439
Answers
701–800
Answers and Explanations
Because 5 + 3x – x
2
≥ 5 – 2x on the interval [0, 5], the shell height is given by
(5 + 3x – x
2
) – (5 – 2x). Therefore, the integral to find the volume is
2
5 3
5 2
2
5
2 5
3
4
2
0
5
2
3
0
5
3
4
π
π
π
x
x x
x dx
x x dx
x
x
+
−
(
) − − )
(
=
−
(
)
=
−





∫
∫
(
)
 
=
−






=
0
5
4
4
2 5
3
5
4
625
6
π
π
701.
175
4
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.
Here’s the region being rotated about the line x = –1: