Steward – Application of Integration

6 APPLICATIONS OF

INTEGRATION

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

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 Q into a large number of

small parts. We next approximate each small part by a quantity of the form f 共x i *兲 ⌬x and thus approximate Q by a Riemann sum. Then we take the limit and express Q as an integral. Finally we evaluate the integral using the Fundamental Theorem of Calculus or the Midpoint Rule.

6.1 AREAS BETWEEN CURVES

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

y=ƒ

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

Consider the region S that lies between two curves y 苷 f 共x兲 and y 苷 t共x兲 and be- tween the vertical lines x苷a and x苷b , where f and are continuous functions and S t

f 共x兲 艌 t共x兲 for all in x 关a, b兴 . (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 ith strip by a rectangle with base ⌬x and height

y=©

f 共x *兲 ⫺ t共x i i *兲 . (See Figure 2. If we like, we could take all of the sample points to be right endpoints, in which case x i *苷x i .) The Riemann sum

FIGURE 1 n

S=s(x, y) | a¯x¯b, ©¯y¯ƒd

i苷1 兺

关 f 共x *兲 ⫺ t共x i *兲兴 ⌬x i

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

_g ( x * i )

(a) Typical rectangle

(b) Approximating rectangles

This approximation appears to become better and better as nl⬁ . Therefore we define the area A of the region as the limiting value of the sum of the areas of these approxi- S mating rectangles.

1 A 苷 lim 兺 关 f 共x

We recognize the limit in (1) as the definite integral of f⫺ t . Therefore we have the fol- lowing formula for area.

2 The area A of the region bounded by the curves y 苷 f 共x兲, y 苷 t共x兲 , and the lines x苷a , x苷b , where and are continuous and f t f 共x兲 艌 t共x兲 for all in x 关a, b兴 , is

A苷 b y

a 关 f 共x兲 ⫺ t共x兲兴 dx

Notice that in the special case where t 共x兲 苷 0 , S is the region under the graph of f and our general definition of area (1) reduces to our previous definition (Definition 2 in Section 5.1).

CHAPTER 6 APPLICATIONS OF INTEGRATION

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

y=ƒ

A 苷 关area under y 苷 f 共x兲兴 ⫺ 关area under y 苷 t共x兲兴

苷 y a f 共x兲 dx ⫺ y a t共x兲 dx 苷 y a 关 f 共x兲 ⫺ t共x兲兴 dx

y=©

EXAMPLE 1 Find the area of the region bounded above by

y苷e x , bounded below by

y苷x , and bounded on the sides by x 苷 0 and x 苷 1.

FIGURE 3

A= x ƒ dx- © dx y苷e and the j jj

b b SOLUTION The region is shown in Figure 4. The upper boundary curve is

a jj j a lower boundary curve is

y苷x x . So we use the area formula (2) with f 共x兲 苷 e , t 共x兲 苷 x ,

a 苷 0, and : b苷1

1 A苷 1 y

In Figure 4 we drew a typical approximating rectangle with width ⌬x as a reminder of

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

y=x Îx

for an area, it’s helpful to sketch the region to identify the top curve , the bottom curve y T

y B , and a typical approximating rectangle as in Figure 5. Then the area of a typical rect-

1 angle is 共y T ⫺y B 兲 ⌬x and the equation

FIGURE 4

A 苷 lim b

n ⬁ 兺 共y T ⫺y B 兲 ⌬x 苷 y a 共y T ⫺y B l 兲 dx

i苷1

summarizes the procedure of adding (in a limiting sense) the areas of all the typical y T -y B rectangles.

Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Figure 3 y the right-hand boundary reduces to a point. In the next example both of the side bound-

B Îx

aries reduce to a point, so the first step is to find a and b.

V EXAMPLE 2 Find the area of the region enclosed by the parabolas y苷x 2 and

y 苷 2x ⫺ x 2 .

FIGURE 5

SOLUTION We first find the points of intersection of the parabolas by solving their equa- tions simultaneously. This gives x 2 苷 2x ⫺ x 2 , or 2x 2 ⫺ 2x 苷 0 . Thus 2x 共x ⫺ 1兲 苷 0 ,

so x苷0 or 1. The points of intersection are 共0, 0兲 and 共1, 1兲 .

y T =2x-≈

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

The area of a typical rectangle is

and the region lies between x苷0 and x苷1 . So the total area is

FIGURE 6

1 共2x ⫺ 2x 2 1 共x ⫺ x A苷 2 y

0 兲 dx 苷 2 y 0 兲 dx

2 x 1 ⫺ x 苷2 3

SECTION 6.1 AREAS BETWEEN CURVES

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

y 苷 x兾sx 2 ⫹1 and

y苷x 4 ⫺ x.

SOLUTION If we were to try to find the exact intersection points, we would have to solve the equation

苷x 4 sx ⫺x

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

y= x

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 x ⬇ 1.18 . (If greater accuracy is required, we could use Newton’s method or a

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

y=x$-x

between the curves is

y 0 冋 sx 2 ⫹1 册

To integrate the first term we use the subsitution

u苷x 2 ⫹1 . Then du 苷 2x dx , and

when . x 苷 1.18, we have u ⬇ 2.39 So

A 1.18 ⬇

1 2.39 du

2 y ⫺ 共x 1 4 ⫺x 兲 dx

su

2.39 x 5 1.18 x 2

苷 su ] 1 ⫺ ⫺

共1.18兲 5 ⫹ 2 共1.18兲 苷 s2.39 ⫺ 1 ⫺

⬇ 0.785 M

√ (mi/h)

EXAMPLE 4 Figure 8 shows velocity curves for two cars, A and B, that start side by side

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

50 A the Midpoint Rule to estimate it.

40 SOLUTION We know from Section 5.4 that the area under the velocity curve A represents

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

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

10 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

0 2 4 6 8 10 12 14 16 t

second . 5280 共1 mi兾h 苷

3600 ft 兾s兲

(seconds)

FIGURE 8

0 2 4 6 8 10 12 14 16 v A 0 34 54 67 76 84 89 92 95

v B 0 21 34 44 51 56 60 63 65 v A ⫺ v B 0 13 20 23 25 28 29 29 30

CHAPTER 6 APPLICATIONS OF INTEGRATION

We use the Midpoint Rule with n苷4 intervals, so that ⌬t 苷 4 . The midpoints of the intervals are t 1 苷2 , t 2 苷6 , t 3 苷 10 , and t 4 苷 14 . We estimate the distance between the

cars after 16 seconds as follows:

y 16

共 v ⫺ 0 v A B 兲 dt ⬇ ⌬t 关13 ⫹ 23 ⫹ 28 ⫹ 29兴

苷 4共93兲 苷 372 ft

If we are asked to find the area between the curves y 苷 f 共x兲 and y 苷 t共x兲 where

y=©

f 共x兲 艌 t共x兲 for some values of but x t 共x兲 艌 f 共x兲 for other values of , then we split the x

S¡ S™

S£

given region into several regions , S S 1 S 2 , . . . with areas , A 1 A 2 , . . . as shown in Figure 9. We then define the area of the region to be the sum of the areas of the smaller regions S

y=ƒ

S 1 , S 2 , . . . , that is, A苷A 1 ⫹A 2 ⫹⭈⭈⭈ . Since

再 t共x兲 ⫺ f 共x兲 when t共x兲 艌 f 共x兲

f 共x兲 ⫺ t共x兲 when f 共x兲 艌 t共x兲

ⱍ f 共x兲 ⫺ FIGURE 9 t共x兲 ⱍ 苷

we have the following expression for A.

