Volumes by Cylindrical Shells - Stewart Calculus

Volumes by Cylindrical Shells

y Some volume problems are very difficult to handle by the methods of Section 6.2. For
y=2≈-˛ instance, let’s consider the problem of finding the volume of the solid obtained by rotating
about the y-axis the region bounded by y ෇ 2x 2 Ϫ x 3 and y ෇ 0. (See Figure 1.) If we slice
1 xR=? perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer
xL=? radius of the washer, we would have to solve the cubic equation y ෇ 2x 2 Ϫ 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:

0 2x V ෇ V2 Ϫ V1
FIGURE 1
෇ ␲ r 2 h Ϫ ␲r 2 h ෇ ␲ ͑r 2 Ϫ r 21͒h
r 2 1 2

෇ ␲ ͑r2 ϩ r1͒͑r2 Ϫ r1͒h

Îr ෇ 2␲ r2 ϩ r1 h͑r2 Ϫ r1͒
2
r¡
r™

If we let ⌬r ෇ r2 Ϫ r1 (the thickness of the shell) and r ෇ 1 ͑r2 ϩ r1͒ (the average radius
2

h of the shell), then this formula for the volume of a cylindrical shell becomes

1 V ෇ 2␲rh ⌬r

FIGURE 2 and it can be remembered as

V ෇ [circumference][height][thickness]

Now let S be the solid obtained by rotating about the y-axis the region bounded by
y ෇ f ͑x͒ [where f ͑x͒ ജ 0], y ෇ 0, x ෇ a, and x ෇ b, where b Ͼ a ജ 0. (See Figure 3.)

y y
y=ƒ y=ƒ

0a bx 0a bx

FIGURE 3

We divide the interval ͓a, b͔ into n subintervals ͓xiϪ1, xi ͔ of equal width ⌬x and let xi be
the midpoint of the ith subinterval. If the rectangle with base ͓xiϪ1, xi ͔ and height f ͑ xi ͒ is
rotated about the y-axis, 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

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:

nn

͚ ͚V Ϸ Vi ෇ 2␲xi f ͑xi ͒ ⌬x
i෇1 i෇1

1

2 ■ VOLUMES BY CYLINDRICAL SHELLS

y This approximation appears to become better as n l ϱ. But, from the definition of an inte-
y=ƒ gral, we know that

n
͚ ylim 2␲xi f ͑ xi ͒ ⌬x ෇ b 2␲x f ͑x͒ dx
n l ϱ i෇1 a

0a Thus, the following appears plausible:

bx

xi-1 x–i xi 2 The volume of the solid in Figure 3, obtained by rotating about the y-axis the
y region under the curve y ෇ f ͑x͒ from a to b, is

y=ƒ V ෇ yb 2␲x f ͑x͒ dx where 0 ഛ a Ͻ b
a

0a b x The argument using cylindrical shells makes Formula 2 seem reasonable, but later we
FIGURE 4 will be able to prove it. (See Exercise 47.)
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 ͑x͒, and thickness ⌬x or dx:

yb ͑2␲x͒ ͓ f ͑x͔͒ dx
a
circumference height

y

x ƒ ƒ
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 y-axis.

y EXAMPLE 1 Find the volume of the solid obtained by rotating about the y-axis the region
bounded by y ෇ 2x 2 Ϫ x 3 and y ෇ 0.
x 2≈-˛
x SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumfer-
ence 2␲x, and height f ͑x͒ ෇ 2x 2 Ϫ x 3. So, by the shell method, the volume is
FIGURE 6
y yV ෇ 2 ͑2␲x͒͑2x 2 Ϫ x 3 ͒ dx ෇ 2␲ 2 ͑2x 3 Ϫ x 4 ͒ dx
2x 0 0

[ ] ( )෇ 2␲1 4 1 x5 2 32 16
2 x Ϫ 5 0 ෇ 2␲ 8 Ϫ 5 ෇ 5 ␲

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

y

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

FIGURE 7

VOLUMES BY CYLINDRICAL SHELLS ■ 3

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 y-axis the region
y=≈ between y ෇ x and y ෇ x 2.
0
FIGURE 8 shell SOLUTION The region and a typical shell are shown in Figure 8. We see that the shell has
height=x-≈ radius x, circumference 2␲x, and height x Ϫ x 2. So the volume is

xx y yV ෇ 1 ͑2␲x͒͑x Ϫ x 2 ͒ dx ෇ 2␲ 1 ͑x 2 Ϫ x 3 ͒ dx

00

ͫ ͬ෇ 2␲ 1␲
x3 x4 ෇
Ϫ
3 40 6

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.

y EXAMPLE 3 Use cylindrical shells to find the volume of the solid obtained by rotating
shell height=1-¥ about the x-axis the region under the curve y ෇ sx from 0 to 1.

