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Unformatted text preview: 5E06(pp 374383) 1/17/06 4:15 PM Page 374 CHAPTER 6 The volume of a sphere is the
limit of sums of volumes of
approximating cylinders. A pplications of Integration 5E06(pp 374383) 1/17/06 4:15 PM Page 375 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 x *
i x and thus approxi mate 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 y y=ƒ S
0 a b x y=© 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 S that lies between two curves y f x and y t x and between the vertical lines x a and x b, where f and t are continuous functions and
fx
t x for all x in a, b . (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 x and height
f x*
t x* . (See Figure 2. If we like, we could take all of the sample points to be right
i
i
endpoints, in which case x* x i .) The Riemann sum
i FIGURE 1 n S=s(x, y)  a¯x¯b, ©¯y¯ƒd f x*
i t x*
i x i1 is therefore an approximation to what we intuitively think of as the area of S.
y y f (x * )
i Guess the area of an island.
Resources / Module 7
/ Areas
/ Start of Areas 0 a f (x *)g(x *)
i
i Îx
F IGURE 2 x b _g(x *)
i 0 a b x x*
i (a) Typical rectangle (b) Approximating rectangles This approximation appears to become better and better as n l . Therefore, we deﬁne
the area A of S as the limiting value of the sum of the areas of these approximating
rectangles.
n 1 A f x*
i lim nl t x*
i x i1 375 5E06(pp 374383) ❙❙❙❙ 376 1/17/06 4:15 PM Page 376 CHAPTER 6 APPLICATIONS OF INTEGRATION We recognize the limit in (1) as the deﬁnite integral of f
following formula for area. t. Therefore, we have the f x , y t x , and the
2 The area A of the region bounded by the curves y
lines x a, x b, where f and t are continuous and f x
t x for all x in a, b , is
y y A y=ƒ b a fx t x dx y=©
S 0 a b x Notice that in the special case where t x
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).
In the case where both f and t are positive, you can see from Figure 3 why (2) is true: FIGURE 3
b A b A=j ƒ dxj © dx
a a area under y y b a y y f x dx b a area under y t x dx y b fx a y
x=1 x, and bounded on the sides by x x2 1 y=x y A y yT
yT yB a FIGURE 5 x2 1 1 x2
2 y x dx
x
0 1 x2 0 1
3 1
2 x
1 x2
x2 1 and
1, 1 dx
5
6 In Figure 4 we drew a typical approximating rectangle with width x 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 yT , the bottom curve
yB , and a typical approximating rectangle as in Figure 5. Then the area of a typical rectangle is yT yB x and the equation FIGURE 4 yB x. So we use the area formula (2) with f x x3
3 x 1 1 0 Îx 0 1, bounded below by 1. SOLUTION The region is shown in Figure 4. The upper boundary curve is y the lower boundary curve is y
tx
x, a 0, and b 1: x=0 0 and x tx
t x dx E XAMPLE 1 Find the area of the region bounded above by y
y=≈+1 0 fx Îx
b x n A lim nl yT
i1 yB x y b a yT yB d x summarizes the procedure of adding (in a limiting sense) the areas of all the typical
rectangles.
Notice that in Figure 5 the lefthand boundary reduces to a point, whereas in Figure 3
the righthand boundary reduces to a point. In the next example both of the side boundaries 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 y y 2x x 2 and 2 x. SOLUTION We ﬁrst ﬁnd the points of intersection of the parabolas by solving their equations simultaneously. This gives x 2 2 x x 2, or 2 x 2 2 x 0. Thus 2 x x 1
0,
so x 0 or 1. The points of intersection are 0, 0 and 1, 1 . 5E06(pp 374383) 1/17/06 4:16 PM Page 377 S ECTION 6.1 AREAS BETWEEN CURVES ❙❙❙❙ 377 We see from Figure 6 that the top and bottom boundaries are yT=2x≈
y
(1 , 1) 2x x2 yB yT x and x2 yB The area of a typical rectangle is
yB=≈ yT Îx
x (0, 0) and the region lies between x FIGURE 6 0 and x y A 1 x3
3 1
2 2 1 2x 2 dx x2
2 2 x2 x 1. So the total area is 2x 0 x2 2x x 2 dx 2y x
0 1 0 1
3 1
3 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 x sx 2 1 and y x4 x. S OLUTION If we were to try to ﬁnd the exact intersection points, we would have to solve
the equation x
sx 2
1.5
y= x
œ„„„„„
≈+1 _1 2
y=x $x
_1 x4 1 x 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 x 1.18. (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
A y x
sx 2 1.18 0 FIGURE 7 x4 1 x2 To integrate the ﬁrst term we use the subsitution u
when x 1.18, we have u 2.39. So
A 1
2 y 2.39 1 su du
su y
x5
5 2.39 1 s2.39
0.785 1.18 0 1 x x4
x2
2 1.18
5 5 dx
1. Then du x dx
1.18 0 1.18
2 2 2 x dx, and 5E06(pp 374383) 378 ❙❙❙❙ 1/17/06 4:16 PM Page 378 CHAPTER 6 APPLICATIONS OF INTEGRATION √ (mi/h) 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. 60 A 50
40
30 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 1 mi h 5280 ft s .
3600 B 20
10
0 2 4 6 8 10 12 14 16 t
(seconds) F IGURE 8 t 0 2 4 6 8 10 12 14 16 vA 0 34 54 67 76 84 89 92 95 vB 0 21 34 44 51 56 60 63 65 0 13 20 23 25 28 29 29 30 vA vB We use the Midpoint Rule with n 4 intervals, so that t 4. The midpoints of the
intervals are t1 2, t2 6, t3 10, and t4 14. We estimate the distance between the
cars after 16 seconds as follows: y 16 0 vA vB dt t 13
4 93 y S£ S™
y=© 0 a 28 29 372 ft If we are asked to ﬁnd the area between the curves y f x and y t x where
fx
t x for some values of x but t x
f x for other values of x, then we split the
given region S into several regions S1 , S2 , . . . with areas A1 , A2 , . . . as shown in Figure 9.
We then deﬁne the area of the region S to be the sum of the areas of the smaller regions
. Since
S1 , S2 , . . . , that is, A A1 A2 y=ƒ
S¡ 23 b x fx FIGURE 9 fx
tx tx tx
fx when f x
when t x tx
fx we have the following expression for A.
3 x The area between the curves y
b is
A y y =Ł x A¡ b a fx x=0 t x dx π
2 x EXAMPLE 5 Find the area of the region bounded by the curves y x 0, and x (since 0 x sin x, y cos x, 2. SOLUTION The points of intersection occur when sin x
FIGURE 10 a and When evaluating the integral in (3), however, we must still split it into integrals corresponding to A1 , A2 , . . . . A™ π
4 y t x and between x y=Ã x π
x=
2 0 f x and y cos x, that is, when x
2). The region is sketched in Figure 10. Observe that cos x 4
sin x 5E06(pp 374383) 1/17/06 4:17 PM Page 379 S ECTION 6.1 AREAS BETWEEN CURVES when 0
area is x 4 but sin x
A y 2 0 y 4 0 cos x when 4 x cos x sin x d x A1 sin x d x y 379 A2 cos x ❙❙❙❙ sin x
1
s2 4 cos x
1
s2 2 s2 2 sin x 4 cos x 0 0 2. Therefore, the required cos x d x sin x 1 0 2
4 1
s2 1 1
s2 2 In this particular example we could have saved some work by noticing that the region
is symmetric about x
4 and so
A 2 A1 2y 4 cos x 0 sin x d x Some regions are best treated by regarding x as a function of y. If a region is bounded
by curves with equations x f y , x t y , y c, and y d, where f and t are continuous and f y
t y for c y d (see Figure 11), then its area is y A d fy c y t y dy
y x=g(y)
y=d d d xR xL
Îy Îy x=f ( y)
c xR xL
y=c c 0 0 x FIGURE 11 y FIGURE 12 If we write x R for the right boundary and x L for the left boundary, then, as Figure 12
illustrates, we have (5, 4) 4 A 1
xL= 2 ¥3 (_1, _2) x
_2 y d c xR xL dy Here a typical approximating rectangle has dimensions x R xR=y+1
0 x EXAMPLE 6 Find the area enclosed by the line y x x L and y. 1 and the parabola y 2 2x SOLUTION By solving the two equations we ﬁnd that the points of intersection are 1, 2 and 5, 4 . We solve the equation of the parabola for x and notice from
Figure 13 that the left and right boundary curves are
FIGURE 13 xL 1
2 y2 3 xR y 1 6. 5E06(pp 374383) ❙❙❙❙ 380 1/17/06 4:18 PM Page 380 CHAPTER 6 APPLICATIONS OF INTEGRATION We must integrate between the appropriate yvalues, y y A 4 xR 2
4 y(
2 1
6 1
2 64 4 y [y xL dy
y2
8 ( 16 4
3 3) dy y3
3 1
2 4) dy 8) 2 4. Thus (1 y2
2 1 2 y 2 and y 4 y2
2 4y
2 18 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 A1 and A2 in Figure 14. The method we used in
Example 6 is much easier.
y y= œ„„„„„
2x+6 (5, 4) A™
y=x1
3 0 A¡ x
(_1, _2) y=_ œ„„„„„
2x+6 FIGURE 14  6.1
1–4 Exercises
5–26  Sketch the region enclosed by the given curves. Decide
whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then ﬁnd the area
of the region. Find the area of the shaded region.  1. y y 2. y=5x≈ (6, 12) 5. y 8. y x y=2x
y=≈4x x y 3. y=1 sx x ■ ■ ■ ■ ■ ■ ■ ■ x ■ ■ y 3 x x3 x
3 sx
2 x, y x2 x, y 3x sx , y 1
2 8 2 3 17. x x 32 y 12 16. y
■ 2, y y2 15. y x=2y¥ x 3, x 2, 14. y x 1, x4 y sx, 13. y
x=œy
„ x, x x2 y x, x 2, 9 y 1 12. y (_3, 3) ■ y 11. y x=¥4y x=¥1 x 2, 1, 10. y y 4. x, 9. y y=x sin x, 7. y 0 x 6. y (4, 4) x x,
2 2y , x x, x
2 x, y y 1 6 9
x 3, x 3 2 5E06(pp 374383) 1/17/06 4:18 PM Page 381 S ECTION 6.1 AREAS BETWEEN CURVES y2 18. 4x 12, x 2 y t 2 19. x 1 y, 20. y sin x 2, 21. y cos x, y sin 2 x, x 0, x 22. y sin x, y sin 2 x, x 0, x 2 23. y cos x, y 24. y x, 26. y
■ 27–28 x x sin ■ x 2 t vC vK 0
20
32
46
54
62 0
22
37
52
61
71 6
7
8
9
10 69
75
81
86
90 80
86
93
98
102 40. The widths (in meters) of a kidneyshaped swimming pool 0, 2x y x, vK 2 x, x ■ ■ y
x ■ were measured at 2meter intervals as indicated in the ﬁgure.
