445
Answers
701–800
Answers and Explanations
Here’s the region being rotated about the line y = 1:
Begin by isolating x in each equation to get y
2
= x and y
1/3
= x. Set these expressions
equal to each other to find the points of intersection, which give you the limits of
integration. Cube both sides of the equation and factor to solve for y:
y
y
y
y
y
y
y y
y
2
13
6
6
5
0
1 0
0 1
=
=
− =
−
(
) =
= ,
Note that y
1/3
> y
2
for a point in the interval (0, 1), so the shell height is y
1/3
– y
2
. Also
note that the region being rotated is below the line of rotation y = 1, so in the interval
(0, 1), the distance from y to the line of rotation is (1 – y), making the shell radius (1 – y).
Therefore, the integral to find the volume using cylindrical shells is
2
1
2
2 3
4
3
3
7
1 3
2
0
1
1 3
2
43
3
0
1
4 3
3
π
π
π
(
)
−
−
(
)
=
− −
+
(
)
=
−
−
∫
∫
y y
y dy
y
y
y
y dy
y
y
y y
y
7 3
4
0
1
4 3
3
7 3
4
4
2 3
4
1
1
3
3
7
1
1
4
10
21
+






=
− −
+






=
π
π
708.
16
3
π
To find the volume of a solid obtained by rotating about the x-axis a region to the left of
the curve x = f
 
(y) and to the right of the y-axis from y = c to y = d with cylindrical shells,
use the integral
V
yf y dy
c
d
= ∫ 2π ( )
More generically, you can use the formula
d
V
y
c
d
= ∫
2π (
) (
)
shell radius shell height