. Thus

M

We could have found the area in Example 6 by integrating with respect to

x

instead of

y

, but the calculation is much more involved. It would have meant splitting the region in

two and computing the areas labeled

and

in Figure 14. The method we used in

Example 6 is

much

easier.

A

2

A

1

苷

⫺

16

共

64 兲

⫹

8 ⫹

16 ⫺

(

4

3

⫹

2 ⫺

8 )

苷

18 苷

⫺

1

2 冉

y

3

3 冊

⫹

y

2

2 ⫹

4 y

册

⫺2 4

苷

y

4

⫺2

(

⫺

1₂

y

2

_⫹

_y

_⫹

4 )

dy

苷

y

4

⫺2

[共

y

⫹

₁

_兲

⫺

(

1 2

y

2

_⫺

3 )]

dy

A

_苷

y

4

⫺2

共x

R

⫺

x

L

兲

dy

y

_苷

4 y

_苷

⫺

₂

y

x

R

苷

y

⫹

1 x

L

苷

12

y

2

_⫺

3 共

5, 4

兲

共

⫺

_1,

⫺

₂

_兲

y

2

苷

2 x

⫹

6 y

_苷

x

⫺

₁

V

⌬

_y

x

R

⫺

x

L

A

_苷

y

d

c

共x

R

⫺

x

L

兲

dy

x

L

x

R

x c

d y

0

y=d

x=g(y) x=f(y)

y=c Îy

F I G U R E 1 1

0 x

y

c d

xR xL

xR-xL Îy

F I G U R E 1 2

A

_苷

y

d

c

关

f

共

y兲

⫺

t

_共y兲兴

_dy

c

艋

y

艋

d

f

共y兲

艌

t

_共y兲

t

f

y

_苷

d

y

_苷

c

x

_苷

t

共y兲

x

_苷

f

共

y兲

x y

_2 4

0

(_1, _2)

(5, 4)

xR=y+1 1

2 xL= ¥-3

F I G U R E 1 3

⫺3

(5, 4)

(_1, _2) y=x-1

A¡

y=_ 2x+6 A™ y= 2x+6œ„„„„„

œ„„„„„

F I G U R E 1 4

0 x

(7)

,

29 –30 Use calculus to find the area of the triangle with the given

vertices.

, ,

30. , ,

31–32 Evaluate the integral and interpret it as the area of a

region. Sketch the region.

31.

32.

33 – 34 _{Use the Midpoint Rule with} _{to approximate the}

area of the region bounded by the given curves.

33. , ,

34. _, _,

;

35 –38 _{Use a graph to find approximate -coordinates of the points} of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.

35. _,

1– 4 Find the area of the shaded region.

1. 2.

4.

5 – 28 _{Sketch the region enclosed by the given curves. Decide}

whether to integrate with respect to xor y. Draw a typical approx-imating rectangle and label its height and width. Then find the area of the region.

(8)

(c) Which car is ahead after two minutes? Explain.

(d) Estimate the time at which the cars are again side by side.

46. The figure shows graphs of the marginal revenue function

and the marginal cost function for a manufacturer. [Recall from Section 4.7 that and represent the revenue and cost when units are manufactured. Assume that and are measured in thousands of dollars.] What is the meaning of the area of the shaded region? Use the Midpoint Rule to estimate the value of this quantity.

;

47. _{The curve with equation} _{is called}

Tschirn-hausen’s cubic. If you graph this curve you will see that part of the curve forms a loop. Find the area enclosed by the loop.

48. _{Find the area of the region bounded by the parabola} _,

the tangent line to this parabola at , and the -axis.

49. Find the number such that the line divides the region

bounded by the curves and into two regions

with equal area.

50. (a) Find the number such that the line bisects the

area under the curve ,

(b) Find the number such that the line bisects the area in part (a).

Find the values of such that the area of the region bounded

by the parabolas and is 576.

52. Suppose that . For what value of is the area of

the region enclosed by the curves , ,

and equal to the area of the region enclosed by the

curves , , and ?

For what values of do the line and the curve enclose a region? Find the area of the region. y苷x兾共x2⫹1兲

39. Use a computer algebra system to find the exact area

enclosed by the curves and .

40. Sketch the region in the -plane defined by the inequalities

, and find its area.

41. Racing cars driven by Chris and Kelly are side by side at the

start of a race. The table shows the velocities of each car (in miles per hour) during the first ten seconds of the race. Use the Midpoint Rule to estimate how much farther Kelly travels than Chris does during the first ten seconds.

42. The widths (in meters) of a kidney-shaped swimming pool

were measured at 2-meter intervals as indicated in the figure. Use the Midpoint Rule to estimate the area of the pool.

43. A cross-section of an airplane wing is shown. Measurements

of the height of the wing, in centimeters, at 20-centimeter

intervals are , , , , , , , , ,

, and . Use the Midpoint Rule to estimate the area of the wing’s cross-section.

44. _{If the birth rate of a population is} _people

per year and the death rate is people per

year, find the area between these curves for . What does this area represent?

Two cars, A and B, start side by side and accelerate from rest. The figure shows the graphs of their velocity functions. (a) Which car is ahead after one minute? Explain. (b) What is the meaning of the area of the shaded region?

(9)

VOLUMES

In trying to find the volume of a solid we face the same type of problem as in finding areas.

We have an intuitive idea of what volume means, but we must make this idea precise by

using calculus to give an exact definition of volume.

We start with a simple type of solid called a

cylinder

(or, more precisely, a

right

cylin-der

). As illustrated in Figure 1(a), a cylinder is bounded by a plane region

, called the

base

, and a congruent region

in a parallel plane. The cylinder consists of all points on

line segments that are perpendicular to the base and join

to

. If the area of the base is

and the height of the cylinder (the distance from

to

) is , then the volume

of the

cylinder is defined as

In particular, if the base is a circle with radius , then the cylinder is a circular cylinder with

volume

[see Figure 1(b)], and if the base is a rectangle with length and width

, then the cylinder is a rectangular box (also called a

rectangular parallelepiped

) with

volume [see

Figure

1(c)].

For a solid

S

that isn’t a cylinder we first “cut”

S

into pieces and approximate each piece

by a cylinder. We estimate the volume of

S

by adding the volumes of the cylinders. We

arrive at the exact volume of

S

through a limiting process in which the number of pieces

becomes large.

We start by intersecting

S

with a plane and obtaining a plane region that is called a

cross-section

of

Let

be the area of the cross-section of in a plane

perpen-dicular to the -axis and passing through the point , where

. (See Figure 2.

Think of slicing with a knife through and computing the area of this slice.) The

cross-sectional area

will vary as increases from to .

F I G U R E 2

y

x

0 a x b

A(b)

A

P_x P P

b

a

x

A共x兲

x

S

a

艋

x

艋

b

x

P

x

S

A共x兲

S

.

F I G U R E 1

h

B¡ B™

h

r

h

l

(a) Cylinder

V=Ah (b) Circular cylinderV=πr@h (c) Rectangular boxV=lwh w

V

_苷

l

w

h

w

l

V

_苷

␲

_r

2

_h

r

V

_苷

Ah

V

h

B

2

B

1

A

B

2

B

1

B

2

(10)

Let’s divide

S

into

n

“slabs” of equal width

by using the planes

,

, . . . to slice

the solid. (Think of slicing a loaf of bread.) If we choose sample points

in

, we

can approximate the th slab

(the part of that lies between the planes

and

) by

a cylinder with base area

and “height”

. (See Figure 3.)

FIGURE 3

The volume of this cylinder is

, so an approximation to our intuitive

concep-tion of the volume of the th slab

is

Adding the volumes of these slabs, we get an approximation to the total volume (that is,

what we think of intuitively as the volume):

This approximation appears to become better and better as

. (Think of the slices as

becoming thinner and thinner.) Therefore, we

define

the volume as the limit of these sums

as

. But we recognize the limit of Riemann sums as a definite integral and so we

have the following definition.

DEFINITION OF VOLUME

_{Let be a solid that lies between}

_and

_{. If the}

cross-sectional area of in the plane , through

x

and perpendicular to the

x

-axis,

is

, where

is a continuous function, then the

volume

of is

When we use the volume formula

, it is important to remember that

is the area of a moving cross-section obtained by slicing through perpendicular to

the -axis.

Notice that, for a cylinder, the cross-sectional area is constant:

for all . So our

definition of volume gives

; this agrees with the formula

EXAMPLE 1

_{Show that the volume of a sphere of radius is}

_.

SOLUTION

_{If we place the sphere so that its center is at the origin (see Figure 4), then the}

plane

P

x

intersects the sphere in a circle whose radius (from the Pythagorean Theorem)

V

_苷

43

␲

r

3

r

V

_苷

Ah

.

V

_苷

x

b

a

A

dx

苷

A共b

⫺

a兲

x

A共x兲

苷

A

x

A共x兲

V

_苷

x

b a

A共x兲

dx

V

_苷

lim

nl⬁

兺

n

i苷1

A共x

i

*

兲

⌬

x

苷

y

b

a

A共x兲

dx

S

A

A共x兲

P

x

S

x

_苷

b

x

_苷

a

S

n

l

⬁

n

l

⬁

V

⬇

兺

n

i苷1

A共x

i

*

兲

⌬

x

V共S

i

兲 ⬇

A共x

i

*

兲

⌬

x

S

i

A共x

*

i

兲

⌬

x

y

0 a=x¸ ⁄ ¤ ‹ x¢xx x∞ xß x¶=b x xi-1 xi

y

0 _x_* x

i Îx

S

a b

⌬

_x

A共x

*

i

兲

P

x_i

P

x_i⫺1

S

i

关x

i⫺1

,

x

i

兴

x

i

*

P

x₂

P

x₁

⌬

_x

F I G U R E 4

y

_r r x

NIt can be proved that this definition is

inde-pendent of how is situated with respect to the -axis. In other words, no matter how we slice with parallel planes, we always get the same answer for .V

S x

(11)

is

. So the cross-sectional area is

Using the definition of volume with

and

, we have

(The integrand is even.)

M

Figure 5 illustrates the definition of volume when the solid is a sphere with radius

. From the result of Example 1, we know that the volume of the sphere is

. Here the slabs are circular cylinders, or

disks

, and the three parts of

Fig-ure 5 show the geometric interpretations of the Riemann sums

when

n

_苷

5, 10, and 20 if we choose the sample points

to be the midpoints . Notice

that as we increase the number of approximating cylinders, the corresponding Riemann

sums become closer to the true volume.

EXAMPLE 2

_{Find the volume of the solid obtained by rotating about the}

_x

_{-axis the}

region under the curve

from 0 to 1. Illustrate the definition of volume by

sketch-ing a typical approximatsketch-ing cylinder.

SOLUTION

_{The region is shown in Figure 6(a). If we rotate about the}

x

-axis, we get the

solid shown in Figure 6(b). When we slice through the point

x

, we get a disk with radius

. The area of this cross-section is

and the volume of the approximating cylinder (a disk with thickness

) is

A共x兲

⌬

x

_苷

␲

_x

⌬

x

⌬

x

A共x兲

苷

␲

(

s

x

)

2

苷

␲

x

s

x

y

_苷

s

x

V

(a) Using 5 disks, VÅ4.2726 (b) Using 10 disks, VÅ4.2097 (c) Using 20 disks, VÅ4.1940

F I G U R E 5 Approximating the volume of a sphere with radius 1

x

i

x

i

*

兺

n i苷1

A

共

x

i

兲

⌬

x

苷

兺

n

i苷1

␲

_共

₁

2

_⫺

_x

i2

兲

⌬

x

4

3

␲

⬇

4.18879

r

_苷

1 苷

43

␲

r

3

苷

2 ␲

冋

r

2

x

⫺

x

3

3 册

0

r

苷

2 ␲

冉

r

3

⫺

r

3

3 冊

苷

2 ␲

y

r

0

共

r

2

_⫺

_x

2

_兲

_dx

V

_苷

y

r

⫺r

A

共

x

兲

dx

苷

y

r

⫺r

␲

_共

_r

2

_⫺

_x

2

_兲

_dx

b

_苷

r

a

_苷

⫺

r

A

共

x

兲

_苷

␲

_y

2

苷

␲

共

r

2

⫺

x

2

兲

y

_苷

s

r

2

_⫺

_x

2

Visual 6.2A shows an animation of Figure 5.