Use the Midpoint Rule to estimate the area of the pool. 3
2
■ ■ ■ ■ ■ ■ Use calculus to ﬁnd the area of the triangle with the given vertices.
27. 0, 0 , 2, 1 , ■ ■ 29–30 6.2 1, 6 2, 28. 0, 5 , y ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 x3 ■ 30. x dx
■ sin 2 32. y
■ ■ ■ x 4, 3 ■ y 4 0 ■ x, ■ cos 2 y 3 s16
■ sx 2 ■ ■ x dx
■ ■ ■ y x, ■ x 4, x ■ ■ 0 x 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?
(c) Which car is ahead after two minutes? Explain.
(d) Estimate the time at which the cars are again side by side.
√ 1 0
■ ■ ■ ■ ■ A ■ B ; 33–36  Use a graph to ﬁnd approximate xcoordinates of the
points of intersection of the given curves. Then ﬁnd (approximately) the area of the region bounded by the curves. 33. y x 2, 34. y 4 x, 35. y sx 36. y x4 ■ CAS ■ y 0 2 cos x y x y x2 1, y x sin x 2 ■ 1 2 t (min) 42. The ﬁgure shows graphs of the marginal revenue function R
3 3x
1, ■ 4.8 41. Two cars, A and B, start side by side and accelerate from rest. 4 to approximate the
31–32  Use the Midpoint Rule with n
area of the region bounded by the given curves.
31. y 5.6 5.0 4.8 ■  1 7
.2 6.8 5, 1 2, ■ Evaluate the integral and interpret it as the area of a
region. Sketch the region. 29. ■ ■ ■ ■ ■ ■ ■ ■ and the marginal cost function C for a manufacturer. [Recall
from Section 4.8 that R x and C x represent the revenue and
cost when x units are manufactured. Assume that R and C 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. 37. Use a computer algebra system to ﬁnd the exact area enclosed by the curves y x5 6x 3 4x and y x. 38. Sketch the region in the xyplane deﬁned by the inequalities x 2y 2 381 2x 2 2y ■  y 1 1 x, y 25. y y vC 0
1
2
3
4
5 2 x ❙❙❙❙ 0, 1 x y y Rª(x)
3 0 and ﬁnd its area. 39. 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. 2
1
0 C ª(x) 50 100 x 5E06(pp 374383) 382 ❙❙❙❙ 1/17/06 4:19 PM Page 382 CHAPTER 6 APPLICATIONS OF INTEGRATION 2
; 43. The curve with equation y x 2 x 3 is called Tschirnhausen’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. 44. Find the area of the region bounded by the parabola y 48. Suppose that 0 2. For what value of c is the area of
c
the region enclosed by the curves y cos x, y cos x c ,
and x 0 equal to the area of the region enclosed by the
curves y cos x c , x
, and y 0? x 2, the ■ tangent line to this parabola at 1, 1 , and the xaxis.
45. Find the number b such that the line y bounded by the curves y
equal area. x 2 and y  6.2 x2 c 2 and y c2 ■ ■ ■ ■ ■ ■ ■ ■  Sketch the region bounded by the given curves and ﬁnd
the area of the region. 49. y a bisects the area 1 x, 50. y b bisects the area 2 ■ x,
■ 1 x 2, y sin x, 51. y 47. Find the values of c such that the area of the region enclosed by the parabolas y ■ 49–51 2 under the curve y 1 x , 1 x 4.
(b) Find the number b such that the line y
in part (a). ■ The following exercises are intended only for those who have
already covered Chapter 7. b divides the region
4 into two regions with 46. (a) Find the number a such that the line x ■ x y e, y 2x ■ ■ x 2 x 2 0, x 2 1
■ ■ ■ ■ ■ ■ ■ ■ 52. For what values of m do the line y x 2 is 576. y x x2 m x and the curve
1 enclose a region? Find the area of the region. 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 cylinder). As illustrated in Figure 1(a), a cylinder is bounded by a plane region B1, called the
base, and a congruent region B2 in a parallel plane. The cylinder consists of all points on
line segments that are perpendicular to the base and join B1 to B2 . If the area of the base is
A and the height of the cylinder (the distance from B1 to B2 ) is h, then the volume V of the
cylinder is deﬁned as
V Ah
In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with
volume V
r 2h [see Figure 1(b)], and if the base is a rectangle with length l and width
w, then the cylinder is a rectangular box (also called a rectangular parallelepiped ) with
volume V lwh [see Figure 1(c)]. B™
h
h
B¡ h
w r
l FIGURE 1 (a) Cylinder
V=Ah (b) Circular cylinder
V=πr@h (c) Rectangular box
V=l wh 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. 5E06(pp 374383) 1/17/06 4:19 PM Page 383 SECTION 6.2 VOLUMES ❙❙❙❙ 383 We start by intersecting S with a plane and obtaining a plane region that is called
a crosssection of S. Let A x be the area of the crosssection of S in a plane Px perpendicular to the xaxis and passing through the point x, where a x b. (See Figure 2.
Think of slicing S with a knife through x and computing the area of this slice.) The crosssectional area A x will vary as x increases from a to b.
y Watch an animation of Figure 2.
Resources / Module 7
/ Volumes
/ Volumes by CrossSection Px S
A(x)
A(b) A(a) 0 FIGURE 2 x a x b Let’s divide S into n “slabs” of equal width x by using the planes Px1 , Px 2 , . . . to slice
the solid. (Think of slicing a loaf of bread.) If we choose sample points x* in x i 1, x i , we
i
can approximate the ith slab Si (the part of S that lies between the planes Px i 1 and Px i ) by
a cylinder with base area A x* and “height” x. (See Figure 3.)
i
y y Îx S 0 FIGURE 3 a xi1 x * xi
i b x 0 ⁄ a=x¸ The volume of this cylinder is A x*
i
tion of the volume of the i th slab Si is ¤ ‹ x¢ x∞ xß x¶=b x x, so an approximation to our intuitive concepA x*
i V Si x Adding the volumes of these slabs, we get an approximation to the total volume (that is,
what we think of intuitively as the volume):
n A x*
i V x i1 This approximation appears to become better and better as n l . (Think of the slices as
becoming thinner and thinner.) Therefore, we deﬁne the volume as the limit of these sums
as n l . But we recognize the limit of Riemann sums as a deﬁnite integral and so we
have the following deﬁnition. 5E06(pp 384393) 384 ❙❙❙❙ 1/17/06 4:10 PM Page 384 CHAPTER 6 APPLICATIONS OF INTEGRATION Definition of Volume Let S be a solid that lies between x a and x b. If the crosssectional area of S in the plane Px , through x and perpendicular to the xaxis, is
A x , where A is a continuous function, then the volume of S is  It can be proved that this deﬁnition is independent of how S is situated with respect to the
xaxis. In other words, no matter how we slice S
with parallel planes, we always get the same
answer for V . n V A x*
i lim nl y x b a i1 A x dx When we use the volume formula V xab A x d x it is important to remember that
A x is the area of a moving crosssection obtained by slicing through x perpendicular to
the xaxis.
Notice that, for a cylinder, the crosssectional area is constant: A x
A for all x. So our
denition of volume gives V xab A d x A b a ; this agrees with the formula V A h.
E XAMPLE 1 Show that the volume of a sphere of radius r is
4
3 V
y SOLUTION If we place the sphere so that its center is at the origin (see Figure 4), then the
plane Px intersects the sphere in a circle whose radius (from the Pythagorean Theorem)
is y sr 2 x 2. So the crosssectional area is
r _r r3 0 y2 Ax y
x r x Using the deﬁnition of volume with a
V y
2 r
r y r r2 0 FIGURE 4
2 2
4
3 Watch an animation of Figure 5.
Resources / Module 7
/ Volumes
/ Volumes (a) Using 5 disks, VÅ4.2726 r
4
3 y A x dx rx
r r
r r2
r and b r2 (The integrand is even.) r 2
0 r, we have x 2 dx x 2 dx
x3
3 x2 r3 r3
3 3 Figure 5 illustrates the deﬁnition of volume when the solid is a sphere with radius
1. From the result of Example 1, we know that the volume of the sphere is
4.18879. Here the slabs are circular cylinders, or disks, and the three parts of Figure 5 (b) Using 10 disks, VÅ4.2097 FIGURE 5 Approximating the volume of a sphere with radius 1 (c) Using 20 disks, VÅ4.1940 5E06(pp 384393) 1/17/06 4:10 PM Page 385 SECTION 6.2 VOLUMES ❙❙❙❙ 385 show the geometric interpretations of the Riemann sums
n n A xi 12 x i1 x i2 x i1 when n 5, 10, and 20 if we choose the sample points x* to be the midpoints xi . Notice
i
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 xaxis the region under the curve y sx from 0 to 1. Illustrate the deﬁnition of volume by sketching a
typical approximating cylinder.
