437
Answers
601–700
Answers and Explanations
To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance
from y to the line of rotation. To find the shell height, let f
(y) be the curve that bounds
the region on the right and let g(y) be the curve that bounds the region on the left; the
shell height is given by f
(y) – g(y). Note that the limits of integration often correspond
to points of intersection.
Notice that if x = 0, then y =
− =
9 0 3, so the limits of integration are y = 0 (given)
and y = 3. Solving for x in the equation y
x
=
−
9
gives you x = 9 – y
2
so that the shell
height is (9 – y
2
). Therefore, the integral to find the volume is
2
9
2 9
2
1
4
2 9
2
3 1
4
3
81
2
2
0
3
2
4
0
3
2
4
π
π
π
π
y
y dy
y
y
−
(
) =
−
(
)
=
−
(
)
=
∫
699.
16
3
π
The region is being rotated about the line x = 2, 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.
Find the limits of integration by setting the functions equal to each other and solving for x:
1
0
1
1
0
1 1
2
−
=
−
+
=
= −
x
x
x
x
(
)(
)
,
Because the line of rotation x = 2 is to the right of the region being rotated, the shell
radius is (2 – x). Therefore, to find the volume of revolution using cylindrical shells,
you use the following integral:
2
2
1
2
2
2
2 1
4
2
3
1
2
2
1
1
3
2
1
1
4
3
2
π
π
π
(
)
−
−
(
)
=
−
− +
(
)
=
−
−
+
−
−
∫
∫
x
x dx
x
x
x
dx
x
x
x 2 2
2
1
4
2
3
1
2
2
1
4
2
3
1
2
2
16
3
1
1
x )
(
=
− − +
(
) − + − −
(
)
=
−
π
π
Answers
601–700
Answers and Explanations
To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance
from y to the line of rotation. To find the shell height, let f
(y) be the curve that bounds
the region on the right and let g(y) be the curve that bounds the region on the left; the
shell height is given by f
(y) – g(y). Note that the limits of integration often correspond
to points of intersection.
Notice that if x = 0, then y =
− =
9 0 3, so the limits of integration are y = 0 (given)
and y = 3. Solving for x in the equation y
x
=
−
9
gives you x = 9 – y
2
so that the shell
height is (9 – y
2
). Therefore, the integral to find the volume is
2
9
2 9
2
1
4
2 9
2
3 1
4
3
81
2
2
0
3
2
4
0
3
2
4
π
π
π
π
y
y dy
y
y
−
(
) =
−
(
)
=
−
(
)
=
∫
699.
16
3
π
The region is being rotated about the line x = 2, 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.
Find the limits of integration by setting the functions equal to each other and solving for x:
1
0
1
1
0
1 1
2
−
=
−
+
=
= −
x
x
x
x
(
)(
)
,
Because the line of rotation x = 2 is to the right of the region being rotated, the shell
radius is (2 – x). Therefore, to find the volume of revolution using cylindrical shells,
you use the following integral:
2
2
1
2
2
2
2 1
4
2
3
1
2
2
1
1
3
2
1
1
4
3
2
π
π
π
(
)
−
−
(
)
=
−
− +
(
)
=
−
−
+
−
−
∫
∫
x
x dx
x
x
x
dx
x
x
x 2 2
2
1
4
2
3
1
2
2
1
4
2
3
1
2
2
16
3
1
1
x )
(
=
− − +
(
) − + − −
(
)
=
−
π
π