(12)

The solid lies between

and

, so its volume is

M

EXAMPLE 3

_{Find the volume of the solid obtained by rotating the region bounded by}

,

, and about

the

-axis.

SOLUTION

_{The region is shown in Figure 7(a) and the resulting solid is shown in}

Figure 7(b). Because the region is rotated about the

y

-axis, it makes sense to slice the

solid perpendicular to the

y

-axis and therefore to integrate with respect to

y

. If we slice

at height

y

, we get a circular disk with radius

x

, where

. So the area of a

cross-section through

y

is

and the volume of the approximating cylinder pictured in Figure 7(b) is

Since the solid lies between

y

_苷

0 and

y

_苷

8, its volume is

M

F I G U R E 7

y=8

x=0

y=˛ or

(a)

0

(x, y) Îy

(b) x

y

0 x

y

8

x=œ„y3

苷

␲

[

35

y

5兾3

]

0 8

苷

96 ␲

5 V

_苷

y

8

0

A

共

y

兲

dy

苷

y

8

0

␲

_y

2兾3

_dy

A

共

y

兲

⌬

y

_苷

␲

_y

2兾3

_⌬

_y

A

共

y

兲

苷

␲

x

2

苷

␲

(

s

3

y

)

2

苷

␲

y

2兾3

x

_苷

s

3

_y

y

x

_苷

0 y

_苷

8 y

_苷

x

3 V

F I G U R E 6 (a)

x

0 x

y

y=œ„œ

1

œ„ œ

(b)

Îx

0 x

y

1

V

_苷

y

1

0

A

共

x

兲

dx

苷

y

1

0

␲

_{x dx}

_苷

␲

x

2

2 册

0 1

苷

␲

2 x

_苷

1 x

_苷

0

NDid we get a reasonable answer in

Example 2? As a check on our work, let’s replace the given region by a square with base

and height . If we rotate this square, we get a cylinder with radius , height , and volume . We computed that the given solid has half this volume. That seems about right.

␲_ⴢ₁2

ⴢ1苷␲

1 1 1

(13)

EXAMPLE 4

_{The region}

_{enclosed by the curves}

_and

_{is rotated about the}

-axis. Find the volume of the resulting solid.

SOLUTION

_{The curves}

_and

_{intersect at the points}

_and

_{. The region}

between them, the solid of rotation, and a cross-section perpendicular to the -axis are

shown in Figure 8. A cross-section in the plane

has the shape of a

washer

(an annular

ring) with inner radius

and outer radius , so we find the cross-sectional area by

sub-tracting the area of the inner circle from the area of the outer circle:

Therefore we have

M

EXAMPLE 5

_{Find the volume of the solid obtained by rotating the region in Example 4}

about the line

.

SOLUTION

_{The solid and a cross-section are shown in Figure 9. Again the cross-section is}

a washer, but this time the inner radius is

and the outer radius is

.

F I G U R E 9

0

y=2 y=2

4

2

x 1

x

y=≈ y=x

y

x x

2-≈

≈

2-x

x

2 ⫺

x

2

2 ⫺

x

y

_苷

2

F I G U R E 8

(1, 1)

y=≈ y=x

᏾

( b)

x y

0

(a) (c)

x ≈

A(x)

x y

(0, 0)

苷

␲

冋

x

3

3 ⫺

x

5

5 册

₀

1

苷

2 ␲

15 V

_苷

y

1

0

A

共

x

兲

dx

苷

y

1

0

␲

_共

x

2

_⫺

_x

4

_兲

_dx

A

共

x

兲

苷

␲

x

2

⫺

␲

共

x

2

兲

2

苷

␲

共

x

2

⫺

x

4

兲

x

2

P

x

共

1, 1

兲

共

0, 0

兲

y

_苷

x

2

y

_苷

x

y

_苷

x

2

y

_苷

x

᏾

Visual 6.2B shows how solids of revolution are formed.

(14)

The cross-sectional area is

and so the volume of is

M

The solids in Examples 1–5 are all called

solids of revolution

because they are obtained

by revolving a region about a line. In general, we calculate the volume of a solid of

revo-lution by using the basic defining formula

and we find the cross-sectional area

or

in one of the following ways:

N

If the cross-section is a disk (as in Examples 1–3), we find the radius of the

disk (in terms of

x

or

y

) and use

N

If the cross-section is a washer (as in Examples 4 and 5), we find the inner

radius

and outer radius

from a sketch (as in Figures 8, 9, and 10) and

compute the area of the washer by subtracting the area of the inner disk from

the area of the outer disk:

The next example gives a further illustration of the procedure.

F I G U R E 1 0

r_in

rout

A

_苷

␲

_共

_{outer radius}

_兲

2

_⫺

_␲

_共

inner radius

兲

2

r

out

r

in

A

_苷

␲

_共

_radius

_兲

2

A共

y兲

A共x兲

V

_苷

y

b

a

A共x兲

dx

or

V

苷

y

d

c

A共y兲

dy

苷

8 ␲

15 苷

␲

冋

x

5

5 ⫺

5 x

3

3 ⫹

4 x

2

2 册

0 1

苷

␲

y

1

0

共

x

4

_⫺

5 x

2

_⫹

4 x

兲

dx

苷

␲

y

1

0

关共

2 ⫺

_x

2

_兲

2

_⫺

_共

2 ⫺

_x

_兲

2

_兴

_dx

V

_苷

y

1

0

A

共

x

兲

dx

S

(15)

EXAMPLE 6

_{Find the volume of the solid obtained by rotating the region in Example 4}

about the line

.

SOLUTION

_{Figure 11 shows a horizontal cross-section. It is a washer with inner radius}

and outer radius

, so the cross-sectional area is

The volume is

M

We now find the volumes of three solids that are

not

solids of revolution.

EXAMPLE 7

_{Figure 12 shows a solid with a circular base of radius 1. Parallel}

cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid.

SOLUTION

_{Let’s take the circle to be}

_{. The solid, its base, and a typical}

cross-section at a distance from the origin are shown in Figure 13.

F I G U R E 1 3

y y

60° 60° _B

A

C

œ„ œ3y

(c) A cross-section

A B(x, y) y=œ„„„„„„≈

(b) Its base x y

0 y

x

(a) The solid 0

A B

1

_1 x

y C

F I G U R E 1 2

Computer-generated picture of the solid in Example 7

y

x

2

_⫹

_y

2

苷

1

F I G U R E 1 1 x=_1

y y

x 0

x=œ„y y

x=y y

1 1+y 1+œ„

苷

␲

冋

4 y

3兾2

3 ⫺

y

2

2 ⫺

y

3

3 册

0 1

苷

␲

2 苷

␲

y

1

0

(

2 s

y

⫺

y

⫺

y

2

₎

_dy

苷

␲

y

1

0

[(

1 ⫹

s

_y

)

2

_⫺

_共

1 ⫹

_y

_兲

2

]

_dy

V

_苷

y

1

0

A

共

y

兲

dy

苷

␲

(

1 ⫹

s

y

)

2

⫺

␲

共

1 ⫹

y

兲

2

A

共

y

兲

苷

␲

共

outer radius

兲

2

⫺

␲

共

inner radius

兲

2

1 ⫹

s

_y

1 ⫹

_y

x

_苷

⫺

₁

Visual 6.2C shows how the solid in Figure 12 is generated.

(16)

Since lies on the circle, we have

and so the base of the triangle

is

. Since the triangle is equilateral, we see from Figure 13(c) that its

height is

. The cross-sectional area is therefore

and the volume of the solid is

M

EXAMPLE 8

_{Find the volume of a pyramid whose base is a square with side and}

whose height is .

SOLUTION

_{We place the origin}

_{at the vertex of the pyramid and the -axis along its}

cen-tral axis as in Figure 14. Any plane

that passes through and is perpendicular to the

-axis intersects the pyramid in a square with side of length , say. We can express in

terms of by observing from the similar triangles in Figure 15 that

and so

. [Another method is to observe that the line

has slope

and

so its equation is

.] Thus the cross-sectional area is

The pyramid lies between

and

, so its volume is

M

We didn’t need to place the vertex of the pyramid at the origin in Example 8.

