Part II: The Answers
434
Answers
601–700
Here’s the region being rotated about the x-axis:
Begin by finding the limits of integration. To do so, find the points of intersection of the
two curves by setting the expressions equal to each other and solving for y:
0 5
0
5
0 5
2
3
2
=
−
=
−
=
y
y
y
y
y
(
)
,
Note that 5y
2
– y
3
≥ 0 on the interval [0, 5] so that the shell height is 5y
2
– y
3
– 0, or
simply 5y
2
– y
3
. Therefore, the integral to find the volume is
2
5
2
5
2
5
4
5
2
5 5
0
5
2
3
3
4
0
5
4
5
0
5
π
π
π
π
y y
y dy
y
y dy
y
y
∫
∫
−
(
) =
−
(
)
=
−
=
( )
4 4
5
4
5
5
625
2
−
=
( )
π
695.
16π
The region is being rotated about the line x = 4, 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
434
Answers
601–700
Here’s the region being rotated about the x-axis:
Begin by finding the limits of integration. To do so, find the points of intersection of the
two curves by setting the expressions equal to each other and solving for y:
0 5
0
5
0 5
2
3
2
=
−
=
−
=
y
y
y
y
y
(
)
,
Note that 5y
2
– y
3
≥ 0 on the interval [0, 5] so that the shell height is 5y
2
– y
3
– 0, or
simply 5y
2
– y
3
. Therefore, the integral to find the volume is
2
5
2
5
2
5
4
5
2
5 5
0
5
2
3
3
4
0
5
4
5
0
5
π
π
π
π
y y
y dy
y
y dy
y
y
∫
∫
−
(
) =
−
(
)
=
−
=
( )
4 4
5
4
5
5
625
2
−
=
( )
π
695.
16π
The region is being rotated about the line x = 4, 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