1 SOLUTION This problem was solved using disks in Example 2 in Section 6.2. To use shells
we relabel the curve y ෇ sx (in the figure in that example) as x ෇ y 2 in Figure 9. For
rotation about the x-axis we see that a typical shell has radius y, circumference 2␲ y, and
height 1 Ϫ y 2. So the volume is

y x=1 y yV ෇ 1 ͑2␲ y͒͑1 Ϫ y 2 ͒ dy ෇ 2␲ 1 ͑ y Ϫ y 3 ͒ dy
x=¥ shell 00
radius=y
ͫ ͬ෇ 2␲
0 1x y2 y4 1 ␲
Ϫ෇
2 40 2

FIGURE 9 In this problem the disk method was simpler.

EXAMPLE 4 Find the volume of the solid obtained by rotating the region bounded by
y ෇ x Ϫ x 2 and y ෇ 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 Ϫ x, circumference 2␲ ͑2 Ϫ x͒, and height x Ϫ x 2.

y y
y=x-≈
x=2

0 x 0 123 4x
x 2-x
FIGURE 10

The volume of the given solid is

y yV ෇ 1 2␲ ͑2 Ϫ x͒͑x Ϫ x 2 ͒ dx ෇ 2␲ 1 ͑x 3 Ϫ 3x 2 ϩ 2x͒ dx
00

ͫ ͬ෇ 2␲ 1␲
x4 Ϫ x3 ϩ x2 ෇
4 02

4 ■ VOLUMES BY CYLINDRICAL SHELLS

Exercises

A Click here for answers. S Click here for solutions. 15–20 Use the method of cylindrical shells to find the volume
generated by rotating the region bounded by the given curves about
1. Let S be the solid obtained by rotating the region shown in the specified axis. Sketch the region and a typical shell.
the figure about the y-axis. Explain why it is awkward to use 15. y ෇ x 2, y ෇ 0, x ෇ 1, x ෇ 2; about x ෇ 1
slicing to find the volume V of S. Sketch a typical approxi- 16. y ෇ x 2, y ෇ 0, x ෇ Ϫ2, x ෇ Ϫ1; about the y-axis
mating shell. What are its circumference and height? Use shells 17. y ෇ x 2, y ෇ 0, x ෇ 1, x ෇ 2; about x ෇ 4
to find V. 18. y ෇ 4x Ϫ x 2, y ෇ 8x Ϫ 2x 2; about x ෇ Ϫ2
19. y ෇ sx Ϫ 1, y ෇ 0, x ෇ 5; about y ෇ 3
y 20. y ෇ x 2, x ෇ y 2; about y ෇ Ϫ1

