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Some volume problems are very difficult to handle by the methods of Section 6.2. For
instance, let’s consider the problem of finding the volume of the solid obtained by rotating
about the -axis the region bounded by and. (See Figure 1.) If we slice
perpendicular to the y -axis, we get a washer. But to compute the inner radius and the outer
radius of the washer, we would have to solve the cubic equation for x in
terms of y ; that’s not easy.
Fortunately, there is a method, called the method of cylindrical shells , that is easier to
use in such a case. Figure 2 shows a cylindrical shell with inner radius , outer radius ,
and height. Its volume is calculated by subtracting the volume of the inner cylinder
from the volume of the outer cylinder:
If we let (the thickness of the shell) and (the average radius
of the shell), then this formula for the volume of a cylindrical shell becomes
and it can be remembered as
Now let be the solid obtained by rotating about the -axis the region bounded by
[where ], and , where. (See Figure 3.)
We divide the interval into n subintervals of equal width and let be
the midpoint of the i th subinterval. If the rectangle with base and height is
rotated about the y -axis, then the result is a cylindrical shell with average radius , height
, and thickness (see Figure 4), so by Formula 1 its volume is
Therefore, an approximation to the volume of is given by the sum of the volumes of
these shells:
V (^)
n
i 1
Vi (^)
n
i 1
2 xi f xi x
V S
Vi 2 xi f xi x
f xi x
xi
x i 1, x i f xi
a , b x i 1, x i x xi
FIGURE 3
a b x
y
0
y=ƒ
x
y
(^0) a b
y=ƒ
y f x f x 0 y 0, x a , x b b a 0
S y
V [circumference][height][thickness]
1 V 2 rh r
r r 2 r 1 r 12 r 2 r 1
r 2 r 1
h r 2 r 1
r 2 r 1 r 2 r 1 h
r^22 h r^21 h r^22 r^21 h
V V 2 V 1
V 2
h V V 1
r 1 r 2
y 2 x^2 x^3
y y 2 x^2 x^3 y 0
FIGURE 1
FIGURE 2
r r¡ r™
Îr
h
y
0 2 x
1
y=2≈-˛
xL =? x (^) R =?
1
Volumes by Cylindrical Shells
This approximation appears to become better as. But, from the definition of an inte-
gral, we know that
Thus, the following appears plausible:
The volume of the solid in Figure 3, obtained by rotating about the y -axis the
region under the curve from a to b , is
The argument using cylindrical shells makes Formula 2 seem reasonable, but later we
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 , height , and thickness or :
This type of reasoning will be helpful in other situations, such as when we rotate about
lines other than the y -axis.
EXAMPLE 1 Find the volume of the solid obtained by rotating about the -axis the region
bounded by and.
SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x , circumfer-
ence , and height. So, by the shell method, the volume is
It can be verified that the shell method gives the same answer as slicing.
FIGURE 7
y
x
2 [ 12 x^4 15 x^5 ] 0
2 2 (8 325 ) 165
V (^) y
2 0
2 x 2 x^2 x^3 dx 2 (^) y
2 0
2 x^3 x^4 dx
2 x f x 2 x^2 x^3
y 2 x^2 x^3 y 0
y
FIGURE 5
2πx Îx
ƒ
y
x x
x ƒ
circumference height
y ^2 x ^ f^ x ^ dx
b a
2 x f x x dx
V (^) y where 0 a b
b a
2 x f x dx
y f x
lim
n l
n
i 1
2 xi f xi x (^) y
b a
2 xf x dx
n l
2 ■ VOLUMES BY CYLINDRICAL SHELLS
FIGURE 4
x
y
(^0) a (^) b
y=ƒ
x– (^) i
(^0) a b x
y
x (^) i-1 x^ i
y=ƒ
FIGURE 6
y
x
x
x 2
■ ■ (^) Figure 7 shows a computer-generated picture of the solid whose volume we computed in Example 1.
