Part II: The Answers
432
Answers
601–700
692.
216π
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 y-axis:
Begin by finding the limits of integration by setting the functions equal to each other
and solving for x:
2
4
0
6
0
6
0 6
2
2
x x
x
x
x
x x
x
=
−
=
−
=
−
=
(
)
,
Because 2x ≥ x
2
– 4x on the interval [0, 6], the shell height is given by 2x – (x
2
– 4x).
Therefore, the integral to find the volume is
2
2
4
2
6
2 2
4
2
2
0
6
2
3
0
6
3
4
0
6
π
π
π
x x x
x dx
x x dx
x
x
−
−
(
)




=
−
(
)
=
−






=
∫
∫
π π
π
2 6
6
4
216
3
4
( )
( )
−






=