We did so merely to make the equations simple. If, instead, we had placed the center of the

base at the origin and the vertex on the positive -axis, as in Figure 16, you can verify that

y

N OT E

苷

L

2

h

2

x

3

3 册

₀ h

苷

L

2

_h

3 V

_苷

y

h

0

A

共

x

兲

dx

苷

y

h

0

L

2

h

2

x

2

_dx

x

_苷

h

x

_苷

0

O

x h

F I G U R E 1 4

s L

O

P

F I G U R E 1 5

x y

x

h

A

共

x

兲

苷

s

2

苷

L

2

h

2

x

2

y

_苷

Lx

兾共

2 h

兲

L

兾共

2 h

兲

OP

s

_苷

Lx

兾

h

x

h

苷

s

兾

2 L

兾

2 苷

s

L

x

s

x

P

x

O

h

L

V

苷

2 y

1

0

s

3 共

1 ⫺

x

2

_兲

_dx

苷

2 s

3 冋

x

⫺

x

3

3 册

₀

1

苷

4 s

3

3 V

_苷

y

1

⫺1

A

共

x

兲

dx

苷

y

1

⫺1

s

3 共

1 ⫺

x

2

_兲

_dx

A

共

x

兲

苷

12

ⴢ

2 s

1 ⫺

x

2

ⴢ

s

3 s

1 ⫺

x

2

苷

s

3 共

1 ⫺

x

2

兲

s

3 y

_苷

s

3 s

1 ⫺

_x

2

ⱍ

AB

ⱍ

_苷

2 s

1 ⫺

x

2

ABC

y

_苷

s

1 ⫺

x

2

B

h

0

y

F I G U R E 1 6

(17)

we would have obtained the integral

EXAMPLE 9

_{A wedge is cut out of a circular cylinder of radius 4 by two planes. One}

plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle

of 30 along a diameter of the cylinder. Find the volume of the wedge.

SOLUTION

_{If we place the -axis along the diameter where the planes meet, then the}

base of the solid is a semicircle with equation

,

. A

cross-section perpendicular to the -axis at a distance from the origin is a triangle

,

as shown in Figure 17, whose base is

and whose height is

. Thus the cross-sectional area is

and the volume is

For another method see Exercise 64.

M

苷

128

1–18 _{Find the volume of the solid obtained by rotating the region}

bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

(18)

44.

45. A CAT scan produces equally spaced cross-sectional views of

a human organ that provide information about the organ other-wise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39, and 0. Use the Midpoint Rule to estimate the volume of the liver.

46. A log 10 m long is cut at 1-meter intervals and its

cross-sectional areas (at a distance from the end of the log) are listed in the table. Use the Midpoint Rule with to esti-mate the volume of the log.

47. (a) If the region shown in the figure is rotated about the

-axis to form a solid, use the Midpoint Rule with to estimate the volume of the solid.

(b) Estimate the volume if the region is rotated about the -axis. Again use the Midpoint Rule with .

48. _{(a) A model for the shape of a bird’s egg is obtained by}

rotating about the -axis the region under the graph of

Use a CAS to find the volume of such an egg.

(b) For a Red-throated Loon, , , ,

and . Graph and find the volume of an egg of this species.

49 – 61 _{Find the volume of the described solid .}

A right circular cone with height and base radius

50. _{A frustum of a right circular cone with height , lower base}

radius , and top radius

19 –30 _{Refer to the figure and find the volume generated by}

rotating the given region about the specified line.

19. _about 20. _about

31–36 Set up, but do not evaluate, an integral for the volume of

the solid obtained by rotating the region bounded by the given curves about the specified line.

31.

;

37–38 Use a graph to find approximate -coordinates of the points of intersection of the given curves. Then use your calcula-tor to find (approximately) the volume of the solid obtained by rotating about the -axis the region bounded by these curves.

37. _,

38.