3 The area between the curves y 苷 f 共x兲 and y 苷 t共x兲 and between x苷a and x苷b is

A苷 b y

a ⱍ f 共x兲 ⫺ t共x兲 ⱍ dx

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

sponding to A 1 , A 2 ,....

V EXAMPLE 5 Find the area of the region bounded by the curves y 苷 sin x , y 苷 cos x , x苷0 , and . x 苷 ␲兾2 SOLUTION The points of intersection occur when sin x 苷 cos x , that is, when x 苷 ␲兾4

(since 0艋x艋␲ 兾2 ). The region is sketched in Figure 10. Observe that cos x 艌 sin x

y =cos x

y=sin x

when 0艋x艋␲ 兾4 but sin x 艌 cos x when ␲ 兾4 艋 x 艋 ␲兾2 . Therefore the required area is

A¡ A™

2 A苷 y 0 ⱍ

␲ x=0 兾2 x= π cos x ⫺ sin x

ⱍ dx 苷 A

1 ⫹A 2

苷 y 0 共cos x ⫺ sin x兲 dx ⫹ y ␲ 兾4 共sin x ⫺ cos x兲 dx

␲ 苷 兾2 [ sin x ⫹ cos x ]

0 ⫹ [ ⫺cos x ⫺ sin x ] ␲ 兾4

FIGURE 10

冉 s2 s2 冊 冉 s2 s2 冊

In this particular example we could have saved some work by noticing that the region is symmetric about x 苷 ␲兾4 and so

␲ A 苷 2A 兾4

1 苷2 y 0 共cos x ⫺ sin x兲 dx M

SECTION 6.1 AREAS BETWEEN CURVES

Some regions are best treated by regarding x as a function of y. If a region is bounded by curves with equations x 苷 f 共y兲 , x 苷 t共y兲 , y苷c , and y苷d , where and are contin- f t uous and f 共y兲 艌 t共y兲 for c艋y艋d (see Figure 11), then its area is

A苷 d y

c 关 f 共y兲 ⫺ t共y兲兴 dy

y=d

Îy

x=g(y)

x=f(y)

If we write x R for the right boundary and x L for the left boundary, then, as Figure 12 illustrates, we have

A苷 d y

c 共x R ⫺x L 兲 dy

Here a typical approximating rectangle has dimensions x R ⫺x L and ⌬y .

V EXAMPLE 6

Find the area enclosed by the line y苷x⫺1 and the parabola

y 2 苷 2x ⫹ 6 .

1 SOLUTION x By solving the two equations we find that the points of intersection are L = ¥-3 2 共⫺1, ⫺2兲 and 共5, 4兲 . We solve the equation of the parabola for x and notice from

x R =y+1

Figure 13 that the left and right boundary curves are

2 y ⫺3

x R 苷y⫹1

We must integrate between the appropriate -values, y y 苷 ⫺2 and y苷4 . Thus

A苷 4 y 共x

共y ⫹ 1兲 ⫺ ( 2 y 2 ⫺3

) ] dy

2 y ⫹y⫹4 ) dy

y= 2x+6 œ„„„„„

y=x-1

0 1 ⫺3 4 A¡ x 苷⫺ 6 共64兲 ⫹ 8 ⫹ 16 ⫺ ( 3 ⫹2⫺8 )

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

y=_ 2x+6 œ„„„„„

y , but the calculation is much more involved. It would have meant splitting the region in two and computing the areas labeled A 1 and A 2 in Figure 14. The method we used in

CHAPTER 6 APPLICATIONS OF INTEGRATION

6.1 EXERCISES

1– 4 Find the area of the shaded region.

21. x苷1⫺y 2 , x苷y 2 ⫺1

1. y y=5x-≈ 2.

22. y 苷 sin共␲x兾2兲 , y=œ„„„„ y苷x x+2

23. y 苷 cos x , y 苷 sin 2x , x苷0 , x 苷 ␲兾2 (4, 4)

x=2

24. y 苷 cos x , y 苷 1 ⫺ cos x , 0艋x艋␲ y=x

y= 1 2 25. y苷x , y 苷 2兾共x 2 ⫹1 兲

x+1

26. y苷

ⱍ 2 x ⱍ , y苷x ⫺2

x=¥-4y

x=¥-2 y=1 28. y 苷 3x 2 , y 苷 8x 2 , 4x ⫹ y 苷 4 , x艌 0

x=e y x 29 –30 Use calculus to find the area of the triangle with the given

y=_1

vertices.

x=2y-¥

5 – 28 Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approx-

imating rectangle and label its height and width. Then find the 31–32 Evaluate the integral and interpret it as the area of a area of the region.

region. Sketch the region.

5. y 苷 x ⫹ 1, y 苷 9 ⫺ x 2 , ␲ x 苷 ⫺1, x 苷 2 31. y 兾2

0 ⱍ sin x ⫺ cos 2x ⱍ dx

6. y 苷 sin x, y 苷 e x , x 苷 0, x 苷 ␲兾2

7. 2 32. 4 , y 0 ⱍ sx ⫹ 2 ⫺x ⱍ y苷x dx y苷x

8. y苷x 2 ⫺ 2x, y苷x⫹4 9. y 苷 1兾x, y 苷 1兾x 2 , x苷2

33 – 34 Use the Midpoint Rule with n苷4 to approximate the

area of the region bounded by the given curves. y 苷 1 ⫹ sx , y 苷 共3 ⫹ x兲兾3

共␲x兾4兲 11. ,

y 苷 cos 2 0艋x艋1 , y 2 苷x

, 35 –38 , 0 艋 x 艋 2␲ ; Use a graph to find approximate -coordinates of the points x

y 苷 cos x y 苷 2 ⫺ cos x of intersection of the given curves. Then find (approximately) the

15. y 苷 tan x , y 苷 2 sin x , ⫺ ␲ 兾3 艋 x 艋 ␲兾3 area of the region bounded by the curves. 16. 3

y 苷 x sin共x y苷x 2 ⫺ x, y 苷 3x 35. 兲 , y苷x 4

17. y 苷 sx , y苷 1 2 x , x苷9

36. y苷e x , y苷2⫺x 2

37. y 苷 3x 2 ⫺ 2x , y苷x 3 ⫺ 3x ⫹ 4 19. x 苷 2y 2 , x苷4⫹y 2

18. y苷8⫺x 2 , y苷x 2 , x 苷 ⫺3 , x苷3

y 苷 x cos x y苷x 10

20. 4x ⫹ y 苷 12 , x苷y

SECTION 6.1 AREAS BETWEEN CURVES

CAS 39. Use a computer algebra system to find the exact area (c) Which car is ahead after two minutes? Explain.

enclosed by the curves y苷x 5 ⫺ 6x 3 ⫹ 4x and y苷x .