SOLUTION The region is shown in Figure 6(a). If we rotate about the xaxis, 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 crosssection is (sx )2 Ax
 Did 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
0, 1 and height 1. If we rotate this square,
we get a cylinder with radius 1, height 1, and
12 1
volume
. We computed that the
given solid has half this volume. That seems
about right. x and the volume of the approximating cylinder (a disk with thickness x) is
Ax
The solid lies between x 0 and x y V 1 0 x xx 1, so its volume is y A x dx 1 0 x2
2 x dx y 1 2 0 y y=œ„
x œ„
x
See a volume of revolution being formed.
Resources / Module 7
/ Volumes
/ Volumes of Revolution 0 1 x 0 x 1 x Îx (b) (a) FIGURE 6 EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by y x 3, y 8, and x 0 about the yaxis. 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 yaxis, it makes sense to slice the
solid perpendicular to the yaxis and therefore to integrate with respect to y. If we slice
3
at height y, we get a circular disk with radius x, where x sy. So the area of a crosssection through y is Ay x2 (sy )2
3 y2 3 5E06(pp 384393) 386 ❙❙❙❙ 1/17/06 4:11 PM Page 386 CHAPTER 6 APPLICATIONS OF INTEGRATION and the volume of the approximating cylinder pictured in Figure 7(b) is
Ay
Since the solid lies between y y2 3 y y 0 and y
V y 8 0 [ 8, its volume is y A y dy
3
5 y5 3 8 0 y 2 3 dy 96
5 8
0 y y y=8 8 Îy x y (x, y) x=0
y=˛
or
x= Œ„
y
0 F IGURE 7 0 x (a) x (b) EXAMPLE 4 The region enclosed by the curves y
xaxis. Find the volume of the resulting solid. x and y x 2 is rotated about the x and y x 2 intersect at the points 0, 0 and 1, 1 . The region
between them, the solid of rotation, and a crosssection perpendicular to the xaxis are
shown in Figure 8. A crosssection in the plane Px has the shape of a washer (an annular
ring) with inner radius x 2 and outer radius x, so we ﬁnd the crosssectional area by subtracting the area of the inner circle from the area of the outer circle:
SOLUTION The curves y Ax
y x2 x2 2 x2 x4 y
(1, 1) A(x) y=x
y=≈
≈
x (0, 0) FIGURE 8 (a) x 0 ( b) x (c) 5E06(pp 384393) 1/17/06 4:11 PM Page 387 SECTION 6.2 VOLUMES ❙❙❙❙ 387 Therefore, we have y V Module 6.2/6.3 illustrates the formation
and computation of volumes using disks
and washers. 1 0 y A x dx
x3
3 1 x2 0 1 x5
5 x 4 dx 2
15 0 EXAMPLE 5 Find the volume of the solid obtained by rotating the region in Example 4 about the line y 2. SOLUTION The solid and a crosssection are shown in Figure 9. Again the crosssection is
a washer, but this time the inner radius is 2 x and the outer radius is 2 x 2. The
crosssectional area is y
4 Ax x2 2 2 2 x 2 and so the volume of S is
y=2 V 2 y 1 0 y 1 0 y=≈ y=x
0 x x 1 y A x dx
x4 1 0 5x 2 2 x2 2 x 2 x5
5 4x dx 2 5 x3
3 dx
4 x2
2 1 0 8
15 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 revolution by using the basic deﬁning formula
V y b a A x dx or V y d c A y dy and we ﬁnd the crosssectional area A x or A y in one of the following ways:
y=2 ■ If the crosssection is a disk (as in Examples 1–3), we ﬁnd the radius of the
disk (in terms of x or y) and use
A 2x
2≈ ■ x ≈
x x 2 If the crosssection is a washer (as in Examples 4 and 5), we ﬁnd the inner
radius r in and outer radius rout from a sketch (as in Figures 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 FIGURE 9 outer radius rin
rout FIGURE 10 radius 2 inner radius 2 5E06(pp 384393) 388 ❙❙❙❙ 1/17/06 4:11 PM Page 388 CHAPTER 6 APPLICATIONS OF INTEGRATION The next example gives a further illustration of the procedure.
EXAMPLE 6 Find the volume of the solid obtained by rotating the region in Example 4
about the line x
1.
SOLUTION Figure 11 shows a horizontal crosssection. It is a washer with inner radius 1 y and outer radius 1 sy, so the crosssectional area is Ay 2 outer radius (1 inner radius sy )2 1 2 2 y The volume is
V y 1 0 A y dy
1 y [(1 sy )2 1 y (2 sy y 4y 3 2
3 y2
2 y3
3 dy y 2 ) dy 0 1 0 y 2 1 2 0 y 1+œ„
y
1+y
1
x=œ„
y
y
x=y
x 0 FIGURE 11 x=_1 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 crosssections perpendicular to the base are equilateral triangles. Find the volume of the solid. y FIGURE 12 Computergenerated picture
of the solid in Example 7 x 5E06(pp 384393) 1/17/06 4:12 PM Page 389 SECTION 6.2 VOLUMES ❙❙❙❙ 389 S OLUTION Let’s take the circle to be x 2 y 2 1. The solid, its base, and a typical crosssection at a distance x from the origin are shown in Figure 13.
y y B(x, y) 1≈
y=œ„„„„„„ C C y
B
_1 0 0 1 x x A A A
FIGURE 13 3
œ„ y x (a) The solid (b) Its base 60° 60°
y y B (c) A crosssection Since B lies on the circle, we have y s1 x 2 and so the base of the triangle ABC
is AB
2 s1 x 2. Since the triangle is equilateral, we see from Figure 13(c) that its
height is s3 y s3 s1 x 2. The crosssectional area is therefore
1
2 Ax 2 s1 x 2 s3 s1 x2 s3 1 x2 and the volume of the solid is
V y 1
1 y A x dx
1 1 s3 1 x 2 dx 2 y s3 1
0 Resources / Module 7
/ Volumes
/ Start of Mystery of the
Topless Pyramid 1 x 2 dx
x3
3 2 s3 x 1 0 4 s3
3 EXAMPLE 8 Find the volume of a pyramid whose base is a square with side L and whose height is h.
SOLUTION We place the origin O at the vertex of the pyramid and the xaxis along its central axis as in Figure 14. Any plane Px that passes through x and is perpendicular to the
xaxis intersects the pyramid in a square with side of length s, say. We can express s in
terms of x by observing from the similar triangles in Figure 15 that x
h s2
L2 s
L and so s L x h. [Another method is to observe that the line OP has slope L 2h and
so its equation is y L x 2h .] Thus, the crosssectional area is
Ax L2 2
x
h2 s2
y y P x
O h s
x O x x h FIGURE 14 FIGURE 15 L 5E06(pp 384393) 390 ❙❙❙❙ 1/17/06 4:12 PM Page 390 CHAPTER 6 APPLICATIONS OF INTEGRATION y The pyramid lies between x 0 and x h y V h 0 A x dx 2 h 3 Lx
h2 3 y NOTE 0 h, so its volume is x y h 0 L2 2
x dx
h2 2 Lh
3 0 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 yaxis, as in Figure 16, you can verify that
we would have obtained the integral
■ FIGURE 16 y V h 0 L2
h
h2 y 2 dy L2h
3 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 xaxis along the diameter where the planes meet, then the
base of the solid is a semicircle with equation y s16 x 2, 4 x 4. A crosssection perpendicular to the xaxis at a distance x from the origin is a triangle ABC,
as shown in Figure 17, whose base is y s16 x 2 and whose height is
BC
y tan 30
s16 x 2 s3. Thus, the crosssectional area is
1
2 Ax 1
s16
s3 x2 s16 16 x 2
2 s3 x2 and the volume is
V y 4
4 1
s3 y A x dx y 4 0 4
4 16 x 2
dx
2 s3 x 2 dx 16 1
16 x
s3 x3
3 4 0 128
3 s3
For another method see Exercise 62. C C
0 y A
4 FIGURE 17 x B y=œ„„„„„„
16≈ 30°
A y B 5E06(pp 384393) 1/17/06 4:13 PM Page 391 S ECTION 6.2 VOLUMES  6.2 27. 3 about OA 29. 3 about AB ■ 1. y x 2, x 2. x 2y 3. y 1 x, x 4. y sx 1, y
2, x 0; about the xaxis 1, x x 2, 0 6. x y 7. y x 2, y 2 2, y
2, x 2, y x y 2, x x, x 1, x tan 3 x, y 11. y x, y sx; about y 12. y x 2, y 4; about y
1; about y 2 13. y x,y
2 14. y 1x,y 0; about the yaxis 0, x 15. x y,x 1; about x x, y sx; about x x 2, x y 2; about x 18. y x, y 1 ■ ■ 0, x ■ ■ ■ CAS 1 2, x sin x, 0 x ; about y y ■ ■ 1 ■ ■ ■  Refer to the ﬁgure and ﬁnd the volume generated by
rotating the given region about the speciﬁed line. x 2, 3; about x
1 ■ 2 4 ■ ■ 10
1
2 2
x ; about x
■ ■ 5
■ ■ ■ ■ T¡
y=˛
x 19. 1 about OA 20. 1 about OC 21. 1 about AB 22. 1 about BC 23. 2 about OA 24. 2 about OC 25. 2 about AB 26. 2 about BC x 4, 38. y 1
■ ■ ■ ■ y
■ x3 3x
■ ■ ■  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. sin2 x, y
x 2 0, 0 2 x, y ; about y x 1 x 4 ; about y x cos 2 ■
■ 41–44 ■  ■ ■ ■ ■ ■ ■ ■ ■ ■ Each integral represents the volume of a solid. Describe the solid. y
y
■ 2 y ■ T™ sx
■ 39–40 44. y=œ„
x y
■ 43. B(1, 1) A(1, 0) 16; about x y 6, y ■ 41. y O 1, x 3y 40. y ■ 19–30 T£ 2 ■ ■ C(0, 1) 2 , 8x ■ 39. y 4; about x ■ 1 0, y
2 ■ 2 17. y ■  Use a graph to ﬁnd approximate xcoordinates of the
points of intersection of the given curves. Then ﬁnd (approximately) the volume of the solid obtained by rotating about the
xaxis the region bounded by these curves. 1 16. y ■ ; 37–38 3; about y 2 0; about y ; about y 37. y 1, x 1, x 4 x 36. 2 x 4 4 ■ sin x, 0 35. x 1; about the xaxis 1 1, y about BC ■ 0, y ■ x 2 3, x ■ x 34. y 2y; about the yaxis 10. y ■ 33. y 0; about the yaxis 1, x ■ 32. y 0; about the yaxis x; about the xaxis sec x, y ■ about OC 3 30. ■ 31. y 0; about the xaxis 4, x 3  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. 0; about the xaxis
5, y ■ 28. 31–36 0; about the xaxis 0y 1, x 5. y 9. y 2 391 Exercises 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. 8. y ❙❙❙❙ 0
1 0 0 cos2x dx y4
2 y 42. 5 2 y dy y 8 dy
1
■ cos x
■ 2 12 dx
■ ■ ■ ■ ■ ■ ■ ■ 45. A CAT scan produces equally spaced crosssectional views of a human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a
human liver shows crosssections spaced 1.5 cm apart. The
liver is 15 cm long and the crosssectional 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 1meter intervals and its cross sectional areas A (at a distance x from the end of the log) are 5E06(pp 384393) 392 ❙❙❙❙ 1/17/06 4:13 PM Page 392 CHAPTER 6 APPLICATIONS OF INTEGRATION listed in the table. Use the Midpoint Rule with n
estimate the volume of the log.
