James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 6.3 Volumes by Cylindrical Shells 451
these shells:
V < o n
V i − o n
2x i f sx i d Dx
i−1 i−1
This approximation appears to become better as n l `. But, from the definition of an
inte gral, we know that
Thus the following appears plausible:
lim
n l ` on 2x i f sx i d Dx − y b
2x f sxd dx
i−1
a
2 The volume of the solid in Figure 3, obtained by rotating about the y-axis the
region under the curve y − f sxd from a to b, is
V − y b
2x f sxd dx
a
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 7.1.73).
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 2x, height f sxd, and thickness Dx or dx:
y
y b
a
s2xd f f sxdg dx
circumference height thickness
x
ƒ
ƒ
FIGURE 5
x
x
2πx
Îx
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 2 x 3 and y − 0.
SOLUtion From the sketch in Figure 6 we see that a typical shell has radius x, circumference
2x, and height f sxd − 2x 2 2 x 3 . So, by the shell method, the volume is
x
x
2≈-˛
2
x
V − y 2
s2xd s2x 2 2 x 3 d dx
0
circumference height thickness
−2 y 2
0
s2x 3 2 x 4 d dx − 2 f 1 2 x 4 2 1 5 x 5 g 0
2
FIGURE 6
− 2(8 2 32
5 ) − 16
5
It can be verified that the shell method gives the same answer as slicing.
n
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