39 – 40 Use a computer algebra system to find the exact volume

of the solid obtained by rotating the region bounded by the given curves about the specified line.

39. , , ;

40. _, _;

41– 44 Each integral represents the volume of a solid. Describe

(19)

62. _{The base of is a circular disk with radius . Parallel}

cross-sections perpendicular to the base are isosceles triangles with height and unequal side in the base.

(a) Set up an integral for the volume of .

(b) By interpreting the integral as an area, find the volume of .

(a) Set up an integral for the volume of a solid torus(the donut-shaped solid shown in the figure) with radii and . (b) By interpreting the integral as an area, find the volume of

the torus.

64. _{Solve Example 9 taking cross-sections to be parallel to the line}

of intersection of the two planes.

65. (a) Cavalieri’s Principle states that if a family of parallel planes

gives equal cross-sectional areas for two solids and , then the volumes of and are equal. Prove this principle. (b) Use Cavalieri’s Principle to find the volume of the oblique

cylinder shown in the figure.

66. Find the volume common to two circular cylinders, each with

radius , if the axes of the cylinders intersect at right angles.

Find the volume common to two spheres, each with radius , if the center of each sphere lies on the surface of the other sphere.

68. A bowl is shaped like a hemisphere with diameter 30 cm. A

ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of centimeters. Find the vol-ume of water in the bowl.

69. _{A hole of radius is bored through a cylinder of radius}

at right angles to the axis of the cylinder. Set up, but do not evaluate, an integral for the volume cut out.

R⬎_r r

h

r

67.

r

h

r

S2

S1

S2

S1

r R

R r

63.

S S

h

r S

A cap of a sphere with radius and height

52. A frustum of a pyramid with square base of side , square top

of side , and height

What happens if ? What happens if ?

53. _{A pyramid with height and rectangular base with dimensions}

and

54. A pyramid with height and base an equilateral triangle with

side (a tetrahedron)

55. _{A tetrahedron with three mutually perpendicular faces and}

three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm

56. _{The base of is a circular disk with radius . Parallel}

cross-sections perpendicular to the base are squares.

The base of is an elliptical region with boundary curve . Cross-sections perpendicular to the -axis are isosceles right triangles with hypotenuse in the base.

58. The base of is the triangular region with vertices ,

, and . Cross-sections perpendicular to the -axis are equilateral triangles.

59. The base of is the same base as in Exercise 58, but

cross-sections perpendicular to the -axis are squares.

60. _{The base of is the region enclosed by the parabola}

and the -axis. Cross-sections perpendicular to the -axis are squares.

61. _{The base of is the same base as in Exercise 60, but}

cross-sections perpendicular to the -axis are isosceles triangles with height equal to the base.

x S

y

x y苷1⫺x2

S

x S

y

共0, 1兲共1, 0兲

共0, 0兲

S

x 9x2_⫹

4y2

苷36

S

57.

r S

a

a a

a

h 2b

b

h

a苷0

a苷b a

b

h a

b

r h

h r

(20)

constant. Show that the radius of each end of the barrel is , where .

(b) Show that the volume enclosed by the barrel is

72. Suppose that a region has area and lies above the -axis.

When is rotated about the -axis, it sweeps out a solid with volume . When is rotated about the line (where is a positive number), it sweeps out a solid with volume . Express V2in terms of V1, , and .k A

V2

k y苷⫺k

᏾

V1

x

᏾

x A

᏾

V苷13␲h

(

2R

2_⫹_r2_⫺2 5d

2

)

d苷ch2兾4

r苷R⫺d

70. _{A hole of radius is bored through the center of a sphere of}

radius . Find the volume of the remaining portion of the sphere.

71. _{Some of the pioneers of calculus, such as Kepler and Newton,}

were inspired by the problem of finding the volumes of wine barrels. (In fact Kepler published a book Stereometria doliorum in 1715 devoted to methods for finding the volumes of barrels.) They often approximated the shape of the sides by parabolas. (a) A barrel with height and maximum radius is

con-structed by rotating about the -axis the parabola ,⫺_h兾2艋x艋_h兾2, where is a positive c y苷R⫺cx2

x

R h

R⬎r r