4 ■ VOLUMES BY CYLINDRICAL SHELLS
Exercises
1. Let be the solid obtained by rotating the region shown in the figure about the -axis. Explain why it is awkward to use slicing to find the volume of. Sketch a typical approxi- mating shell. What are its circumference and height? Use shells to find. 2. Let be the solid obtained by rotating the region shown in the figure about the -axis. Sketch a typical cylindrical shell and find its circumference and height. Use shells to find the volume of. Do you think this method is preferable to slicing? Explain.
3–7 Use the method of cylindrical shells to find the volume gen- erated by rotating the region bounded by the given curves about the -axis. Sketch the region and a typical shell.
3. , , , 4. , , 5. 6. , 7. , ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 8. Let be the volume of the solid obtained by rotating about the -axis the region bounded by and. Find both by slicing and by cylindrical shells. In both cases draw a dia- gram to explain your method.
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 -axis. Sketch the region and a typical shell.
9.
10.
11. , , 12. , 13. ,
14. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
x y 3, x 4 y 1 ^2
y 4 x^22 x y 6
x 4 y^2 y^3 x 0
y x^3 y 8 x 0
x s y , x 0, y 1
x 1 y^2 , x 0, y 1, y 2
x
y y s x y x^2 V
V
y 4 x 2 ^2 y x^2 4 x 7
y 3 2 x x^2 x y 3
y e x
2 , y 0, x 0, x 1
y x^2 y 0 x 1
y 1 x y 0 x 1 x 2
y
0 x
y
œ„π
y=sin{≈}
S
y
S
0 x
y
1
y=x(x-1)@
V
V S
y
S
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. , ; about 16. , ; about the -axis 17. , ; about 18. , ; about 19. , ; about 20. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
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 curves about the specified axis. 21.
22. , ; about 23. , ; about 24. about 25. about 26. about ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 27. Use the Midpoint Rule with to estimate the volume obtained by rotating about the -axis the region under the curve ,. 28. If the region shown in the figure is rotated about the -axis to form a solid, use the Midpoint Rule with to estimate the volume of the solid.
29–32 Each integral represents the volume of a solid. Describe the solid.
29.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
y
4 0
2 x cos x sin x dx
y
1 02 ^ ^3 ^ y ^1 ^ y
(^2) dy
2 (^) y
2 0
y 1 y^2
dy
y
3 0
2 x^5 dx
0 x
y
1
1
2
3
4
5
2 3 4 5 6 7 8 9 10 11 12
n 5
y
y tan x 0 x 4
y
n 4
x^2 y^2 7, x 4; y 5
x ssin y , 0 y , x 0; y 4
y 1 1 x^2 , y 0, x 0, x 2; x 2
y x^4 y sin x 2 x 1
y x y 4 x x^2 x 7
y ln x , y 0, x 2; about the y -axis
y x^2 , x y^2 ; about y 1
y s x 1 y 0, x 5 y 3
y 4 x x^2 y 8 x 2 x^2 x 2
y x^2 y 0, x 1, x 2 x 4
y x^2 y 0, x 2, x 1 y
y x^2 y 0, x 1, x 2 x 1
A Click here for answers. S Click here for solutions.
VOLUMES BY CYLINDRICAL SHELLS ■ 5
; 33–34^ Use a graph to estimate the^ -coordinates of the points of
intersection of the given curves. Then use this information to esti- mate the volume of the solid obtained by rotating about the -axis the region enclosed by these curves.
33. , 34. , ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 35–36 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. 35. , , ; about 36. , , ; about ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 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. 37. , ; about the -axis 38. , ; about the -axis 39. , ; about 40. , ; about 41. ; about the -axis 42. ; about the -axis ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 43–45 Use cylindrical shells to find the volume of the solid. 43. A sphere of radius 44. The solid torus (a donut-shaped solid with radii and ) shown in the figure
r
R
R r
r
x^2 y 1 ^2 1 x
x^2 y 1 ^2 1 y
x 1 y^4 x 0 x 2
y 5 y x 4 x x 1
y x^2 3 x 2 y 0 y
y x^2 x 2 y 0 x
y x^3 sin x y 0 0 x x 1
y sin^2 x y sin 4 x 0 x x 2
CAS
y x^4 y 3 x x^3
y 0 y x x^2 x^4
y
x 45. A right circular cone with height and base radius ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
46. Suppose you make napkin rings by drilling holes with different diameters through two wooden balls (which also have different diameters). You discover that both napkin rings have the same height , 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 through the center of a sphere of radius and express the answer in terms of. 47. We arrived at Formula 2, , by using cylindrical shells, but now we can use integration by parts to prove it using the slicing method of Section 6.2, at least for the case where is one-to-one and therefore has an inverse func- tion. Use the figure to show that
Make the substitution and then use integration by parts on the resulting integral to prove that.