x (m)  x (m) A (m2 ) 0
1
2
3
4
5
47–59 A (m2 )
0.68
0.65
0.64
0.61
0.58
0.59 6
7
8
9
10 5 to 0.53
0.55
0.52
0.50
0.48 53. A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm,
and 5 cm
54. The base of S is a circular disk with radius r. Parallel cross sections perpendicular to the base are squares.
55. The base of S is an elliptical region with boundary curve 9x 2 4y 2 36. Crosssections perpendicular to the xaxis
are isosceles right triangles with hypotenuse in the base.
x, y x 2 y 1 .
Crosssections perpendicular to the yaxis are equilateral
triangles. 56. The base of S is the parabolic region Find the volume of the described solid S. 47. A right circular cone with height h and base radius r 57. S has the same base as in Exercise 56, but crosssections perpendicular to the yaxis are squares. 48. A frustum of a right circular cone with height h, lower base 58. The base of S is the triangular region with vertices 0, 0 , radius R, and top radius r 3, 0 , and 0, 2 . Crosssections perpendicular to the yaxis
are semicircles. r 59. S has the same base as in Exercise 58, but crosssections h perpendicular to the yaxis are isosceles triangles with height
equal to the base. R
■ 49. A cap of a sphere with radius r and height h ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 60. The base of S is a circular disk with radius r. Parallel cross sections perpendicular to the base are isosceles triangles with
height h and unequal side in the base.
(a) Set up an integral for the volume of S.
(b) By interpreting the integral as an area, ﬁnd the volume of S. h
r 61. (a) Set up an integral for the volume of a solid torus (the 50. A frustum of a pyramid with square base of side b, square top donutshaped solid shown in the ﬁgure) with radii r and R.
(b) By interpreting the integral as an area, ﬁnd the volume of
the torus. of side a, and height h
a
R
r 62. Solve Example 9 taking crosssections to be parallel to the line b What happens if a of intersection of the two planes. b ? What happens if a 0? 51. A pyramid with height h and rectangular base with dimensions b and 2b
52. A pyramid with height h and base an equilateral triangle with side a (a tetrahedron) 63. (a) Cavalieri’s Principle states that if a family of parallel planes gives equal crosssectional areas for two solids S1 and S2 ,
then the volumes of S1 and S2 are equal. Prove this principle.
(b) Use Cavalieri’s Principle to ﬁnd the volume of the oblique
cylinder shown in the ﬁgure. h a
a a r 5E06(pp 384393) 1/17/06 4:13 PM Page 393 SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS 64. Find the volume common to two circular cylinders, each with ❙❙❙❙ 393 68. A hole of radius r is bored through the center of a sphere of radius r, if the axes of the cylinders intersect at right angles. radius R
sphere. r. Find the volume of the remaining portion of the 69. 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 h and maximum radius R is constructed by rotating about the xaxis the parabola
y R c x 2, h 2 x h 2, where c is a positive
constant. Show that the radius of each end of the barrel is
r R d, where d ch 2 4.
(b) Show that the volume enclosed by the barrel is 65. Find the volume common to two spheres, each with radius r, if the center of each sphere lies on the surface of the other sphere.
66. 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 h centimeters. Find the volume of water in the bowl.
67. A hole of radius r is bored through a cylinder of radius R V 2
5 d2) has area A and lies above the xaxis.
When is rotated about the xaxis, it sweeps out a solid with
volume V1. When is rotated about the line y
k (where k
is a positive number), it sweeps out a solid with volume V2.
Express V2 in terms of V1, k, and A. r Volumes by Cylindrical Shells y y=2≈˛
1 xL=? xR=? 0 2 x Some volume problems are very difﬁcult to handle by the methods of the preceding section. For instance, let’s consider the problem of ﬁnding the volume of the solid obtained
by rotating about the yaxis the region bounded by y 2 x 2 x 3 and y 0. (See Figure 1.)
If we slice perpendicular to the yaxis, 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
y 2 x 2 x 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 r1, outer radius r2,
and height h. Its volume V is calculated by subtracting the volume V1 of the inner cylinder
from the volume V2 of the outer cylinder: FIGURE 1 r r2 70. Suppose that a region at right angles to the axis of the cylinder. Set up, but do not
evaluate, an integral for the volume cut out.  6.3 h (2R 2 1
3 V V2
rh r™ 2
h r2h
1 r2 Îr r¡ V1
2
2 r1 r2 r2 r1
2 r2
2 r2 h
1 r1 h h r2 r1 If we let r r2 r1 (the thickness of the shell) and r 1 r2 r1 (the average radius
2
of the shell), then this formula for the volume of a cylindrical shell becomes
1
FIGURE 2 V 2 rh r 5E06(pp 394403) 394 ❙❙❙❙ 1/17/06 4:05 PM Page 394 CHAPTER 6 APPLICATIONS OF INTEGRATION y and it can be remembered as
y=ƒ 0 a V x b [circumference][height][thickness] Now let S be the solid obtained by rotating about the yaxis 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 x i 1, x i of equal width x and let xi be
the midpoint of the i th subinterval. If the rectangle with base x i 1, x i and height f xi is
rotated about the yaxis, then the result is a cylindrical shell with average radius xi , height
f xi , and thickness x (see Figure 4), so by Formula 1 its volume is y y=ƒ Vi 2 xi f xi x Therefore, an approximation to the volume V of S is given by the sum of the volumes of
these shells:
n 0 a b n V x Vi
i1 2 xi f xi x i1 This approximation appears to become better as n l . But, from the deﬁnition of an integral, we know that FIGURE 3 n lim y y=ƒ 0 2 xi f xi nl a b
–
x i1 x i x i1 y b a 2 x f x dx Thus, the following appears plausible: x 2 The volume of the solid in Figure 3, obtained by rotating about the yaxis the
region under the curve y f x from a to b, is xi V y b a 2 x f x dx where 0 a b y y=ƒ 0 a b The argument using cylindrical shells makes Formula 2 seem reasonable, but later we
will be able to prove it (see Exercise 65 in Section 8.1).
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 2 x, height f x , and thickness x or dx :
x y b 2x fx circumference height a FIGURE 4 dx y ƒ ƒ x
x x 2πx Îx FIGURE 5 This type of reasoning will be helpful in other situations, such as when we rotate about
lines other than the yaxis.
EXAMPLE 1 Find the volume of the solid obtained by rotating about the yaxis the region
bounded by y 2 x 2 x 3 and y 0. 5E06(pp 394403) 1/17/06 4:06 PM Page 395 S ECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS y ❙❙❙❙ 395 SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumfer 2x 2 ence 2 x, and height f x
x y V 2≈˛ 2 x 2x 2 0 [ 2 2x x 2 x 3. So, by the shell method, the volume is 1
2 1
5 x4 x 3 dx x5 2
0 2 (8 2 y 2 0 32
5 2x 3 ) x 4 dx 16
5 It can be veriﬁed that the shell method gives the same answer as slicing.
FIGURE 6 y  Figure 7 shows a computergenerated
picture of the solid whose volume we computed
in Example 1.
x FIGURE 7 NOTE 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 x in terms of y. However, in other examples
the methods of the preceding section may be easier. y y=x EXAMPLE 2 Find the volume of the solid obtained by rotating about the yaxis the region y=≈ between y shell
height=x≈
0 ■ x 2. x and 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 x x V y 1 0 FIGURE 8 x3
3 2
Module 6.2/6.3 compares shells with
disks and washers. x 2 dx 2xx 2 y 1 0 x2 x 3 dx 1 x4
4 6 0 As the following example shows, the shell method works just as well if we rotate about
the xaxis. We simply have to draw a diagram to identify the radius and height of a shell.