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

Sketch the region in the xy -plane defined by the inequalities

x⫺ 2y 2 艌0 , 1⫺x⫺ ⱍ y ⱍ 艌0 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. 0 1 2 t (min)

46. The figure shows graphs of the marginal revenue function R⬘ 0 0 0 6 69 80 and the marginal cost function C⬘ for a manufacturer. [Recall

1 20 22 7 75 86 from Section 4.7 that R 共x兲 and C 共x兲 represent the revenue and 2 32 37 8 81 93 cost when units are manufactured. Assume that and x R C are 3 46 52 9 86 98 measured in thousands of dollars.] What is the meaning of the

area of the shaded region? Use the Midpoint Rule to estimate 5 62 71 the value of this quantity.

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

Rª(x)

were measured at 2-meter intervals as indicated in the figure.

Use the Midpoint Rule to estimate the area of the pool.

2 1 Cª(x)

; 47. The curve with equation y 2 苷x 2 共x ⫹ 3兲 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. 43. A cross-section of an airplane wing is shown. Measurements

48. Find the area of the region bounded by the parabola 2 , of the height of the wing, in centimeters, at 20-centimeter

y苷x the tangent line to this parabola at 共1, 1兲 , and the -axis. x

intervals are 5.8 , 20.3 , 26.7 , 29.0 , 27.6 , 27.3 , 23.8 , 20.5 , 15.1 ,

8.7 , and 2.8 . Use the Midpoint Rule to estimate the area of 49. Find the number such that the line b y苷b divides the region the wing’s cross-section.

bounded by the curves y苷x 2 and y苷4 into two regions

with equal area.

50. (a) Find the number such that the line a x苷a bisects the area under the curve y 苷 1兾x 2 , 1 艋 x 艋 4.

200 cm

(b) Find the number such that the line b y苷b bisects the

area in part (a).

44. If the birth rate of a population is b 共t兲 苷 2200e 0.024 t people

51. Find the values of such that the area of the region bounded c

by the parabolas y苷x 2 ⫺c people per 2 and y苷c 2 ⫺x 2 is 576. year, find the area between these curves for 0 艋 t 艋 10 . What

per year and the death rate is d 共t兲 苷 1460e 0.018t

52. Suppose that 0⬍c⬍␲ 兾2 . For what value of is the area of c does this area represent?

the region enclosed by the curves y 苷 cos x , y 苷 cos共x ⫺ c兲 ,

and x苷0 equal to the area of the region enclosed by the Two cars, A and B, start side by side and accelerate from rest.

curves , y 苷 cos共x ⫺ c兲 x苷␲ , and y苷0 ? The figure shows the graphs of their velocity functions.

(a) Which car is ahead after one minute? Explain. 53. For what values of m do the line y 苷 mx and the curve (b) What is the meaning of the area of the shaded region?

y 苷 x兾共x 2 ⫹1 兲 enclose a region? Find the area of the region.

CHAPTER 6 APPLICATIONS OF INTEGRATION

6.2 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 B 1 , called the base, and a congruent region B 2 in a parallel plane. The cylinder consists of all points on line segments that are perpendicular to the base and join B 1 to B 2 . If the area of the base is

A and the height of the cylinder (the distance from B 1 to B 2 ) is , then the volume h V of the

cylinder is defined as

V 苷 Ah

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

V 苷 ␲r 2 h [see Figure 1(b)], and if the base is a rectangle with length and width l w , then the cylinder is a rectangular box (also called a rectangular parallelepiped ) with

volume [see V苷l w h Figure 1(c)].

(a) Cylinder

(b) Circular cylinder

(c) Rectangular box

V=Ah

V=πr@h

V=lwh

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

be the area of the cross-section of in a plane S P x perpen- dicular to the -axis and passing through the point , where x x a艋x艋b . (See Figure 2. Think of slicing with a knife through and computing the area of this slice.) The cross- S x

cross-section of S . Let A 共x兲

sectional area A 共x兲 will vary as increases from to . x a b

A A(b)

SECTION 6.2 VOLUMES

Let’s divide S into n “slabs” of equal width ⌬x by using the planes P x 1 , P x 2 , . . . to slice the solid. (Think of slicing a loaf of bread.) If we choose sample points x i * in 关x i⫺ 1 ,x i 兴 , we can approximate the th slab i S i (the part of that lies between the planes S P x i⫺ 1 and P x i ) by

a cylinder with base area A 共x i *兲 and “height” ⌬x . (See Figure 3.)

x∞ xß x¶=b x

FIGURE 3

The volume of this cylinder is A 共x i *兲 ⌬x , so an approximation to our intuitive concep-

tion of the volume of the th slab i S i is

V 共S i 兲 ⬇ A共x *兲 ⌬x i

Adding the volumes of these slabs, we get an approximation to the total volume (that is, what we think of intuitively as the volume):

V ⬇ 兺 A 共x *兲 ⌬x i

i苷1

This approximation appears to become better and better as nl⬁ . (Think of the slices as becoming thinner and thinner.) Therefore, we define the volume as the limit of these sums as nl⬁ . But we recognize the limit of Riemann sums as a definite integral and so we have the following definition.

N It can be proved that this definition is inde-

DEFINITION OF VOLUME Let be a solid that lies between S x苷a and x苷b . If the

pendent of how is situated with respect to S

cross-sectional area of in the plane , through x and perpendicular to the x-axis, S P x

the -axis. In other words, no matter how we x

共x兲

is A , where A is a continuous function, then the volume of is S

slice with parallel planes, we always get the S

same answer for . V

V 苷 lim n ⬁ l

i苷1 兺 A 共x

i *兲 ⌬x 苷 y a A 共x兲 dx

V苷 b x

When we use the volume formula

a A 共x兲 dx , it is important to remember that

A 共x兲 is the area of a moving cross-section obtained by slicing through perpendicular to x the -axis. x

Notice that, for a cylinder, the cross-sectional area is constant: A for all . So our x

V苷 x a A dx 苷 A共b ⫺ a兲 ; this agrees with the formula V 苷 Ah.

definition of volume gives

EXAMPLE 1

3 ␲r 3 . SOLUTION If we place the sphere so that its center is at the origin (see Figure 4), then the

Show that the volume of a sphere of radius is 4 r V苷

CHAPTER 6 APPLICATIONS OF INTEGRATION

is y 苷 sr 2 ⫺x 2 . So the cross-sectional area is

A 2 2 共x兲 苷 ␲y 2 苷 ␲ 共r ⫺x 兲

Using the definition of volume with a 苷 ⫺r and b苷r , we have

2 V苷 2 y A

共x兲 dx 苷 y ␲ ⫺r ⫺x ⫺r 共r 兲 dx

2 苷 2␲ 2 y

0 共r ⫺x 兲 dx

(The integrand is even.)

