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

444 Chapter 6 Applications of Integration
Example 6 Find the volume of the solid obtained by rotating the region in
Example 4 about the line x − 21.
SOLUtion Figure 11 shows a horizontal cross-section. It is a washer with inner radius
1 1 y and outer radius 1 1 sy , so the cross-sectional area is
Asyd − souter radiusd 2 2 sinner radiusd 2
The volume is
− (1 1 sy ) 2 2 s1 1 yd 2
V − y 1
Asyd dy − y 1
0
− y 1
0
0 fs1 1 sy d 2 2 s1 1 yd 2 g dy
3y2
4y
s2sy 2 y 2 y 2 d dy − F
3
2 y 2
2 2 y 3
1
3
G0
− 2
y
1+œ„
y
1+y
1 y
y
x=y
x=œ„y
FIGURE 11
x=_1
0
x
n
We now find the volumes of three solids that are not solids of revolution.
y
Example 7 Figure 12 shows a solid with a circular base of radius 1. Parallel crosssections
perpendicular to the base are equilateral triangles. Find the volume of the solid.
SOLUtion Let’s take the circle to be x 2 1 y 2 − 1. The solid, its base, and a typical
cross-section at a distance x from the origin are shown in Figure 13.
Since B lies on the circle, we have y − s1 2 x 2 and so the base of the triangle ABC
is | AB | − 2y − 2s1 2 x 2 . Since the triangle is equilateral, we see from Figure 13(c)
FIGURE 12
Computer-generated picture
of the solid in Example 7
TEC Visual 6.2C shows how the
solid in Figure 12 is generated.
x
_1 0 x
1 x
A
C
B
y
y=œ„„„„„„ 1-≈
y
0
B(x, y)
A
y
x
x
A
C
œ„3y
60° 60°
y y
B
(a) The solid
(b) Its base
(c) A cross-section
FIGURE 13
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