y
0 a b x
c
d
x=a
x=b
x=g(y) y=ƒ
V x ab 2 x f x dx
y f x
V b^2 d a^2 c (^) y
d c
t y ^2 dy
t
f
V x ab 2 x f x dx
h
h
r R
h
h r
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 = g 1 (y) and x = g 2 (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
R (^) b 0
[g 1 (y)]^2 − [g 2 (y)]^2
dy.
Using shells, we find that a typical approximating shell has radius x, so its circumference is 2 πx. Its height is y, that
is, x(x − 1)^2. So the total volume is
V =
R 1
0 2 πx
x(x − 1)^2
dx = 2π
R 1
0
x^4 − 2 x^3 + x^2
dx = 2π
x^5 5
x^4 4
x^3 3
0
π 15
3. V =
Z 2
1
2 πx ·
x
dx = 2π
Z 2
1
1 dx
= 2π [x]^21 = 2π(2 − 1) = 2π
5. V =
R 1
0 2 πxe
−x^2 dx. Let u = x (^2).
Thus, du = 2x dx, so
V = π
R 1
0 e
−u (^) du = π£−e−u¤^1 0
= π(1 − 1 /e)
VOLUMES BY CYLINDRICAL SHELLS ■ 7
Solutions: Volumes by Cylindrical Shells
7. The curves intersect when 4(x − 2)^2 = x^2 − 4 x + 7 ⇔ 4 x^2 − 16 x + 16 = x^2 − 4 x + 7 ⇔
3 x^2 − 12 x + 9 = 0 ⇔ 3(x^2 − 4 x + 3) = 0 ⇔ 3(x − 1)(x − 3) = 0, so x = 1 or 3.
V = 2π
R 3
1
x
x^2 − 4 x + 7
− 4(x − 2)^2
dx = 2π
R 3
1
x(x^2 − 4 x + 7 − 4 x^2 + 16x − 16)
dx
= 2π
R 3
1
x(− 3 x^2 + 12x − 9)
dx = 2π(−3)
R 3
1 (x
(^3) − 4 x (^2) + 3x) dx = − 6 π£^1 4 x
3 x
2 x
2 ¤^3
1
= − 6 π
4 −^ 36 +^
27 2
4 −^
4 3 +^
3 2
= − 6 π
= − 6 π
= 16π
9. V =
R 2
1 2 πy
1 + y^2
dy = 2π
R 2
1
y + y^3
dy = 2π
2 y
4 y
4 ¤^2
1 = 2π
2 +^
1 4
= 2π
4
= 212 π
11. V = 2π
Z 8
0
[y( 3
y − 0)] dy
= 2π
Z 8
0
y^4 /^3 dy = 2π
h 3 7 y
7 / 3 i^8 0
=^6 π 7
(8^7 /^3 ) =^6 π 7
(2^7 ) =^768 π 7
8 ■ VOLUMES BY CYLINDRICAL SHELLS
21. V =
R 2
1 2 πx^ ln^ x dx^ 23.^ V^ =^
R 1
0 2 π[x^ −^ (−1)]
sin π 2 x − x^4
dx
25. V =
R (^) π 0 2 π(4^ −^ y)^
sin y dy
27. ∆x =
π/ 4 − 0 4
π 16
V =
R (^) π/ 4 0 2 πx^ tan^ x dx^ ≈^2 π^ ·^
π 16
¡ (^) π 32 tan^
π 32 +^
3 π 32 tan^
3 π 32 +^
5 π 32 tan^
5 π 32 +^
7 π 32 tan^
7 π 32
R 3
0 2 πx
(^5) dx = 2π R^3 0 x(x
(^4) ) dx. The solid is obtained by rotating the region 0 ≤ y ≤ x (^4) , 0 ≤ x ≤ 3 about the
y -axis using cylindrical shells.