EXAMPLE 3 Use cylindrical shells to ﬁnd the volume of the solid obtained by rotating about the xaxis the region under the curve y
y SOLUTION This problem was solved using disks in Example 2 in Section 6.2. To use shells
we relabel the curve y s x (in the ﬁgure in that example) as x y 2 in Figure 9. For
rotation about the xaxis we see that a typical shell has radius y, circumference 2 y, and
height 1 y 2. So the volume is shell height=1¥ 1
y x=¥ 0 FIGURE 9 x=1 shell
radius=y
1 s x from 0 to 1. V
x y 1 0 2 2y1
y2
2 y 2 dy
y4
4 2 1 0 In this problem the disk method was simpler. 2 y 1 0 y y 3 dy 5E06(pp 394403) ❙❙❙❙ 396 1/17/06 4:06 PM Page 396 CHAPTER 6 APPLICATIONS OF INTEGRATION EXAMPLE 4 Find the volume of the solid obtained by rotating the region bounded by y x 2 and y x 0 about the line x 2. SOLUTION Figure 10 shows the region and a cylindrical shell formed by rotation about the line x 2. It has radius 2 x, circumference 2 y 2 x 2. x , and height x y x=2 y=x≈ 0 0 x 1 2 x FIGURE 10 3 4 x 2x The volume of the given solid is
V y 1 0 2  6.3 2 2 x 2 dx xx
x 3 x3 3x 2 2x dx x 2 2 0 Exercises
4. y y x 2, y 0, 5. y the ﬁgure about the yaxis. Explain why it is awkward to use
slicing to ﬁnd the volume V of S. Sketch a typical approximating shell. What are its circumference and height? Use shells
to ﬁnd V . 2 0 x x, 6. y 3 7. y
■ 0 1 ﬁgure about the yaxis. Sketch a typical cylindrical shell and
ﬁnd its circumference and height. Use shells to ﬁnd the volume
of S. Do you think this method is preferable to slicing? Explain. 2, 9. x y y
x ■ y 2, 1
sy,
3 2 ■ x 0 3
4x 7 ■ 12. x 4y 0 œπ
„ 3–7 x  Use the method of cylindrical shells to ﬁnd the volume generated by rotating the region bounded by the given curves about the
yaxis. Sketch the region and a typical shell. x 2 y y ■ ■ ■ ■ ■ ■ 1 x 0 8,
3 y, 14. x y 3,
■ x
y 1, y 2 0 2x 2 ■ 0, 0, y
2 4x , ■ x x x, 13. y 1, x 4, 9–14  Use the method of cylindrical shells to ﬁnd the volume of
the solid obtained by rotating the region bounded by the given
curves about the xaxis. Sketch the region and a typical shell. 11. y
y=sin { ≈} x x, ■ 10. x y 0, y yaxis the region bounded by y sx and y x 2. Find V both
by slicing and by cylindrical shells. In both cases draw a diagram to explain your method. x 2. Let S be the solid obtained by rotating the region shown in the y 2, 2 4x 1 2 2x ■ x 8. Let V be the volume of the solid obtained by rotating about the y=x(x1)@ 1 x, 1 0 1 x4
4 1. Let S be the solid obtained by rotating the region shown in 3. y y 2 6 x
■ 4 y
■ 1
■ 2
■ ■ ■ 15–20 ■ ■ ■  Use the method of cylindrical shells to ﬁnd the volume
generated by rotating the region bounded by the given curves about
the speciﬁed axis. Sketch the region and a typical shell. 15. y x 2, y 0, x 1, x 2; about x 1 5E06(pp 394403) 1/17/06 4:07 PM Page 397 S ECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS x 2, y 16. y 0, x 2 17. y x,y 2, x 0, x 1, x 2 1; about the yaxis
2; about x 18. y 4x x,y 8x 2 x ; about x 19. y sx 1, y 0, x 5; about y 20. y x 2, x ■ ■ y 2; about y ■ ■ ■  Use a graph to estimate the xcoordinates of the points of
intersection of the given curves. Then use this information to estimate the volume of the solid obtained by rotating about the yaxis
the region enclosed by these curves. 2 33. y ■ ■ ■ ■ ■  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 axis. 21. y sin x, y 0, x 22. y x, y 23. y 4 x,y 24. y 11 25. x 4x x ; about x sin y2 0, x y 7, x 0, x ,x 0; about y 4; about y 2 y ■ ■ ■ ■ ■ ■ 4 ■ ■ ■ 4 to estimate the volume
obtained by rotating about the yaxis the region under the curve
y tan x, 0 x
4. x sin x, y
■ ■ x2 x x2 3x 39. y 5, y 40. x 5 to estimate the y 1
1 2 3 4 5 6 7 31. y Each integral represents the volume of a solid. Describe y 2 dy 2 1 y2 3 0 1 x
■ ■ 32.
■ y 0
■ ■ 1
■ ■ ■ ■ y,x
y 1 y 1 2 42. x ■ ■  5x 0; about the yaxis
9; about x 0; about x 1 2 1; about the yaxis
1; about the xaxis
■ ■ ■ ■ ■ ■ ■ ■ ■ Use cylindrical shells to ﬁnd the volume of the solid. y1 ■ ■ ■ ■ ■ ■ y2 dy
h 4 ; about x 2 2 x cos x
■ ■ ■ sin x d x
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ diameters through two wooden balls (which also have different
diameters). You discover that both napkin rings have the same
height h, as shown in the ﬁgure.
(a) Guess which ring has more wood in it.
(b) Check your guess: Use cylindrical shells to compute the
volume of a napkin ring created by drilling a hole with
radius r through the center of a sphere of radius R and
express the answer in terms of h. 2 x 5 dx 0 ; about x x 46. Suppose you make napkin rings by drilling holes with different 9 10 11 12 x 8 the solid. y x2 2 ■ 0 30. 2 ■ 45. A right circular cone with height h and base radius r 2 3 ■ 44. The solid torus of Exercise 61 in Section 6.2 3 0 ■ 43. A sphere of radius r 4 y ■ 0; about the xaxis 2, y 2 43–45 5 29. ■ 2, y 4 1 ■  0, 0 ■ 41. x 2 28. If the region shown in the ﬁgure is rotated about the yaxis to 29–32 ■ ■ 27. Use the Midpoint Rule with n form a solid, use the Midpoint Rule with n
volume of the solid. ■ sin4 x, 0 3 37. y
■ ■ 37–42  The region bounded by the given curves is rotated about
the speciﬁed axis. Find the volume of the resulting solid by any
method. 5
■ ■ sin2 x, y 38. y ■ x3 3x ■ x4  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. ■ 2; about x x2 x 35–36 36. y 1 y ■ 35. y 7 x 2 ; about x x2 , y CAS 3 ; about the yaxis 2 ssin y, 0 26. x 2 2,x x 4, ■ ■ 21–26 0, 34. y 3 1
■ 397 ; 33–34 4 2 ❙❙❙❙ ■ ■ 5E06(pp 394403) 398 ❙❙❙❙ 1/17/06 4:07 PM Page 398 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 t , 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:
F 1 m d 2s
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 kg m s2 ). Thus, a force
of 1 N acting on a mass of 1 kg produces an acceleration of 1 m s2. In the U. S. Customary
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 d that the object moves:
2 W Fd work force distance If F is measured in newtons and d in meters, then the unit for W is a newtonmeter, which
is called a joule (J). If F is measured in pounds and d in feet, then the unit for W is a footpound (ftlb), which is about 1.36 J.
EXAMPLE 1 (a) How much work is done in lifting a 1.2kg 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 t 9.8 m s2.
(b) How much work is done in lifting a 20lb 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 mt
1.2 9.8
11.76 N
and then Equation 2 gives the work done as
W
(b) Here the force is given as F
W Fd 11.76 0.7 8.2 J 20 lb, so the work done is
Fd 20 6 120 ftlb Notice that in part (b), unlike part (a), we did not have to multiply by t because we
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 xaxis in the positive direction,
from x a to x b, and at each point x between a and b a force f x acts on the object,
where f is a continuous function. We divide the interval a, b into n subintervals with endpoints x 0 , x 1, . . . , x n and equal width x. We choose a sample point x* in the i th subi
interval x i 1, x i . Then the force at that point is f x* . If n is large, then x is small, and
i 5E06(pp 394403) 1/17/06 4:07 PM Page 399 SECTION 6.4 WORK ❙❙❙❙ 399 since f is continuous, the values of f don’t change very much over the interval x i 1, x i .
In other words, f is almost constant on the interval and so the work Wi that is done in moving the particle from x i 1 to x i is approximately given by Equation 2:
f x*
i Wi x Thus, we can approximate the total work by
n f x*
i W 3 x i1 It seems that this approximation becomes better as we make n larger. Therefore, we deﬁne
the work done in moving the object from a to b as the limit of this quantity as n l .
Since the right side of (3) is a Riemann sum, we recognize its limit as being a deﬁnite integral and so
n W 4 f x*
i lim nl y x b a i1 f x dx EXAMPLE 2 When a particle is located a distance x feet from the origin, a force of x2
2 x pounds acts on it. How much work is done in moving it from x SOLUTION frictionless
surface x 0 (a) Natural position of spring
ƒ=kx y W 3 1 x2 3 x3
3 2x dx x2
1 1 to x 3? 50
3 The work done is 16 2 ftlb.
3
In the next example we use a law from physics: Hooke’s Law states that the force
required to maintain a spring stretched x units beyond its natural length is proportional
to x :
fx
kx
where k is a positive constant (called the spring constant). Hooke’s Law holds provided
that x is not too large (see Figure 1). 0 x x (b) Stretched position of spring
FIGURE 1 Hooke’s Law EXAMPLE 3 A force of 40 N is required to hold a spring that has been stretched from 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?
SOLUTION According to Hooke’s Law, the force required to hold the spring stretched x
meters beyond its natural length is f x
k x. When the spring is stretched from 10 cm
to 15 cm, the amount stretched is 5 cm 0.05 m. This means that f 0.05
40, so 0.05k
Thus, f x 40 k 40
0.05 800 800 x and the work done in stretching the spring from 15 cm to 18 cm is
W y 0.08 0.05 800 x dx 400 0.08 2 x2
800
2
0.05 2 0.08 0.05 1.56 J 5E06(pp 394403) 400 ❙❙❙❙ 1/17/06 4:08 PM Page 400 CHAPTER 6 APPLICATIONS OF INTEGRATION EXAMPLE 4 A 200lb cable is 100 ft long and hangs vertically from the top of a tall building. How much work is required to lift the cable to the top of the building? 0 x i* SOLUTION Here we don’t have a formula for the force function, but we can use an argument similar to the one that led to Deﬁnition 4.
