431
Answers
601–700
Answers and Explanations
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.
Because y = 1 corresponds to one of the limits of integration, find the other limit of
integration by solving the equation 0 = y
1/3
, which gives you 0 = y. Then find the volume
using cylindrical shells:
2
2
2 3
7
2 3
7
0
6
7
1 3
0
1
4 3
0
1
7 3
0
1
π
π
π
π
π
y y
dy
y dy
y
( ) =
=
( )
=
−
( )
=
∫
∫
691.
40
3
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 using cylindrical shells, use the
following 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.
The line x = 2 gives you one of the limits of integration, so begin by finding the other
limit of integration by determining where the function intersects the x-axis; 0 = x
2
gives
you 0 = x. Notice that the shell radius is (x + 1), so here’s the integral to find the volume
using cylindrical shells:
2
1
2
2
4
3
2 2
4
2
3
2
0
2
3
2
0
2
4
3
0
2
4
3
π
π
π
π
(
)
x
x dx
x
x dx
x
x
+
=
+
(
)
=
+
=
+
−
∫
∫
( (
)
0 0
40
3
+
= π
Answers
601–700
Answers and Explanations
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.
Because y = 1 corresponds to one of the limits of integration, find the other limit of
integration by solving the equation 0 = y
1/3
, which gives you 0 = y. Then find the volume
using cylindrical shells:
2
2
2 3
7
2 3
7
0
6
7
1 3
0
1
4 3
0
1
7 3
0
1
π
π
π
π
π
y y
dy
y dy
y
( ) =
=
( )
=
−
( )
=
∫
∫
691.
40
3
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 using cylindrical shells, use the
following 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.
The line x = 2 gives you one of the limits of integration, so begin by finding the other
limit of integration by determining where the function intersects the x-axis; 0 = x
2
gives
you 0 = x. Notice that the shell radius is (x + 1), so here’s the integral to find the volume
using cylindrical shells:
2
1
2
2
4
3
2 2
4
2
3
2
0
2
3
2
0
2
4
3
0
2
4
3
π
π
π
π
(
)
x
x dx
x
x dx
x
x
+
=
+
(
)
=
+
=
+
−
∫
∫
( (
)
0 0
40
3
+
= π