y=x(x-1)@ ■■■■■■■■■■■■■

0 1x 21–26 Set up, but do not evaluate, an integral for the volume
of the solid obtained by rotating the region bounded by the given
2. Let S be the solid obtained by rotating the region shown in the curves about the specified axis.
figure about the y-axis. Sketch a typical cylindrical shell and 21. y ෇ ln x, y ෇ 0, x ෇ 2; about the y-axis
find its circumference and height. Use shells to find the volume 22. y ෇ x, y ෇ 4x Ϫ x 2; about x ෇ 7
of S. Do you think this method is preferable to slicing? Explain. 23. y ෇ x 4, y ෇ sin͑␲x͞2͒; about x ෇ Ϫ1
24. y ෇ 1͑͞1 ϩ x 2 ͒, y ෇ 0, x ෇ 0, x ෇ 2; about x ෇ 2
y 25. x ෇ ssin y, 0 ഛ y ഛ ␲, x ෇ 0; about y ෇ 4
26. x 2 Ϫ y 2 ෇ 7, x ෇ 4; about y ෇ 5
y=sin{ ≈}
■■■■■■■■■■■■■
0 œπ„ x
27. Use the Midpoint Rule with n ෇ 4 to estimate the volume
3–7 Use the method of cylindrical shells to find the volume gen- obtained by rotating about the y-axis the region under the curve
erated by rotating the region bounded by the given curves about the y ෇ tan x, 0 ഛ x ഛ ␲͞4 .
y-axis. Sketch the region and a typical shell.
28. If the region shown in the figure is rotated about the y-axis to
3. y ෇ 1͞x, y ෇ 0, x ෇ 1, x ෇ 2 form a solid, use the Midpoint Rule with n ෇ 5 to estimate the
4. y ෇ x 2, y ෇ 0, x ෇ 1 volume of the solid.
5. y ෇ eϪx2, y ෇ 0, x ෇ 0, x ෇ 1
6. y ෇ 3 ϩ 2x Ϫ x 2, x ϩ y ෇ 3 y
7. y ෇ 4͑x Ϫ 2͒2, y ෇ x 2 Ϫ 4x ϩ 7 5
4
■■■■■■■■■■■■■ 3
2
8. Let V be the volume of the solid obtained by rotating about the 1
y-axis the region bounded by y ෇ sx and y ෇ x 2. Find V both
by slicing and by cylindrical shells. In both cases draw a dia- 0 1 2 3 4 5 6 7 8 9 10 11 12 x
gram to explain your method.
29–32 Each integral represents the volume of a solid. Describe
9–14 Use the method of cylindrical shells to find the volume of the solid.
the solid obtained by rotating the region bounded by the given
curves about the x-axis. Sketch the region and a typical shell. y29. 3 2␲x 5 dx
9. x ෇ 1 ϩ y 2, x ෇ 0, y ෇ 1, y ෇ 2 0
10. x ෇ sy, x ෇ 0, y ෇ 1
11. y ෇ x 3, y ෇ 8, x ෇ 0 2␲ 2y
12. x ෇ 4y 2 Ϫ y 3, x ෇ 0 y30.0 1 ϩ y2 dy
13. y ෇ 4x 2, 2x ϩ y ෇ 6
14. x ϩ y ෇ 3, x ෇ 4 Ϫ ͑ y Ϫ 1͒2 y31. 1 2␲ ͑3 Ϫ y͒͑1 Ϫ y2͒ dy
0
■■■■■■■■■■■■■
y32. ␲͞4 2␲ ͑␲ Ϫ x͒͑cos x Ϫ sin x͒ dx
0

■■■■■■■■■■■■■

VOLUMES BY CYLINDRICAL SHELLS ■ 5

; 33–34 Use a graph to estimate the x-coordinates of the points of 45. A right circular cone with height h and base radius r
intersection of the given curves. Then use this information to esti-
mate the volume of the solid obtained by rotating about the y-axis ■■■■■■■■■■■■■
the region enclosed by these curves.
33. y ෇ 0, y ෇ x ϩ x 2 Ϫ x 4 46. Suppose you make napkin rings by drilling holes with different
34. y ෇ x 4, y ෇ 3x Ϫ x 3 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.
CAS 35–36 Use a computer algebra system to find the exact volume (b) Check your guess: Use cylindrical shells to compute the
of the solid obtained by rotating the region bounded by the given volume of a napkin ring created by drilling a hole with
curves about the specified line. radius r through the center of a sphere of radius R and
35. y ෇ sin2 x, y ෇ sin4 x, 0 ഛ x ഛ ␲ ; about x ෇ ␲͞2 express the answer in terms of h.
36. y ෇ x 3 sin x, y ෇ 0, 0 ഛ x ഛ ␲ ; about x ෇ Ϫ1
h
■■■■■■■■■■■■■
47. We arrived at Formula 2, V ෇ xb 2␲x f ͑x͒ dx, by using
37–42 The region bounded by the given curves is rotated about a
the specified axis. Find the volume of the resulting solid by any
method. cylindrical shells, but now we can use integration by parts to
37. y ෇ x 2 ϩ x Ϫ 2, y ෇ 0; about the x-axis
38. y ෇ x 2 Ϫ 3x ϩ 2, y ෇ 0; about the y-axis prove it using the slicing method of Section 6.2, at least for the
39. y ෇ 5, y ෇ x ϩ ͑4͞x͒; about x ෇ Ϫ1
40. x ෇ 1 Ϫ y 4, x ෇ 0; about x ෇ 2 case where f is one-to-one and therefore has an inverse func-
41. x 2 ϩ ͑ y Ϫ 1͒2 ෇ 1; about the y-axis
42. x 2 ϩ ͑ y Ϫ 1͒2 ෇ 1; about the x-axis tion t. Use the figure to show that

