Ensayo

S ECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS

r. Find the volume of the remaining portion of the

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

V

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 ofthe sides by parabolas.
(a) A barrel with height h and maximum radius R is constructed by rotating about the x-axis the parabola
y R c x 2, h 2 x h 2, where c is a positive

6.3
y

y=2≈-˛
1

xL=?

xR=?

0

2

x

1
3

h (2R 2

2
5

r2

d2)

72. Suppose that a region

has area A and lies above the x-axis.
When is rotated about the x-axis, it sweeps out a solid withvolume 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.

VOLUMES BY CYLINDRICAL SHELLS
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 y-axis theregion bounded by y 2 x 2 x 3 and y 0. (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
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 2shows 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:

F I GU RE 1

V

V2

Îr

h

433

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

70. A hole ofradius r is bored through the center of a sphere of

radius R
sphere.

||||

r22 h

V1
r2

r1 r2

r12 h
r1 h

2

r22
r2

r1
2

r12 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
V

1

2 rh r

and it can be remembered as
F IGURE 2V

y

[circumference][height][thickness]

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

y

y=ƒ

y=ƒ

0

F IGURE 3

a

b

x

0

a

b

x

4 34

||||

CHAPTER 6 APPLICATIONS OF INTEGRATION

y

We divide the interval a, b into n subintervals x i 1, x iof equal width x and let x i be
the midpoint of the i th subinterval. If the rectangle with base x i 1, x i and height f x i is
rotated about the y-axis, then the result is a cylindrical shell with average radius x i , height
f x i , and thickness x (see Figure 4), so by Formula 1 its volume is

y=ƒ

0

a

b
–
x i-1 x i

Vi

x

2 xi f xi

x

Therefore an approximation to thevolume V of S is given by the sum of the volumes of
these shells:

xi

n

y

n

V

y=ƒ

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
n

b

x

lim

nl

2 xi f xi

x

i1

y

b

a

2 x f x dx

Thus the following appears plausible:
FI G URE 4
2 The volume of thesolid in Figure 3, obtained by rotating about the y-axis the
region under the curve y f x from a to b, is

V

y

b

a

2 x f x dx

where 0

a

b

The argument using cylindrical shells makes Formula 2 seem reasonable, but later we
will be able to prove it (see Exercise 67 in Section 7.1).
The best way to remember Formula 2 is to think of a typical shell, cut and flattened as
in...