Figure 5 illustrates the definition of volume when the solid is a sphere with radius r苷1 . From the result of Example 1, we know that the volume of the sphere is 4

3 ␲ ⬇ 4.18879 . Here the slabs are circular cylinders, or disks, and the three parts of Fig- ure 5 show the geometric interpretations of the Riemann sums

i苷1 兺

A 2 共x 2

i 兲 ⌬x 苷

i苷1 兺 ␲ 共1 ⫺x

i 兲 ⌬x

TEC Visual 6.2A shows an animation when n 苷 5, 10, and 20 if we choose the sample points x * i to be the midpoints . Notice x i

of Figure 5.

that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume.

(c) Using 20 disks, VÅ4.1940 FIGURE 5 Approximating the volume of a sphere with radius 1

(a) Using 5 disks, VÅ4.2726

(b) Using 10 disks, VÅ4.2097

V EXAMPLE 2 Find the volume of the solid obtained by rotating about the x-axis the region under the curve

y 苷 sx from 0 to 1. Illustrate the definition of volume by sketch-

ing a typical approximating 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 sx . The area of this cross-section is

A 2 共x兲 苷 ␲ ( sx ) 苷␲x

and the volume of the approximating cylinder (a disk with thickness ⌬x ) is

A 共x兲 ⌬x 苷 ␲x ⌬x

SECTION 6.2 VOLUMES

N Did we get a reasonable answer in

The solid lies between x苷0 and x苷1 , so its volume is

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

关0, 1兴 and height . If we rotate this square, 1 V苷 y A 共x兲 dx 苷 y ␲ x dx 苷 ␲

1 1 0 we get a cylinder with radius , height , and 0

volume ␲ⴢ1 2 ⴢ1苷␲ . We computed that the given solid has half this volume. That seems

y=œ„ œ

about right.

V EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by y苷x 3 , y苷8 , and about x苷0 the y -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 x苷s 3 y . So the area of a cross-

section through y is

A 共y兲 苷 ␲x 2

( s 3 y 2 2 苷␲ 兾3 ) 苷␲y

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

A 2 共y兲 ⌬y 苷 ␲y 兾3 ⌬y Since the solid lies between y 苷 0 and y 苷 8, its volume is

V苷 y 0 A 共y兲 dy 苷 y 0 ␲y dy 苷␲ [ 5 y ] 0 苷

x=0 y=˛ or x=œ„y 3

FIGURE 7

(a)

(b)

CHAPTER 6 APPLICATIONS OF INTEGRATION

EXAMPLE 4 The region ᏾ enclosed by the curves

and y苷x 2 y苷x is rotated about the

x -axis. Find the volume of the resulting solid. SOLUTION The curves

y苷x 2 and y苷x intersect at the points 共0, 0兲 and 共1, 1兲 . The region between them, the solid of rotation, and a cross-section perpendicular to the -axis are x

shown in Figure 8. A cross-section in the plane P x has the shape of a washer (an annular ring) with inner radius x 2 and outer radius , so we find the cross-sectional area by sub- x tracting the area of the inner circle from the area of the outer circle:

A 2 ⫺ ␲ 共x 2 兲 2 2 共x兲 苷 ␲x 4 苷 ␲ 共x ⫺x 兲 Therefore we have

V苷 y 0 A 2 4 共x兲 dx 苷 x ⫺ y 0 ␲ 共x ⫺x 兲 dx 苷␲ 苷

1 3 1 1 x 5 2␲

EXAMPLE 5 Find the volume of the solid obtained by rotating the region in Example 4 about the line y苷2 .

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 2⫺x and the outer radius is 2⫺x 2 .

TEC Visual 6.2B shows how solids of

revolution are formed.

y=2

y=2

2-x

2-≈

y=x

y=≈

xx

FIGURE 9

SECTION 6.2 VOLUMES

The cross-sectional area is

A 2 兲 2 共x兲 苷 ␲共2 ⫺ x 2 ⫺ ␲ 共2 ⫺ x兲

and so the volume of is S

V苷 1 y

0 A 共x兲 dx

苷␲ 1 y

0 关共2 ⫺ x 2 兲 2 ⫺ 共2 ⫺ x兲 2 兴 dx

苷␲ 1 y

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

b V苷 d y

a A 共x兲 dx

V苷 y c A 共y兲 dy

or

and we find the cross-sectional area A 共x兲 or A 共y兲 in one of the following ways:

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

A 苷 ␲共radius兲 2

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

A 苷 ␲ 共outer radius兲 2 ⫺ ␲ 共inner radius兲 2

r in r out

FIGURE 10

The next example gives a further illustration of the procedure.

CHAPTER 6 APPLICATIONS OF INTEGRATION

EXAMPLE 6 Find the volume of the solid obtained by rotating the region in Example 4 about the line x 苷 ⫺1 .

SOLUTION Figure 11 shows a horizontal cross-section. It is a washer with inner radius 1⫹y and outer radius 1⫹

, so the cross-sectional area is sy

A 2 ⫺ ␲ 共inner radius兲 共y兲 苷 ␲共outer radius兲 2

苷␲ 2 ( sy ) ⫺ ␲ 共1 ⫹ y兲 2

The volume is

2 V苷 2 y

1 A 1 共y兲 dy

y 0 [ ( 1⫹ sy ) 共1 ⫹ y兲 ] dy

1 4y 3 兾2

( 2 sy ⫺y⫺y 苷␲ 2 ) dy 苷␲

y 1+œ„ y y 1+y 1

x=œ„y

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 x 2 ⫹y 2 . The solid, its base, and a typical cross-

TEC Visual 6.2C shows how the solid

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

in Figure 12 is generated.

y=œ„„„„„„ ≈

B(x, y)