■■■■■■■■■■■■■ yV ෇ ␲b 2d Ϫ ␲a 2c Ϫ d ␲ ͓ t͑ y͔͒2 dy
c
43–45 Use cylindrical shells to find the volume of the solid.
43. A sphere of radius r Make the substitution y ෇ f ͑x͒ and then use integration by
44. The solid torus (a donut-shaped solid with radii R and r)
parts on the resulting integral to prove that V ෇ xb 2␲ x f ͑x͒ dx.
shown in the figure a

R y y=ƒ
r x=g( y)

d

c x=b
x=a bx

0a

6 ■ VOLUMES BY CYLINDRICAL SHELLS 9. 21␲͞2 y
1+¥
Answers x
2 x
S Click here for solutions.
1. Circumference ෇ 2␲x, height ෇ x ͑x Ϫ 1͒2; ␲͞15 1

y 01
x(x-1)@ y

0 1x y
y 0

1

3. 2␲ x 1x 11. 768␲͞7 13. 250␲͞3
y
1 15. 17␲͞6 17. 67␲͞6 19. 24␲
0 x
y 21. x2 2␲x ln x dx
x 1

23. x1 2␲͑x ϩ 1͓͒sin͑␲x͞2͒ Ϫ x 4 ͔ dx
0

25. x␲ 2␲ ͑4 Ϫ y͒ ssin y dy 27. 1.142
0

29. Solid obtained by rotating the region 0 ഛ y ഛ x 4, 0 ഛ x ഛ 3

about the y-axis

31. Solid obtained by rotating the region bounded by
(i) x ෇ 1 Ϫ y 2, x ෇ 0, and y ෇ 0, or (ii) x ෇ y 2, x ෇ 1, and y ෇ 0

about the line y ෇ 3

33. 0, 1.32; 4.05 35. 1 ␲ 3 37. 81␲͞10
32

x 39. 8␲͑3 Ϫ ln 4͒ 41. 4␲͞3 43. 4 ␲r 3 45. 1 ␲r 2h
3 3
x

5. ␲ ͑1 Ϫ 1͞e͒ 7. 16␲

VOLUMES BY CYLINDRICAL SHELLS ■ 7

Solutions: Volumes by Cylindrical Shells

1. If we were to use the “washer” method, we would first have

to locate the local maximum point (a, b) of y = x(x − 1)2

using the methods of Chapter 4. Then we would have to

solve the equation y = x(x − 1)2 for x in terms of y to

obtain the functions x = g1(y) and x = g2(y) shown in the
first figure. This step would be difficult because it involves

the cubic formula. Finally we would find the volume using

V = π Rb © (y)]2 − [g2(y)]2ª dy.
[g1
0

Using shells, we find that a typical approximating shell has radius x, so its circumference is 2πx. Its height is y, that

is, x(x − 1)2. So the total volume is

V = R1 £ − 1)2¤ dx = 2π R1 ¡x4 − 2x3 + x2¢ dx = · x5 x4 + x3 ¸1 = π
2πx x(x 2π 5 −2 30 15
0 0
4

3. V = Z2 2πx · 1 dx = 2π Z2 1 dx
x
1 1

= 2π [x]12 = 2π(2 − 1) = 2π

5. V = R1 2πxe−x2 dx. Let u = x2.

0

Thus, du = 2x dx, so

V = π R1 e−u du = π£−e−u ¤1

0 0

= π(1 − 1/e)

8 ■ VOLUMES BY CYLINDRICAL SHELLS

7. The curves intersect when 4(x − 2)2 = x2 − 4x + 7 ⇔ 4x2 − 16x + 16 = x2 − 4x + 7 ⇔

3x2 − 12x + 9 = 0 ⇔ 3(x2 − 4x + 3) = 0 ⇔ 3(x − 1)(x − 3) = 0, so x = 1 or 3.

