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

Section 6.2 Volumes 441
The solid lies between x − 0 and x − 1, so its volume is
V − y 1
Asxd dx − y 1
x dx − x 2
0
0
2G0
1
− 2
n
y
y=œ„x
y
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 f0, 1g and height 1. If
we rotate this square, we get a cylinder
with radius 1, height 1, and volume
? 1 2 ? 1 − . We computed that the
given solid has half this volume. That
seems about right.
œ„x
0 x 1
x
0
1 x
Îx
FIGURE 6
(a)
(b)
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 y-axis.
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 y-axis, it makes sense to slice the solid
perpendicular to the y-axis (obtaining circular cross-sections) and therefore to integrate
with respect to y. If we slice at height y, we get a circular disk with radius x, where
x − s 3 y . So the area of a cross-section through y is
Asyd − x 2 − (s 3 y ) 2 − y 2y3
and the volume of the approximating cylinder pictured in Figure 7(b) is
Asyd Dy − y 2y3 Dy
Since the solid lies between y − 0 and y − 8, its volume is
V − y 8
Asyd dy − y 8
0
0
y 2y3 dy − f 3 5 y 5y3 g 0
8
−
96
5
y
y
y=8
8
x=0
3
x=œ„y
Îy
x
y (x, y)
y=˛
or
3
x=œ„y
0 x
0 x
FIGURE 7
(a)
(b)
n
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