(a) The solid

(b) Its base

(c) A cross-section

FIGURE 12

FIGURE 13

Computer-generated picture of the solid in Example 7

SECTION 6.2 VOLUMES

Since lies on the circle, we have B y 苷 s1 ⫺ x 2 and so the base of the triangle ABC

is ⱍ AB ⱍ 2 苷 2 s1 ⫺ x . Since the triangle is equilateral, we see from Figure 13(c) that its height is

s3 y 苷 s3 s1 ⫺ x 2 . The cross-sectional area is therefore

A 1 2 共x兲 苷

2 s1 ⫺ x ⴢ 2 s1 ⫺ x 2 ⴢ s3 苷 s3 共1 ⫺ x 2 兲

and the volume of the solid is

1 A 1 V苷 y

共x兲 dx 苷 y ⫺1 s3 共1 ⫺ x 2 兲 dx

苷2 y

V EXAMPLE 8 Find the volume of a pyramid whose base is a square with side and L whose height is . h SOLUTION We place the origin O at the vertex of the pyramid and the -axis along its cen- x tral axis as in Figure 14. Any plane P x that passes through and is perpendicular to the x

x -axis intersects the pyramid in a square with side of length , say. We can express in s s terms of by observing from the similar triangles in Figure 15 that x

s 兾2

h 苷 L 兾2 苷 L

and so s 苷 Lx兾h . [Another method is to observe that the line OP has slope L 兾共2h兲 and so its equation is y 苷 Lx兾共2h兲 .] Thus the cross-sectional area is

The pyramid lies between x苷0 and x苷h , so its volume is

V苷 y 0 A 共x兲 dx 苷 y 0

x dx 苷

N OT E 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

FIGURE 16

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

CHAPTER 6 APPLICATIONS OF INTEGRATION

we would have obtained the integral

V苷 y 0 2 共h ⫺ y兲 dy 苷

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 x base of the solid is a semicircle with equation y 苷 s16 ⫺ x 2 , ⫺4 艋 x 艋 4 . A cross-

section perpendicular to the -axis at a distance from the origin is a triangle x x ABC , as shown in Figure 17, whose base is y 苷 s16 ⫺ x 2 and whose height is

BC ⱍ 2 ⱍ 苷 y tan 30⬚ 苷 s16 ⫺ x 兾 s3 . Thus the cross-sectional area is

A 共x兲 苷 2 s16 ⫺ x 2 1 ⴢ 16 ⫺ x s16 ⫺ x 2

0 s3 苷 2 s3 y

A 4 B y=œ„„„„„„ 16-≈

and the volume is

V苷 y A 共x兲 dx 苷 y ⫺4 dx ⫺4

16 ⫺ x 2

C 2 s3

s3 冋 3 册 0

y 0 共16 ⫺ x 2 兲 dx 苷 16x ⫺

For another method see Exercise 64.

6.2 EXERCISES

1–18 Find the volume of the solid obtained by rotating the region 10. y苷 1 4 x 2 , x苷2 , y苷0 ; about the -axis y bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

11. y苷x , y 苷 sx ; about y苷1

1. y苷2⫺ 1 2 x , y苷0 , x苷1 , x苷2 ; about the -axis x 12. y苷e ⫺x , y苷1 , x苷2 ; about y苷2 2. y苷1⫺x 2 , y苷0 ; about the -axis x

y 苷 1 ⫹ sec x , y苷3 ; about y苷1 y 苷 1兾x , x苷1 , x苷2 , y苷0 ; about the -axis x

4. y 苷 s25 ⫺ x 2 , 14. y苷0 , , about the -axis y 苷 1兾x y苷0 , x苷2 x苷4 ; x , x苷1 , x苷3 ; about y 苷 ⫺1 5. x 苷 2sy , x苷0 , y苷9 ; about the -axis y

15. x苷y 2 , x苷1 ; about x苷1

6. y 苷 ln x , y苷1 , y苷2 , x苷0 ; about the -axis y

16. y苷x , y 苷 sx ; about x苷2

7. y苷x 3 , y苷x , x艌 0 ; about the -axis x

17. y苷x 2 , x苷y 2 ; about x 苷 ⫺1

8. y苷 1 4 x 2 , y苷5⫺x 2 ; about the -axis x 18. y苷x , y苷0 , x苷2 , x苷4 ; about x 苷 1

9. y 2 苷x , x 苷 2y ; about the -axis y

SECTION 6.2 VOLUMES

19 –30 Refer to the figure and find the volume generated by 43. 1 4 8 44. ␲ ␲ 兾2

0 共y ⫺y 兲 dy

␲ y 0 关共1 ⫹ cos x兲 y 2 ⫺1 2 兴 dx

rotating the given region about the specified line. y C(0, 1)

B(1, 1)

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

a human organ that provide information about the organ other- T™ wise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The

y=œ„x

liver is 15 cm long and the cross-sectional areas, in square y=˛ 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. O

T£

T¡

46. A log 10 m long is cut at 1-meter intervals and its cross- sectional areas A (at a distance from the end of the log) are x 19. ᏾ 1 about OA

A(1, 0)

listed in the table. Use the Midpoint Rule with n苷5 to esti- 21. ᏾

20. ᏾ 1 about OC

mate the volume of the log.

1 about AB 22. ᏾ 1 about BC

A ( m 2 2 about OA 2 about OC )

x (m)

x (m) A ( m 2 ) 25. ᏾ 2 about AB 26. ᏾ 2 about BC 0 0.68 6 0.53

1 0.65 7 0.55 27. ᏾ 3 about OA

28. ᏾ 3 about OC

2 0.64 8 0.52 29. ᏾ 3 about AB 30. ᏾ 3 about BC 3 0.61 9 0.50

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

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

x -axis to form a solid, use the Midpoint Rule with n苷4 31. y 苷 tan 3 x , y 苷 1, x 苷 0; about y 苷 1

to estimate the volume of the solid. 32. 4

y 苷 共x ⫺ 2兲 , 8x ⫺ y 苷 16 ; about x 苷 10

33. y苷0 , y 苷 sin x , 0艋x艋␲ ; about y苷1

y苷0 , y 苷 sin x , 0艋x艋␲ ; about y 苷 ⫺2 35. x 2 ⫺y 2 苷1 , x苷3 ; about x 苷 ⫺2

0 2 4 6 8 10 x 36. y 苷 cos x , y 苷 2 ⫺ cos x ,

0 艋 x 艋 2␲ ; about y苷4

(b) Estimate the volume if the region is rotated about the y -axis. Again use the Midpoint Rule with n苷4 .

; 37–38 Use a graph to find approximate -coordinates of the x CAS 48. (a) A model for the shape of a bird’s egg is obtained by

points of intersection of the given curves. Then use your calcula- rotating about the -axis the region under the graph of x tor to find (approximately) the volume of the solid obtained by

rotating about the -axis the region bounded by these curves. x f 共x兲 苷 共ax 3 ⫹ bx 2 ⫹ cx ⫹ d 兲 s1 ⫺ x 2 37. y苷2⫹x 2 cos x ,

Use a CAS to find the volume of such an egg. 38. y 苷 3 sin共x 2 兲, y 苷 e x 兾2 ⫹e ⫺2x

y苷x 4 ⫹x⫹1

(b) For a Red-throated Loon, a 苷 ⫺0.06 , b 苷 0.04 , c 苷 0.1 , and d 苷 0.54 . Graph and find the volume of an egg of f this species.

