Part II: The Answers
442
Answers
701–800
In this example, the limits of integration correspond to the lines x = 0 and x = 3 and the
shell height is given by e
x
−
2
. Therefore, to find the volume of revolution using cylindrical shells, use the following integral:
2
2
0
3
π x e
dx
x
−
( )
∫
To evaluate this integral, use the substitution u = –x
2
so that du = –2x
 
dx, or −
=
1
2
du x dx.
Find the new limits of integration by noting that if x = 3, then u = –9, and if x = 0, then
u = 0. With these new values, the integral becomes
−
=
= ( )
=
−
(
)
=
−




−
−
−
−
∫
∫
π
π
π
π
π
e du
e du
e
e e
e
u
u
u
0
9
9
0
9
0
0
9
9
1 1
705.
25
21
π
The region is being rotated about the line x = –1, which is parallel to the y-axis. To find
the volume of a solid obtained by rotating about the y-axis a region that’s under the
curve y = f
 
(x) and above the x-axis from x = a to x = b with cylindrical shells, use the
integral
V
xf x dx
a
b
= ∫ 2π ( )
More generically, you can use the formula
d
V
x
a
b
= ∫
2π (
) (
)
shell radius shell height
To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance
from x to the line of rotation. To find the shell height, let f
 
(x) be the function that
bounds the region above and let g(x) be the function that bounds the region below; the
shell height is given by f
 
(x) – g(x). Note that the limits of integration often correspond
to points of intersection.