Let’s place the origin at the top of the building and the x axis pointing downward as
in Figure 2. We divide the cable into small parts with length x . If x* is a point in the
i
ith such interval, then all points in the interval are lifted by approximately the same
amount, namely x*. The cable weighs 2 pounds per foot, so the weight of the ith part is
i
2 x . Thus, the work done on the ith part, in footpounds, is Îx 100 2x FIGURE 2 x*
i force x distance 2 x* x
i We get the total work done by adding all these approximations and letting the number
of parts become large (so x l 0 ):
 If we had placed the origin at the bottom of
the cable and the xaxis upward, we would have
gotten y W 100 0 2 100 n W nl x dx x2 y 2 x* x
i lim
i1
100
0 100 0 2 x dx 10,000 ftlb which gives the same answer. EXAMPLE 5 A tank has the shape of an inverted circular cone with height 10 m and base
radius 4 m. It is ﬁlled with water to a height of 8 m. Find the work required to empty
the tank by pumping all of the water to the top of the tank. (The density of water is
1000 kg m3.)
SOLUTION Let’s measure depths from the top of the tank by introducing a vertical coordinate line as in Figure 3. The water extends from a depth of 2 m to a depth of 10 m and
so we divide the interval 2, 10 into n subintervals with endpoints x 0 , x 1, . . . , x n and
choose x* in the i th subinterval. This divides the water into n layers. The ith layer is
i
approximated by a circular cylinder with radius ri and height x. We can compute ri
from similar triangles, using Figure 4, as follows: 4m
0 2m
x*
i
10 m Îx ri ri 10 4
10 x*
i 2
5 ri x*
i 10 Thus, an approximation to the volume of the ith layer of water is
x Vi 4
10
25 ri2 x FIGURE 3 x*
i 2 x and so its mass is
4 mi density
1000 ri 10 volume
4
10
25 x*
i 2 FIGURE 4 160 x*
i 10 2 x The force required to raise this layer must overcome the force of gravity and so
Fi mi t 9.8 160 1570 10x *
i x 10 x*
i 10
2 x*
i 2 x x Each particle in the layer must travel a distance of approximately x*. The work Wi done to
i
raise this layer to the top is approximately the product of the force Fi and the distance x* :
i
Wi Fi x*
i 1570 x* 10
i x*
i 2 x 5E06(pp 394403) 1/17/06 4:08 PM Page 401 S ECTION 6.4 WORK ❙❙❙❙ 401 To ﬁnd the total work done in emptying the entire tank, we add the contributions of each
of the n layers and then take the limit as n l :
n W 1570 x* 10
i lim nl 1570 y 1570  6.4 x*
i 2 x i1 ( 10 2 2048
3 20x 2 100x ) x 3 dx y 10 2 1570 x 10
1570 50 x 2 x 2 dx
20x 3
3 x4
4 10 2 6 3.4 10 J Exercises 1. Find the work done in pushing a car a distance of 8 m while exerting a constant force of 900 N.
2. How much work is done by a weightlifter in raising a 60kg natural length of 30 cm to a length of 42 cm. How much work
is needed to stretch it from 35 cm to 40 cm?
10. If the work required to stretch a spring 1 ft beyond its natural barbell from the ﬂoor to a height of 2 m?
3. A particle is moved along the xaxis by a force that measures
2 10 1 x pounds at a point x feet from the origin. Find the
work done in moving the particle from the origin to a distance
of 9 ft.
4. When a particle is located a distance x meters from the origin, a force of cos x 3 newtons acts on it. How much work is
done in moving the particle from x 1 to x 2? Interpret
your answer by considering the work done from x 1 to
x 1.5 and from x 1.5 to x 2.
5. Shown is the graph of a force function (in newtons) that increases to its maximum value and then remains constant.
How much work is done by the force in moving an object a
distance of 8 m?
F
(N)
30
20
10
0 9. Suppose that 2 J of work is needed to stretch a spring from its length is 12 ftlb, how much work is needed to stretch it 9 in.
beyond its natural length?
11. How far beyond its natural length will a force of 30 N keep the spring in Exercise 9 stretched?
12. If 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another 10 J is needed to stretch it from 12 cm to
14 cm, what is the natural length of the spring?
13–20  Show how to approximate the required work by a
Riemann sum. Then express the work as an integral and evaluate it. 13. A heavy rope, 50 ft long, weighs 0.5 lb ft and hangs over the edge of a building 120 ft high.
(a) How much work is done in pulling the rope to the top of
the building?
(b) How much work is done in pulling half the rope to the top
of the building?
14. A chain lying on the ground is 10 m long and its mass is 80 kg. How much work is required to raise one end of the chain to a
height of 6 m? 2 3 4 5 6 7 8 x (m) 1 15. A cable that weighs 2 lb ft is used to lift 800 lb of coal up a mineshaft 500 ft deep. Find the work done. 6. The table shows values of a force function f x , where x is 16. A bucket that weighs 4 lb and a rope of negligible weight are measured in meters and f x in newtons. Use the Midpoint
Rule to estimate the work done by the force in moving an
object from x 4 to x 20.
x 4 6 8 10 12 14 16 18 20 fx 5 5.8 7.0 8.8 9.6 8.2 6.7 5.2 4.1 7. A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching
it from its natural length to 6 in. beyond its natural length?
8. A spring has a natural length of 20 cm. If a 25N force is required to keep it stretched to a length of 30 cm, how much
work is required to stretch it from 20 cm to 25 cm? used to draw water from a well that is 80 ft deep. The bucket is
ﬁlled with 40 lb of water and is pulled up at a rate of 2 ft s, but
water leaks out of a hole in the bucket at a rate of 0.2 lb s.
Find the work done in pulling the bucket to the top of the well.
17. A leaky 10kg bucket is lifted from the ground to a height of 12 m at a constant speed with a rope that weighs 0.8 kg m. Initially the bucket contains 36 kg of water, but the water leaks at
a constant rate and ﬁnishes draining just as the bucket reaches
the 12 m level. How much work is done?
18. A 10ft chain weighs 25 lb and hangs from a ceiling. Find the work done in lifting the lower end of the chain to the ceiling so
that it’s level with the upper end. 5E06(pp 394403) 402 ❙❙❙❙ 1/17/06 4:08 PM Page 402 CHAPTER 6 APPLICATIONS OF INTEGRATION 19. An aquarium 2 m long, 1 m wide, and 1 m deep is full of 27. When gas expands in a cylinder with radius r, the pressure at water. Find the work needed to pump half of the water out
of the aquarium. (Use the fact that the density of water is
1000 kg m3.)
20. A circular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. How much work is
required to pump all of the water out over the side? (Use the
fact that water weighs 62.5 lb ft 3.)
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ any given time is a function of the volume: P P V . The
force exerted by the gas on the piston (see the ﬁgure) is the
product of the pressure and the area: F
r 2P. Show that the
work done by the gas when the volume expands from volume
V1 to volume V2 is
W y V2 V1 P dV
piston head ■ 21–24  A tank is full of water. Find the work required to pump
the water out of the outlet. In Exercises 23 and 24 use the fact that
water weighs 62.5 lb ft3. 21. 22. 3m V
x 1m
28. In a steam engine the pressure P and volume V of steam satisfy 2m
3m the equation PV 1.4 k , where k is a constant. (This is true for
adiabatic expansion, that is, expansion in which there is no heat
transfer between the cylinder and its surroundings.) Use Exercise 27 to calculate the work done by the engine during a cycle
when the steam starts at a pressure of 160 lb in2 and a volume
of 100 in3 and expands to a volume of 800 in3. 6m 1.5 m
8m 23. 24. 4 ft 5 ft 29. Newton’s Law of Gravitation states that two bodies with masses 1 ft m1 and m2 attract each other with a force
8 ft F semicircle
■ ■ hemisphere ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 25. Suppose that for the tank in Exercise 21 the pump breaks down
after 4.7 10 5 J of work has been done. What is the depth of
the water remaining in the tank?
26. Solve Exercise 22 if the tank is half full of oil that has a density of 920 kg m3.  6.5 where r is the distance between the bodies and G is the gravitational constant. If one of the bodies is ﬁxed, ﬁnd the work
needed to move the other from r a to r b.
30. Use Newton’s Law of Gravitation to compute the work required to launch a 1000kg satellite vertically to an orbit
1000 km high. You may assume that Earth’s mass is
5.98 10 24 kg and is concentrated at its center. Take the radius
of Earth to be 6.37 10 6 m and G 6.67 10 11 N m2 kg2. It is easy to calculate the average value of ﬁnitely many numbers y1 , y2 , . . . , yn : 15 yn
n
But how do we compute the average temperature during a day if inﬁnitely many temperature readings are possible? Figure 1 shows the graph of a temperature function T t , where
t is measured in hours and T in C, and a guess at the average temperature, Tave.
In general, let’s try to compute the average value of a function y f x , a x b.
We start by dividing the interval a, b into n equal subintervals, each with length
*
*
x
b a n. Then we choose points x 1 , . . . , x n in successive subintervals and
yave 10
5 Tave 6 FIGURE 1 m1 m2
r2 Average Value of a Function T 0 G 12 18 24 t y1 y2 5E06(pp 394403) 1/17/06 4:09 PM Page 403 S ECTION 6.5 AVERAGE VALUE OF A FUNCTION ❙❙❙❙ 403 *
*
calculate the average of the numbers f x 1 , . . . , f x n :
f x*
1 f x*
n
n (For example, if f represents a temperature function and n 24, this means that we take
b a n, we can
temperature readings every hour and then average them.) Since x
b a x and the average value becomes
write n
*
f x1 f x*
n
b 1 a b *
f x1 a *
f xn x x x
n 1
b f x*
i a x i1 If we let n increase, we would be computing the average value of a large number of closely
spaced values. (For example, we would be averaging temperature readings taken every
minute or even every second.) The limiting value is
lim nl n 1
b f x*
i a 1 x b i1 a y b a f x dx by the deﬁnition of a deﬁnite integral.
Therefore, we deﬁne the average value of f on the interval a, b as
1 fave b y
a b a f x dx EXAMPLE 1 Find the average value of the function f x
SOLUTION With a 1 and b
fave 1
b
1
3 x 2 on the interval 1 1, 2 . 2 we have y
a b a x3
3 x 1 f x dx 2 1 y 2
1 1 x 2 dx 2 2
1 The question arises: Is there a number c at which the value of f is exactly equal to the
fave? The following theorem says that this is
average value of the function, that is, f c
true for continuous functions.
The Mean Value Theorem for Integrals If f is continuous on a, b , then there exists a number c in a, b such that y b a f x dx fc b a The Mean Value Theorem for Integrals is a consequence of the Mean Value Theorem
for derivatives and the Fundamental Theorem of Calculus. The proof is outlined in Exercise 23. 5E06(pp 404411) 404 ❙❙❙❙ 1/17/06 4:00 PM Page 404 CHAPTER 6 APPLICATIONS OF INTEGRATION The geometric interpretation of the Mean Value Theorem for Integrals is that, for positive functions f , there is a number c such that the rectangle with base a, b and height
f c has the same area as the region under the graph of f from a to b. (See Figure 2 and
the more picturesque interpretation in the margin note.)
y y=ƒ
 You can always chop off the top of a (twodimensional) mountain at a certain height and
use it to ﬁll in the valleys so that the mountain
becomes completely ﬂat. f(c)=fave
0a FIGURE 2 c x b 1 x 2 is continuous on the interval 1, 2 , the Mean Value
Theorem for Integrals says there is a number c in 1, 2 such that
EXAMPLE 2 Since f x y y (2, 5) y=1+≈ 2 x 2 dx 1 1 fc 2 In this particular case we can ﬁnd c explicitly. From Example 1 we know that fave
so the value of c satisﬁes
fc (_1, 2) Therefore 0 FIGURE 3 1 2 fave 2 so c2 1 x 2, 2
c2 Thus, in this case there happen to be two numbers c
work in the Mean Value Theorem for Integrals. fave=2 _1 1 1
1 in the interval 1, 2 that Examples 1 and 2 are illustrated by Figure 3.
EXAMPLE 3 Show that the average velocity of a car over a time interval t1, t2 is the same
as the average of its velocities during the trip.
SOLUTION If s t is the displacement of the car at time t, then, by deﬁnition, the average
velocity of the car over the interval is s
t s t2
t2 s t1
t1 On the other hand, the average value of the velocity function on the interval is
vave 1
t2 t1
1 t2
s t2
t2 t1 y t2 t1 v t dt 1
t2 t1 y t2 t1 s t dt s t2 s t1 s t1
t1 average velocity (by the Net Change Theorem) 5E06(pp 404411) 1/17/06 4:01 PM Page 405 S ECTION 6.5 AVERAGE VALUE OF A FUNCTION  6.5
1–8 2 1. f x x, 1, 1 3. t x cos x, 0, 2 2. f x t s1 t, 6. f sec tan , 0, 7. h x cos x sin x,
31 x 2 s1 x, x 3, 0, 2 r 2, ■ 4 1, 6 ■ ■ 3 2, ■ ■ ■ 2 sin x sin 2 x, ; 12. f x 2x 1 x 2 2, ■ ■ ■ 10. f x 2, 5 ; 11. f x
■ ■ ■ ■ ■ sx, 0, 4 0, 2
■ ■ ■ ■ ■ ■ ■ ■ 8, show that f takes on the
value 4 at least once on the interval 1, 3 . 14. Find the numbers b such that the average value of 2 6x t
12 tance x meters from one end of the rod. What is the average
temperature of the rod?
19. The linear density in a rod 8 m long is 12 s x 1 kg m,
where x is measured in meters from one end of the rod. Find
the average density of the rod. 20. If a freely falling body starts from rest, then its displacement is
given by s 1 tt 2. Let the velocity after a time T be v T . Show
2 0, 3
13. If f is continuous and x1 f x d x fx 14 sin 18. The temperature of a metal rod, 5 m long, is 4 x (in C) at a dis■ (a) Find the average value of f on the given interval.
(b) Find c such that fave f c .
(c) Sketch the graph of f and a rectangle whose area is the same as
the area under the graph of f .
x 50 Find the average temperature during the period from 9 A.M.
to 9 P.M.  9. f x Tt 0, 5 4 ■ 9–12 2 17. In a certain city the temperature (in F) t hours after 9 A.M. was modeled by the function 0, 2 0, 8. h r x 4. t x 2 5. f t ■ 405 Exercises Find the average value of the function on the given interval.  ❙❙❙❙ 3x 2 on the interval 0, b is equal to 3. 15. The table gives values of a continuous function. Use the Mid point Rule to estimate the average value of f on 20, 50 .
x 20 25 30 35 40 45 42 38 31 29 35 48 21. Use the result of Exercise 59 in Section 5.5 to compute the average volume of inhaled air in the lungs in one respiratory
cycle.
22. The velocity v of blood that ﬂows in a blood vessel with radius R and length l at a distance r from the central axis is
P
R2
4l vr 50 fx that if we compute the average of the velocities with respect to
t we get vave 1 v T , but if we compute the average of the veloc2
ities with respect to s we get vave 2 v T .
3 r2 60 16. The velocity graph of an accelerating car is shown. (a) Estimate the average velocity of the car during the ﬁrst
12 seconds.
(b) At what time was the instantaneous velocity equal to the
average velocity? 23. Prove the Mean Value Theorem for Integrals by applying the √
(km / h)
60 Mean Value Theorem for derivatives (see Section 4.2) to the
function F x
xax f t dt. 40 24. If fave a, b denotes the average value of f on the interval a, b and a 20
0 where P is the pressure difference between the ends of the vessel and is the viscosity of the blood (see Example 7 in Section 3.4). Find the average velocity (with respect to r) over the
interval 0 r R. Compare the average velocity with the
maximum velocity. 4 8 12 t (seconds) c
fave a, b b, show that
c
b a
fave a, c
a b
b c
fave c, b
a 5E06(pp 404411) ❙❙❙❙ 406  1/17/06 4:01 PM Page 406 CHAPTER 6 APPLICATIONS OF INTEGRATION 6 Review CONCEPT CHECK ■ 1. (a) Draw two typical curves y f x and y t x , where
fx
t x for a x b. Show how to approximate the
area between these curves by a Riemann sum and sketch
the corresponding approximating rectangles. Then write an
expression for the exact area.
(b) Explain how the situation changes if the curves have
equations x f y and x t y , where f y
t y for
c y d. ■ (b) If S is a solid of revolution, how do you ﬁnd the crosssectional areas?
4. (a) What is the volume of a cylindrical shell? (b) Explain how to use cylindrical shells to ﬁnd the volume of
a solid of revolution.
(c) Why might you want to use the shell method instead of
slicing?
5. Suppose that you push a book across a 6meterlong table by exerting a force f x at each point from x 0 to x 6. What
does x06 f x dx represent? If f x is measured in newtons, what
are the units for the integral? 2. Suppose that Sue runs faster than Kathy throughout a 1500 meter race. What is the physical meaning of the area between
their velocity curves for the ﬁrst minute of the race? 6. (a) What is the average value of a function f on an interval
3. (a) Suppose S is a solid with known crosssectional areas. a, b ?
(b) What does the Mean Value Theorem for Integrals say?
What is its geometric interpretation? Explain how to approximate the volume of S by a Riemann
sum. Then write an expression for the exact volume. ■ 1–6 2 1. y x 2. y 20 x 2, y 3. y 1 2 x, 4. x y 0, 5. y x sin 6. y y 0 y y x 2 y 3y x, 2 2x x ■ ■ ■ ■ ■ ■ ■ 1 9. x 0, x 10. y
11. x y 2, y x 2 2 x y ; about x 1, y
2 y
a,x
about the yaxis ■ ■ ■ ■ 1 2 9 2 x ; about y
a 1 h (where a ■ ■ ; 18. Let 0, h ■ ■ y
(a)
(b)
(c)
(d) 0);
■ ■ ■ ■ 12–14  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 axis. 12. y
13. y cos x, y
3 x,y 0, x
2 3 x ; about y ■ 0; about x
■ ■ 2 ■ ■ ■ ■ ■ ■ 2, x
1 5 2; about the yaxis be the region in the ﬁrst quadrant bounded by the curves
x 3 and y 2 x x 2. Calculate the following quantities.
The area of
The volume obtained by rotating about the xaxis
The volume obtained by rotating about the yaxis be the region bounded by the curves y tan x 2 , x 1,
and y 0. Use the Midpoint Rule with n 4 to estimate the
following.
(a) The area of
(b) The volume obtained by rotating about the xaxis 3; about the yaxis 2 9 8, x 17. Let x 2; about the xaxis
2 x, y 8. x y
(a)
(b)
(c) ■ 7–11  Find the volume of the solid obtained by rotating the
region bounded by the given curves about the speciﬁed axis. 7. y ■ 16. Let 2 ■ ■ bounded by the curves y x and y x 2 about the following
lines:
(a) The xaxis
(b) The yaxis
(c) y 2 sx y2 x 3, y 15. Find the volumes of the solids obtained by rotating the region 12 1 x ■ ■ x2 x 2, sx,
■ 6, ■ 14. y Find the area of the region bounded by the given curves.  ■ EXERCISES 19–22 be the region bounded by the curves y 1 x 2 and
x 1. Estimate the following quantities.
x
The xcoordinates of the points of intersection of the curves
The area of
The volume generated when is rotated about the xaxis
The volume generated when is rotated about the yaxis
6 Each integral represents the volume of a solid. Describe  the solid.
19. y 0 2 2 x cos x dx 20. y 0 2 2 cos2x dx 5E06(pp 404411) 1/17/06 4:01 PM Page 407 C HAPTER 6 REVIEW 21.
22.
■ y 2 2 y4 0
1 y [2
■ ■ 2 ■ 407 29. A tank full of water has the shape of a paraboloid of revolution y 2 dy
x2 0 ❙❙❙❙ (2
■ s x )2 ] dx
■ ■ ■ ■ ■ ■ ■ 23. The base of a solid is a circular disk with radius 3. Find the ; as shown in the ﬁgure; that is, its shape is obtained by rotating
a parabola about a vertical axis.