CAS

39 – 40 Use a computer algebra system to find the exact volume of the solid obtained by rotating the region bounded by the given 49 – 61 Find the volume of the described solid . S

curves about the specified line. 49. A right circular cone with height and base radius h r 39. y 苷 sin 2 x , y苷0 , 0艋x艋␲ ; about y 苷 ⫺1

50. A frustum of a right circular cone with height , lower base h 40. y苷x 1⫺x , y 苷 xe 兾2 ;

radius , and top radius R about y 苷 3 r

41– 44 Each integral represents the volume of a solid. Describe h the solid. 41. ␲ 兾2

0 cos 2 x dx

42. ␲ y

2 y dy

CHAPTER 6 APPLICATIONS OF INTEGRATION

51. A cap of a sphere with radius and height r h 62. The base of is a circular disk with radius . Parallel cross- S r sections perpendicular to the base are isosceles triangles with h height and unequal side in the base. h

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

(b) By interpreting the integral as an area, find the volume of . S 63. (a) Set up an integral for the volume of a solid torus (the

donut-shaped solid shown in the figure) with radii and . r R (b) By interpreting the integral as an area, find the volume of

the torus.

52. A frustum of a pyramid with square base of side , square top b of side , and height a h

b 64. Solve Example 9 taking cross-sections to be parallel to the line of intersection of the two planes.

What happens if a苷b ? What happens if a苷0 ? 65. (a) Cavalieri’s Principle states that if a family of parallel planes

53. A pyramid with height and rectangular base with dimensions h gives equal cross-sectional areas for two solids S 1 and , S 2 b and 2b

then the volumes of S 1 and S 2 are equal. Prove this principle. 54. A pyramid with height and base an equilateral triangle with h (b) Use Cavalieri’s Principle to find the volume of the oblique side (a a tetrahedron)

cylinder shown in the figure.

55. A tetrahedron with three mutually perpendicular faces and 66. Find the volume common to two circular cylinders, each with three mutually perpendicular edges with lengths 3 cm,

radius , if the axes of the cylinders intersect at right angles. r 4 cm, and 5 cm

56. The base of is a circular disk with radius . Parallel cross- S r sections perpendicular to the base are squares. 57. The base of is an elliptical region with boundary curve S

9x 2 ⫹ 4y 2 苷 36 . Cross-sections perpendicular to the -axis x are isosceles right triangles with hypotenuse in the base.

58. The base of is the triangular region with vertices S 共0, 0兲 , 共1, 0兲 , and 共0, 1兲 . Cross-sections perpendicular to the -axis y

are equilateral triangles.

Find the volume common to two spheres, each with radius , if r The base of is the same base as in Exercise 58, but cross- S

the center of each sphere lies on the surface of the other sphere. sections perpendicular to the -axis are squares. x

The base of is the region enclosed by the parabola 68. S A bowl is shaped like a hemisphere with diameter 30 cm. A y苷1⫺x 2 and the -axis. Cross-sections perpendicular to the x

ball with diameter 10 cm is placed in the bowl and water is y -axis are squares.

poured into the bowl to a depth of centimeters. Find the vol- h ume of water in the bowl.

61. The base of is the same base as in Exercise 60, but cross- S sections perpendicular to the -axis are isosceles triangles with x

69. A hole of radius is bored through a cylinder of radius r R⬎r height equal to the base.

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

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS

70. A hole of radius is bored through the center of a sphere of r constant. Show that the radius of each end of the barrel is radius R⬎r . Find the volume of the remaining portion of the

r苷R⫺d , where . d 苷 ch 2 兾4

sphere. (b) Show that the volume enclosed by the barrel is 71. Some of the pioneers of calculus, such as Kepler and Newton,

1 were inspired by the problem of finding the volumes of wine 2 V苷

3 ␲h ( 2R 2 ⫹r 2 ⫺ 5 d 2 )

barrels. (In fact Kepler published a book Stereometria doliorum in 1715 devoted to methods for finding the volumes of barrels.)

72. Suppose that a region ᏾ has area A and lies above the -axis. x They often approximated the shape of the sides by parabolas.

When ᏾ is rotated about the -axis, it sweeps out a solid with x

volume V 1 . When ᏾ is rotated about the line y 苷 ⫺k (where k structed by rotating about the -axis the parabola x

(a) A barrel with height and maximum radius is con- h R

is a positive number), it sweeps out a solid with volume V 2 .

y 苷 R ⫺ cx 2 , ⫺h 兾2 艋 x 艋 h兾2 , where is a positive c Express V 2 in terms of V 1 , , and . k A

6.3 VOLUMES BY CYLINDRIC AL SHELLS

Some volume problems are very difficult to handle by the methods of the preceding sec-

y=2≈-˛

tion. For instance, let’s consider the problem of finding the volume of the solid obtained by rotating about the -axis the region bounded by y

2 ⫺x 1 3 y 苷 2x and y苷0 . (See Figure 1.)

x x =? L

x =? x R

If we slice perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer radius of the washer, we would have to solve the cubic equation

2 ⫺x y 苷 2x 3 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 , outer radius , r 1 r 2 and height . Its volume h V is calculated by subtracting the volume V 1 of the inner cylinder

from the volume V 2 of the outer cylinder:

2 ⫺r 1 (the thickness of the shell) and r苷 2 共r 2 ⫹r 1 兲 (the average radius of the shell), then this formula for the volume of a cylindrical shell becomes

1 V 苷 2␲rh ⌬r

and it can be remembered as

FIGURE 2

V 苷 [circumference][height][thickness] Now let be the solid obtained by rotating about the -axis the region bounded by S y

y 苷 f 共x兲 [where f 共x兲 艌 0 ], y 苷 0, x 苷 a, and x苷b , where b⬎a艌 0 . (See Figure 3.)

y=ƒ

y=ƒ

FIGURE 3

CHAPTER 6 APPLICATIONS OF INTEGRATION y

We divide the interval 关a, b兴 into n subintervals 关x i⫺ 1 ,x i 兴 of equal width ⌬x and let be x i

y=ƒ

the midpoint of the ith subinterval. If the rectangle with base 关x i⫺ 1 ,x i 兴 and height f 共x i 兲 is rotated about the y-axis, then the result is a cylindrical shell with average radius , height x i

f 共x i 兲 , and thickness ⌬x (see Figure 4), so by Formula 1 its volume is

0 0 V i 苷 共2␲ x i 兲关 f 共x i a 兲兴 ⌬x b x

x i-1 x – i x i

Therefore an approximation to the volume V of is given by the sum of the volumes of S

these shells:

i苷1 兺 i苷1 兺

2␲ x i f 共x i 兲 ⌬x

y=ƒ

This approximation appears to become better as nl⬁ . But, from the definition of an inte- gral, we know that

b b lim

n ⬁ 兺 2␲ x i f 共x i

Thus the following appears plausible:

FIGURE 4

2 The volume of the solid in Figure 3, obtained by rotating about the y-axis the region under the curve y 苷 f 共x兲 from a to b, is

V苷 b y

a 2␲x f 共x兲 dx

where 0 艋 a ⬍ b

The argument using cylindrical shells makes Formula 2 seem reasonable, but later we will be able to prove it (see Exercise 67 in Section 7.1).