V = 2π R3 ©x£¡x2 − 4x + ¢ − 4(x − 2)2¤ª dx = 2π R3 £x(x2 − 4x + 7 − 4x2 + 16x − ¤ dx
7 16)
1 1

= 2π R3 £x(−3x2 + 12x − ¤ = 2π(−3) R13(x3 − 4x2 + 3x) dx = £ 1 x4 − 4 x3 + 3 x2 ¤3
9) dx −6π 4 3 2
1 1

£¡ ¢¡ ¢¤ ¡ ¢ ¡¢
= −6π 81 − 36 + 27 − 1 − 4 + 3 = −6π 20 − 36 + 12 + 4 = −6π − 8 = 16π
4 2 4 3 2 3 3

9. V = R21π2£2(π2y+¡14+) −y2¡¢21d+y =14 ¢2¤π=R122π¡¡y + y3¢ dy = £ 1 y2 + 1 y4 ¤2
= 2π 2 4
21π 1
21 ¢ 2
=
4

11. V Z 8√
= 2π [y( 3 y − 0)] dy

0

= 2π Z8 y4/3 dy = h 3 y7/3 i8
2π 7
0 0

= 6π (87/3) = 6π (27) = 768π
7 77

VOLUMES BY CYLINDRICAL SHELLS ■ 9

13. The curves intersect when 4x2 = 6 − 2x ⇔ 2x2 + x − 3 = 0 ⇔ (2x + 3)(x − 1) = 0 ⇔ x = − 3 or 1.
2

Solving the equations for x gives us y = 4x2 ⇒Z x = ± 1 √y and 2x + y = 6 ⇒x = − 1 y + 3.
Z 2π 2 ¢¤ª 2
= 2π
4 © £¡ √y ¢ ¡ √y ¢¤ª 9 © £¡ ¢ ¡ √y dy
y − y− 3 −
V 1 − 1 dy + 1 y + − 1
2 2 2 2
Z04 √ 4
2π (y y ) dy Z
+ 2π 9³ 1 y2 1 y3/2 ´h 2 y5/2 i4 h 1 y3 3 y2 1 y5/2 i9
2 2 5 6 2 5
= − + 3y + dy = 2π 0 + 2π − + + 4

= 21π528¡π250+· 322π¢¡+413523π¢£=¡−1212245530 4 ¢ ¡ 32 ¢¤
= −
+ 243 + 243 − 32 + 24 + 5
2 5 3

π = 250 π
3

15. V = R2 2π(x − 1)x2 dx = £ 1 x4 − 1 x3 ¤2
2π 4 3
1 1

= £¡ 8 ¢ − ¡ 1 − 1 ¢¤ = 17 π
2π 4 − 3 4 6
3

17. V = R2 2π(4 − x)x2 dx = £ 4 x3 − 1 x4 ¤2
2π 3 4
1 1

= 2π £¡ 32 ¢ − ¡4 − 1 ¢¤ = 67 π
3 −4 6
3 4

19. V = R2 2π(3 − y)(5 − x)dy

0

= R2 2π(3 − ¡ − y2 − ¢ dy
y) 5 1
0

= R2 ¡ − 4y − 3y2 + y3¢ dy
2π 12
0

= £ − 2y2 − y3 + 1 y4¤2
2π 12y 4
0

= 2π(24 − 8 − 8 + 4) = 24π

10 ■ VOLUMES BY CYLINDRICAL SHELLS

21. V = R2 2πx ln x dx 23. V = R1 2π[x − ¡ π x − x4¢ dx
(−1)] sin 2
1 0

25. V = Rπ 2π(4 − y) √ y dy
sin
0

27. ∆x = π/4 − 0 = π .
4 16

V = R π/4 2πx tan x dx ≈ 2π · π¡π tan π + 3π tan 3π + 5π tan 5π + 7π tan 7π ¢ ≈ 1.142
32 32 32 32 32 32 32
0 16 32

29. R3 2πx5 dx = 2π R3 x(x4) dx. The solid is obtained by rotating the region 0 ≤ y ≤ x4, 0 ≤ x ≤ 3 about the

0 0

y-axis using cylindrical shells.