(a) If its height is 4 ft and the radius at the top is 4 ft, ﬁnd the
work required to pump the water out of the tank.
(b) After 4000 ftlb of work has been done, what is the depth
of the water remaining in the tank? volume of the solid if parallel crosssections perpendicular to
the base are isosceles right triangles with hypotenuse lying
along the base. 4 ft 24. The base of a solid is the region bounded by the parabolas 4 ft y x 2 and y 2 x 2. Find the volume of the solid if the
crosssections perpendicular to the xaxis are squares with one
side lying along the base.
25. The height of a monument is 20 m. A horizontal crosssection at a distance x meters from the top is an equilateral triangle
with side x 4 meters. Find the volume of the monument.
26. (a) The base of a solid is a square with vertices located at 1, 0 , 0, 1 , 1, 0 , and 0, 1 . Each crosssection perpendicular to the xaxis is a semicircle. Find the volume of
the solid.
(b) Show that by cutting the solid of part (a), we can rearrange
it to form a cone. Thus compute its volume more simply.
27. A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much
work is done in stretching the spring from 12 cm to 20 cm?
28. A 1600lb elevator is suspended by a 200ft cable that weighs 10 lb ft. How much work is required to raise the elevator from
the basement to the third ﬂoor, a distance of 30 ft? t sin t 2 on the 30. Find the average value of the function f t interval 0, 10 .
31. If f is a continuous function, what is the limit as h l 0 of the average value of f on the interval x, x
32. Let h?
2 x , y 0, and x b,
1 be the region bounded by y
where b 0. Let 2 be the region bounded by y x 2, x 0,
and y b 2.
(a) Is there a value of b such that 1 and 2 have the same
area?
(b) Is there a value of b such that 1 sweeps out the same
volume when rotated about the xaxis and the yaxis?
(c) Is there a value of b such that 1 and 2 sweep out the
same volume when rotated about the xaxis?
(d) Is there a value of b such that 1 and 2 sweep out the
same volume when rotated about the yaxis? 5E06(pp 404411) 1/17/06 4:02 PM PROBLEMS
PLUS Page 408 1. (a) Find a positive continuous function f such that the area under the graph of f from 0 to t is At
t 3 for all t 0.
(b) A solid is generated by rotating about the xaxis the region under the curve y f x ,
where f is a positive function and x 0. The volume generated by the part of the curve
from x 0 to x b is b 2 for all b 0. Find the function f .
2. There is a line through the origin that divides the region bounded by the parabola y x x2 and the xaxis into two regions with equal area. What is the slope of that line?
3. The ﬁgure shows a horizontal line y c intersecting the curve y
ber c such that the areas of the shaded regions are equal. y y=8x27˛ 8x 27x 3. Find the num 4. A cylindrical glass of radius r and height L is ﬁlled with water and then tilted until the water
y=c x 0 FIGURE FOR PROBLEM 3 remaining in the glass exactly covers its base.
(a) Determine a way to “slice” the water into parallel rectangular crosssections and then
set up a deﬁnite integral for the volume of the water in the glass.
(b) Determine a way to “slice” the water into parallel crosssections that are trapezoids and
then set up a deﬁnite integral for the volume of the water.
(c) Find the volume of water in the glass by evaluating one of the integrals in part (a) or
part (b).
(d) Find the volume of the water in the glass from purely geometric considerations.
(e) Suppose the glass is tilted until the water exactly covers half the base. In what direction
can you “slice” the water into triangular crosssections? Rectangular crosssections?
Crosssections that are segments of circles? Find the volume of water in the glass. L L r r 5. (a) Show that the volume of a segment of height h of a sphere of radius r is V r 1
3 h 2 3r h (b) Show that if a sphere of radius 1 is sliced by a plane at a distance x from the center in
such a way that the volume of one segment is twice the volume of the other, then x is a
solution of the equation
h FIGURE FOR PROBLEM 5 3x 3 9x 2 0 where 0 x 1. Use Newton’s method to ﬁnd x accurate to four decimal places.
(c) Using the formula for the volume of a segment of a sphere, it can be shown that the depth
x to which a ﬂoating sphere of radius r sinks in water is a root of the equation
x3 3rx 2 4r 3s 0 where s is the speciﬁc gravity of the sphere. Suppose a wooden sphere of radius 0.5 m has
speciﬁc gravity 0.75. Calculate, to fourdecimalplace accuracy, the depth to which the
sphere will sink.
(d) A hemispherical bowl has radius 5 inches and water is running into the bowl at the rate
of 0.2 in3 s.
(i) How fast is the water level in the bowl rising at the instant the water is 3 inches deep?
(ii) At a certain instant, the water is 4 inches deep. How long will it take to ﬁll the bowl? 408 5E06(pp 404411) 1/17/06 4:02 PM Page 409 6. Archimedes’ Principle states that the buoyant force on an object partially or fully submerged in a ﬂuid is equal to the weight of the ﬂuid that the object displaces. Thus, for an object of
density 0 ﬂoating partly submerged in a ﬂuid of density f , the buoyant force is given by
0
F
f t x h A y d y, where t is the acceleration due to gravity and A y is the area of a typical crosssection of the object. The weight of the object is given by
W 0 ty Lh
h A y dy (a) Show that the percentage of the volume of the object above the surface of the liquid is
f 100 0
f (b) The density of ice is 917 kg m3 and the density of seawater is 1030 kg m3. What percentage of the volume of an iceberg is above water?
(c) An ice cube ﬂoats in a glass ﬁlled to the brim with water. Does the water overﬂow when
the ice melts?
(d) A sphere of radius 0.4 m and having negligible weight is ﬂoating in a large freshwater
lake. How much work is required to completely submerge the sphere? The density of the
water is 1000 kg m3.
y=Lh
y=0
L h
y=_h 7. Water in an open bowl evaporates at a rate proportional to the area of the surface of the water. (This means that the rate of decrease of the volume is proportional to the area of the surface.)
Show that the depth of the water decreases at a constant rate, regardless of the shape of the
bowl. y y=2≈ 8. A sphere of radius 1 overlaps a smaller sphere of radius r in such a way that their intersection C is a circle of radius r. (In other words, they intersect in a great circle of the small sphere.)
Find r so that the volume inside the small sphere and outside the large sphere is as large as
possible. y=≈
B P 9. The ﬁgure shows a curve C with the property that, for every point P on the middle curve A y 2 x 2, the areas A and B are equal. Find an equation for C. 10. A paper drinking cup ﬁlled with water has the shape of a cone with height h and semivertical
0 FIGURE FOR PROBLEM 9 x angle (see the ﬁgure). A ball is placed carefully in the cup, thereby displacing some of the
water and making it overﬂow. What is the radius of the ball that causes the greatest volume of
water to spill out of the cup? h
¨ 409 5E06(pp 404411) 1/17/06 4:02 PM Page 410 11. A clepsydra, or water clock, is a glass container with a small hole in the bottom through which water can ﬂow. The “clock” is calibrated for measuring time by placing markings on
the container corresponding to water levels at equally spaced times. Let x f y be continuous on the interval 0, b and assume that the container is formed by rotating the graph of f
about the yaxis. Let V denote the volume of water and h the height of the water level at time t.
(a) Determine V as a function of h.
(b) Show that
dV
dt fh 2 dh
dt (c) Suppose that A is the area of the hole in the bottom of the container. It follows from
Torricelli’s Law that the rate of change of the volume of the water is given by
dV
dt k A sh where k is a negative constant. Determine a formula for the function f such that dh dt is a
constant C. What is the advantage in having dh dt C ?
y
b x=f(y)
h
x 12. A cylindrical container of radius r and height L is partially ﬁlled with a liquid whose volume y is V . If the container is rotated about its axis of symmetry with constant angular speed , then
the container will induce a rotational motion in the liquid around the same axis. Eventually,
the liquid will be rotating at the same angular speed as the container. The surface of the liquid
will be convex, as indicated in the ﬁgure, because the centrifugal force on the liquid particles
increases with the distance from the axis of the container. It can be shown that the surface of
the liquid is a paraboloid of revolution generated by rotating the parabola v L
h 22 r
F IGURE FOR PROBLEM 12 x y h x
2t about the yaxis, where t is the acceleration due to gravity.
(a) Determine h as a function of .
(b) At what angular speed will the surface of the liquid touch the bottom? At what speed will
it spill over the top?
(c) Suppose the radius of the container is 2 ft, the height is 7 ft, and the container and liquid
are rotating at the same constant angular speed. The surface of the liquid is 5 ft below the
top of the tank at the central axis and 4 ft below the top of the tank 1 ft out from the central axis.
(i) Determine the angular speed of the container and the volume of the ﬂuid.
(ii) How far below the top of the tank is the liquid at the wall of the container?
x 3 intersects the curve again at Q, let A be the
area of the region bounded by the curve and the line segment PQ. Let B be the area of the
region deﬁned in the same way starting with Q instead of P. What is the relationship between
A and B ? 13. If the tangent at a point P on the curve y 410 5E06(pp 404411) 1/17/06 4:02 PM Page 411 CAS 14. Suppose we are planning to make a taco from a round tortilla with diameter 8 inches by bend ing the tortilla so that it is shaped as if it is partially wrapped around a circular cylinder. We
will ﬁll the tortilla to the edge (but no more) with meat, cheese, and other ingredients. Our
problem is to decide how to curve the tortilla in order to maximize the volume of food it can
hold.
(a) We start by placing a circular cylinder of radius r along a diameter of the tortilla and
folding the tortilla around the cylinder. Let x represent the distance from the center of the
tortilla to a point P on the diameter (see the ﬁgure). Show that the crosssectional area of
the ﬁlled taco in the plane through P perpendicular to the axis of the cylinder is
Ax r s16 x2 1
2 r 2 sin 2
s16
r x2 and write an expression for the volume of the ﬁlled taco.
(b) Determine (approximately) the value of r that maximizes the volume of the taco. (Use a
graphical approach with your CAS.) x
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