The best way to remember Formula 2 is to think of a typical shell, cut and flattened as in Figure 5, with radius x, circumference , height , and 2␲x f 共x兲 thickness or ⌬x dx :

a 共2␲x兲 关 f 共x兲兴 dx

circumference

height thickness

2πx Îx

FIGURE 5

This type of reasoning will be helpful in other situations, such as when we rotate about lines other than the y-axis.

EXAMPLE 1 Find the volume of the solid obtained by rotating about the -axis the region y bounded by

y 苷 2x 2 ⫺x 3 and y苷0 .

SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumfer- ence 2␲x , and height f 2 共x兲 苷 2x 3 ⫺x . So, by the shell method, the volume is

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS

2 2 V苷 2 y

0 共2␲x兲共2x ⫺x 3 3 ⫺x 兲 dx 苷 2␲ 4 y 0 共2x 兲 dx

2 x ⫺ 1 2 x 32 5 5 16 ] 0 苷 2␲ ( 8⫺ 5 ) 苷 5 ␲

2≈-˛

It can be verified that the shell method gives the same answer as slicing.

FIGURE 6

N Figure 7 shows a computer-generated picture of the solid whose volume we computed in Example 1.

FIGURE 7

N OT E Comparing the solution of Example 1 with the remarks at the beginning of this section, we see that the method of cylindrical shells is much easier than the washer method

for this problem. We did not have to find the coordinates of the local maximum and we did not have to solve the equation of the curve for in terms of . However, in other examples x y the methods of the preceding section may be easier.

V y EXAMPLE 2 Find the volume of the solid obtained by rotating about the -axis the y

y=x

region between

y苷x 2 and y苷x .

y=≈

SOLUTION The region and a typical shell are shown in Figure 8. We see that the shell has

radius x, circumference 2␲x , and height x⫺x 2 . So the volume is

shell

height =x-≈

0 共2␲x兲共x ⫺ x 兲 dx 苷 2␲ y 0 共x ⫺x 0 兲 dx x x

1 2 1 2 V苷 3 y

FIGURE 8

As the following example shows, the shell method works just as well if we rotate about the x-axis. We simply have to draw a diagram to identify the radius and height of a shell.

V EXAMPLE 3 Use cylindrical shells to find the volume of the solid obtained by rotating

from 0 to 1.

shell height =1-¥

about the -axis the region under the curve x

y 苷 sx

1 SOLUTION This problem was solved using disks in Example 2 in Section 6.2. To use shells we relabel the curve

(in the figure in that example) as y 苷 sx 2 x苷y in Figure 9. For

rotation about the x-axis we see that a typical shell has radius y, circumference , and 2␲ y

x= 2 =¥ ¥ ¥ ¥ x=1 shell

height 1⫺y . So the volume is

radius =y

V苷 共2␲ y兲共1 ⫺ y 兲 dy 苷 2␲ 0 共y ⫺ y 兲 dy 苷 2␲

FIGURE 9

In this problem the disk method was simpler.

CHAPTER 6 APPLICATIONS OF INTEGRATION

V EXAMPLE 4 Find the volume of the solid obtained by rotating the region bounded by y苷x⫺x 2 and y苷0 about the line x苷2 .

SOLUTION Figure 10 shows the region and a cylindrical shell formed by rotation about the line

. It has radius 2⫺x , circumference 2␲ 共2 ⫺ x兲 , and height x⫺x x苷2 2 .

y=x-≈

The volume of the given solid is

1 2 V苷 1 y

0 2␲ 共2 ⫺ x兲共x ⫺ x 兲 dx 苷 2␲ y 0 3 ⫺ 3x 共x 2 ⫹ 2x 兲 dx

⫺x ⫹x

6.3 EXERCISES

1. Let be the solid obtained by rotating the region shown in S

4. y苷x 2 , y苷0 , x苷1

the figure about the -axis. Explain why it is awkward to use y

slicing to find the volume V of . Sketch a typical approxi- S

mating shell. What are its circumference and height? Use shells

6. y 苷 3 ⫹ 2x ⫺ x 2 , x⫹y苷3

to find . V

7. y 苷 4共x ⫺ 2兲 2 , y苷x 2 ⫺ 4x ⫹ 7

y y=x(x-1)@

8. Let V be the volume of the solid obtained by rotating about the y -axis the region bounded by y 苷 sx and y苷x 2 . Find V both by slicing and by cylindrical shells. In both cases draw a dia-

0 gram to explain your method.

9 –14 Use the method of cylindrical shells to find the volume of the 2. Let be the solid obtained by rotating the region shown in the S

solid obtained by rotating the region bounded by the given curves figure about the -axis. Sketch a typical cylindrical shell and y

about the -axis. Sketch the region and a typical shell. x find its circumference and height. Use shells to find the volume

of . Do you think this method is preferable to slicing? Explain. S 9. x苷1⫹y 2 , x 苷 0, y 苷 1, y 苷 2 y

10. x 苷 sy , x 苷 0, y 苷 1 11. y苷x 3 , y苷8 , x苷0

y=sin{≈}

3–7 Use the method of cylindrical shells to find the volume gener- 15 – 20 Use the method of cylindrical shells to find the volume gen- ated by rotating the region bounded by the given curves about the

erated by rotating the region bounded by the given curves about the y -axis. Sketch the region and a typical shell.

specified axis. Sketch the region and a typical shell.

15. y 苷 1兾x 4 , y苷0 , x苷1 , x苷2 y苷x , y 苷 0, x 苷 1 ; about x 苷 2

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS

16. y 苷 sx , y 苷 0, x 苷 1 ; about x 苷 ⫺1 ; 33 –34 Use a graph to estimate the -coordinates of the points of x 17. y 苷 4x ⫺ x 2 , y苷3 ; about x苷1

intersection of the given curves. Then use this information and

2 2 18. your calculator to estimate the volume of the solid obtained by

y苷x , y苷2⫺x ; about x苷1 rotating about the -axis the region enclosed by these curves. y 19. y苷x 3 , y苷0 , x苷1 ; about y苷1

33. y苷e x , y 苷 sx ⫹1

20. 2 y苷x 2 ,x苷y ; about y 苷 ⫺1 34. y苷x 3 ⫺x⫹1 , y 苷 ⫺x 4 ⫹ 4x ⫺ 1

35 –36 Use a computer algebra system to find the exact volume of the solid obtained by rotating the region bounded by the given

21– 26 Set up, but do not evaluate, an integral for the volume

CAS

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

curves about the specified line.