31. R1 2π(3 − y)(1 − y2) dy. The solid is obtained by rotating the region bounded by (i) x = 1 − y2, x = 0, and

0

y = 0 or (ii) x = y2, x = 1, and y = 0 about the line y = 3 using cylindrical shells.

33. From the graph, the curves intersect at x = 0 and at x = a ≈ 1.32, with
x + x2 − x4 > 0 on the interval (0, a). So the volume of the solid
obtained by rotating the region about the y-axis is

Z a £ x4)¤ Z a
= 2π x(x
V + x2 − dx = 2π (x2 + x3 − x5) dx

00

= £ 1 x3 + 1 x4 − 1 x6¤a ≈ 4.05
2π 3 4 6
0

Z π/2 £¡ π x¢¡sin2 sin4 ¢¤
= 2π x dx
35. V 2 − x −

0

C=AS 1 π3
32

VOLUMES BY CYLINDRICAL SHELLS ■ 11

37. Use disks:

V = R1 π¡x2 + x − 2¢2 dx = π R1 ¡x4 + 2x3 − 3x2 ¢
− 4x + 4 dx
−2 −2

= π £ 1 x5 + 1 x4 − x3 − 2x2 + 4x¤−1 2 = £¡ 1 + 1 − 1 − 2 + ¢ − ¡ 32 + 8 + 8 − 8 − ¢¤
5 2 π 5 2 4 − 5 8

= ¡ 33 + 3¢ = 81 π
π 5 10
2

39. Use shells:

V = R4 2π[x − (−1)][5 − (x + 4/x)] dx

1

= 2π R4 (x + 1)(5 − x − 4/x) dx

1

= 2π R4 ¡ − x2 − 4 + 5 − x − ¢ dx
5x 4/x
1

= 2π R4 ¡−x2 + 4x + 1 − ¢ dx = £ 1 x3 + 2x2 + x − 4 ln x¤14
4/x 2π − 3
1

= £¡ 64 + 32 + 4 − 4 ¢ − ¡ 1 + 2 + 1 − ¢¤
2π − 3 ln 4 − 3 0

= 2π(12 − 4 ln 4) = 8π(3 − ln 4)

Z 2 ·q ¸2 Z 2 ¡ y2¢ π£y2 − ¤2 µ 8 ¶ 4
π 2y π4 3 3
41. Use disks: V = π 1 − (y − 1)2 dy = − dy = 1 y3 0 = − = π
3
0 0

43. V = 2 Rr 2πx √ − x2 dx = −2π Rr ¡r2 − x2¢1/2(−2x) dx = h · 2 ¡r2 − x2 ¢3/2 ir
= r2 −2π 3
0 0 0

− 4 ¡ − r3 ¢ = 4 π r3
3 π0 3

Z r µ hx ¶ Z r µ x2 ¶ · x3 x2 ¸r = 2πh r2 πr2h
= 2π x− h − x 2πh −
45. V 0r + dx = 2πh 0r + dx = + =
3r 2 0 63

47. Using the formula for volumes of rotation and the figure, we see that

Volume = Rd πb2 dy − Rc πa2 dy − Rd π[g(y)]2 dy = πb2d − πa2c − Rd π[g(y)]2 dy. Let y = f (x), which

0 0 c Rb c

gives dy = f 0(x) dx and g(y) = x, so that V = πb2d − πa2c − π a x2f 0(x) dx. Now integrate

by parts with u = x2, and dv = f 0(x) dx ⇒ du = 2x dx, v = f (x), and

Rb x2 f 0(x) dx = £x2 f (x)¤hab − Rb 2x f (x) dx = b2 f (b) − a2 f (a) − Rb 2x f (x) dx, but f (a) = c and f (b) = d
V = πb2d − πa2c a2 c i
a a a
(x) dx
⇒ − π b2d − − Rb 2xf = Rb 2πxf (x) dx.

a a