R 1
0 2 π(3^ −^ y)(1^ −^ y
(^2) ) dy. The solid is obtained by rotating the region bounded by (i) x = 1 − y (^2) , x = 0, and
y = 0 or (ii) x = y^2 , 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 + x^2 − x^4 > 0 on the interval (0, a). So the volume of the solid obtained by rotating the region about the y-axis is
V = 2π
Z (^) a
0
x(x + x^2 − x^4 )
dx = 2π
Z (^) a
0
(x^2 + x^3 − x^5 ) dx
= 2π
3 x
4 x
6 x
6 ¤a 0 ≈^4.^05
35. V = 2π
Z (^) π/ 2
0
£¡ (^) π 2 −^ x
sin^2 x − sin^4 x
dx
CAS = 1
32 π
3
10 ■ VOLUMES BY CYLINDRICAL SHELLS
37. Use disks:
V =
R 1
− 2 π
x^2 + x − 2
dx = π
R 1
− 2
x^4 + 2x^3 − 3 x^2 − 4 x + 4
dx
= π
5 x
2 x
(^4) − x (^3) − 2 x (^2) + 4x¤^1 − 2 =^ π
5 +^
1 2 −^1 −^ 2 + 4
= π
5 +^
3 2
= 8110 π
39. Use shells:
V =
R 4
1 2 π[x^ −^ (−1)][5^ −^ (x^ + 4/x)]^ dx
= 2π
R 4
1 (x^ + 1)(5^ −^ x^ −^4 /x)^ dx
= 2π
R 4
1
5 x − x^2 − 4 + 5 − x − 4 /x
dx
= 2π
R 4
1
−x^2 + 4x + 1 − 4 /x
dx = 2π
− 13 x^3 + 2x^2 + x − 4 ln x
1 = 2π
− 643 + 32 + 4 − 4 ln 4
= 2π(12 − 4 ln 4) = 8π(3 − ln 4)
41. Use disks: V = π
Z 2
0
·q 1 − (y − 1)^2
dy = π
Z 2
0
2 y − y^2
dy = π
y^2 − 13 y^3
0 =^ π
μ 4 −
π
43. V = 2
R (^) r 0 2 πx^
r^2 − x^2 dx = − 2 π
R (^) r 0
r^2 − x^2
(− 2 x) dx =
h − 2 π · (^23)
r^2 − x^2
¢ 3 / 2 ir 0 = − 43 π
0 − r^3
= 43 πr^3
45. V = 2π
Z (^) r
0
x
μ −
h r
x + h
dx = 2πh
Z (^) r
0
μ −
x^2 r
dx = 2πh
x^3 3 r
x^2 2
¸r
0
= 2πh
r^2 6
πr^2 h 3
Volume =
R (^) d 0 πb
(^2) dy − R^ c 0 πa
(^2) dy − R^ d c π[g(y)]
(^2) dy = πb (^2) d − πa (^2) c − R^ d c π[g(y)]
(^2) dy. Let y = f (x), which
gives dy = f 0 (x) dx and g(y) = x, so that V = πb^2 d − πa^2 c − π
R (^) b a x
(^2) f 0 (x) dx. Now integrate
by parts with u = x^2 , and dv = f 0 (x) dx ⇒ du = 2x dx, v = f (x), and R (^) b a x
(^2) f 0 (x) dx = £x (^2) f(x)¤b a −^
R (^) b a 2 x f^ (x)^ dx^ =^ b
(^2) f (b) − a (^2) f (a) − R^ b a 2 x f^ (x)^ dx, but^ f^ (a) =^ c^ and^ f^ (b) =^ d
⇒ V = πb^2 d − πa^2 c − π
h b^2 d − a^2 c −
R (^) b a 2 xf^ (x)^ dx
i
R (^) b a 2 πxf^ (x)^ dx.
47. Using the formula for volumes of rotation and the figure, we see that
VOLUMES BY CYLINDRICAL SHELLS ■ 11