21. y 苷 ln x, y 苷 0, x 苷 2; about the y-axis 35. y 苷 sin 2 x , y 苷 sin 4 x , 0艋x艋␲ ; about x 苷 ␲兾2 22. ,

y苷x 2 y 苷 4x ⫺ x ; about x苷7 36. y苷x 3 sin x , y苷0 , 0艋x艋␲ ; about x 苷 ⫺1 23. y苷x 4 , y 苷 sin共␲x兾2兲 ; about x 苷 ⫺1

24. y 苷 1兾共1 ⫹ x 2 兲, y 苷 0, x 苷 0, x 苷 2; about x苷2 37– 42 The region bounded by the given curves is rotated about

25. x 苷 ssin y the specified axis. Find the volume of the resulting solid by any , 0 艋 y 艋 ␲, x 苷 0; about y苷4 method.

26. x 2 ⫺y 2 苷 7, x 苷 4; about y苷5 37. y 苷 ⫺x 2 ⫹ 6x ⫺ 8 , y苷0 ; about the -axis y

y 苷 ⫺x 2 ⫹ 6x ⫺ 8 , y苷0 ; about the -axis x Use the Midpoint Rule with n苷5 to estimate the volume

obtained by rotating about the -axis the region under the y 39. y苷5 , y 苷 x ⫹ 共4兾x兲 ; about x 苷 ⫺1 curve , y 苷 s1 ⫹ x 3 0艋x艋1 .

40. x苷1⫺y 4 , x苷0 ; about x苷2 28. If the region shown in the figure is rotated about the -axis to y

41. x 2 ⫹ 共 y ⫺ 1兲 2 苷1 ; about the -axis y form a solid, use the Midpoint Rule with n苷5 to estimate the volume of the solid.

42. x 苷 共 y ⫺ 3兲 2 , x苷4 ; about y苷1 y

43 – 45 Use cylindrical shells to find the volume of the solid. 5

43. A sphere of radius 4 r 44. 3 The solid torus of Exercise 63 in Section 6.2 2

45. A right circular cone with height and base radius h r 1

46. Suppose you make napkin rings by drilling holes with differ- 0 1 234567

8 9 10 11 12 x

ent diameters through two wooden balls (which also have dif- ferent diameters). You discover that both napkin rings have

the same height , as shown in the figure. h Each integral represents the volume of a solid. Describe (a) Guess which ring has more wood in it. the solid. (b) Check your guess: Use cylindrical shells to compute the

y volume of a napkin ring created by drilling a hole with

0 2␲x dx radius through the center of a sphere of radius and r R express the answer in terms of . h

30. 2 2␲ y y

0 1⫹y 2 dy

31. y 1

0 2␲ 共3 ⫺ y兲共1 ⫺ y 2 兲 dy

32. ␲ y 兾4

0 2␲ 共␲ ⫺ x兲共cos x ⫺ sin x兲 dx

CHAPTER 6 APPLICATIONS OF INTEGRATION

6.4 WORK The term work is used in everyday language to mean the total amount of effort required to

perform a task. In physics it has a technical meaning that depends on the idea of a force. Intuitively, you can think of a force as describing a push or pull on an object—for example,

a horizontal push of a book across a table or the downward pull of the earth’s gravity on a ball. In general, if an object moves along a straight line with position function s 共t兲 , then the force F on the object (in the same direction) is defined by Newton’s Second Law of Motion as the product of its mass m and its acceleration:

1 d 2 F苷m s dt 2

In the SI metric system, the mass is measured in kilograms (kg), the displacement in meters (m), the time in seconds (s), and the force in newtons (

N 苷 kg⭈m兾s 2 ). Thus a force of 1 N acting on a mass of 1 kg produces an acceleration of 1 m 兾s . In the US Customary 2

system, the fundamental unit is chosen to be the unit of force, which is the pound. In the case of constant acceleration, the force F is also constant and the work done is defined to be the product of the force F and the distance that the object moves: d

2 W 苷 Fd

work 苷 force ⫻ distance

If is measured in newtons and in meters, then the unit for F d W is a newton-meter, which is called a joule (J). If is measured in pounds and in feet, then the unit for F d W is a foot-

pound (ft-lb), which is about 1.36 J.

V EXAMPLE 1

(a) How much work is done in lifting a 1.2-kg book off the floor to put it on a desk that is 0.7 m high? Use the fact that the acceleration due to gravity is

m 兾s . t 苷 9.8 2 (b) How much work is done in lifting a 20-lb weight 6 ft off the ground?

SOLUTION

(a) The force exerted is equal and opposite to that exerted by gravity, so Equation 1 gives

F 苷 mt 苷 共1.2兲共9.8兲 苷 11.76 N and then Equation 2 gives the work done as

W 苷 Fd 苷 共11.76兲共0.7兲 ⬇ 8.2 J

(b) Here the force is given as

F 苷 20 lb, so the work done is W 苷 Fd 苷 20 ⴢ 6 苷 120 ft-lb

Notice that in part (b), unlike part (a), we did not have to multiply by because we t were given the weight (which is a force) and not the mass of the object.

Equation 2 defines work as long as the force is constant, but what happens if the force is variable? Let’s suppose that the object moves along the -axis in the positive direction, x from x苷a to x苷b , and at each point between and a force x a b f 共x兲 acts on the object, where is a continuous function. We divide the interval f 关a, b兴 into n subintervals with end- points x 0 ,x 1 , ..., x n and equal width ⌬x . We choose a sample point x * i in the th sub- i interval 关x i⫺ 1 ,x i 兴 . Then the force at that point is f 共x *兲 i . If is large, then n ⌬x is small, and

SECTION 6.4 WORK

since is continuous, the values of don’t change very much over the interval f f 关x i⫺ 1 ,x i 兴 . In other words, f is almost constant on the interval and so the work W i that is done in mov- ing the particle from x i⫺ 1 to x i is approximately given by Equation 2:

W i ⬇ f 共x *兲 ⌬x i

Thus we can approximate the total work by

3 W ⬇ 兺 f 共x i *兲 ⌬x

i苷1

It seems that this approximation becomes better as we make larger. Therefore we define n the work done in moving the object from a to b as the limit of this quantity as nl⬁ . Since the right side of (3) is a Riemann sum, we recognize its limit as being a definite inte- gral and so

4 b W 苷 lim 兺 f 共x

i *兲 ⌬x 苷 y a f 共x兲 dx

nl⬁ i苷1

EXAMPLE 2 When a particle is located a distance feet from the origin, a force of x x 2 ⫹ 2x pounds acts on it. How much work is done in moving it from x苷1 to x苷3 ?

W苷 y 1 共x

The work done is 2 16

3 ft-lb.

In the next example we use a law from physics: Hooke’s Law states that the force required to maintain a spring stretched units beyond its natural length is proportional x to : x

f 共x兲 苷 kx

frictionless 0 x

where is a positive constant (called the spring constant). Hooke’s Law holds provided k

surface

that is not too large (see Figure 1). x

(a) Natural position of spring

V EXAMPLE 3

A force of 40 N is required to hold a spring that has been stretched from

ƒ=kx

its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm?