435
Answers
601–700
Answers and Explanations
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.
Begin by finding the points of intersection by setting the functions equal to each other
and solving for x in order to find the limits of integration:
4
0 2
4
0 2
2
0 2
2
2
2
x x
x
x
x
x x
x
−
=
=
−
=
−
=
(
)
,
Note that on the interval [0, 2], you have 4x – x
2
≥ x
2
so that the shell height is
(4x – x
2
) – (x
2
). Also note that the region is to the left of the line of rotation x = 4,
so the shell radius is (4 – x). Therefore, the integral to find the volume is
2
4
4
2
16 12
2
2 8
4
2
2
0
2
2
3
0
2
2
3
π
π
π
(
)
−
−
(
) −
=
−
+
(
)
=
−
∫
∫
x
x x
x dx
x
x
x dx
x
x + +
(
)
=
− +
=
1
2
2 32 32 8
16
4
0
2
x
π
π
(
)
696.
13
6
π
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, the limits of integration correspond to the lines x = 0 and x = 1, the shell height is
(1 + x + x
2
), and the shell radius is x. Therefore, the integral to find the volume is
2
1
2
2
2
3
4
2
2
0
1
2
3
0
1
2
3
4
0
1
π
π
π
π
x
x x dx
x x
x dx
x
x
x
+ +
(
) =
+ +
(
)
=
+
+
=
∫
∫
1 1
2
1
3
1
4
0
13
6
2
3
4
+ + −
=
( )
π
Answers
601–700
Answers and Explanations
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.
Begin by finding the points of intersection by setting the functions equal to each other
and solving for x in order to find the limits of integration:
4
0 2
4
0 2
2
0 2
2
2
2
x x
x
x
x
x x
x
−
=
=
−
=
−
=
(
)
,
Note that on the interval [0, 2], you have 4x – x
2
≥ x
2
so that the shell height is
(4x – x
2
) – (x
2
). Also note that the region is to the left of the line of rotation x = 4,
so the shell radius is (4 – x). Therefore, the integral to find the volume is
2
4
4
2
16 12
2
2 8
4
2
2
0
2
2
3
0
2
2
3
π
π
π
(
)
−
−
(
) −
=
−
+
(
)
=
−
∫
∫
x
x x
x dx
x
x
x dx
x
x + +
(
)
=
− +
=
1
2
2 32 32 8
16
4
0
2
x
π
π
(
)
696.
13
6
π
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, the limits of integration correspond to the lines x = 0 and x = 1, the shell height is
(1 + x + x
2
), and the shell radius is x. Therefore, the integral to find the volume is
2
1
2
2
2
3
4
2
2
0
1
2
3
0
1
2
3
4
0
1
π
π
π
π
x
x x dx
x x
x dx
x
x
x
+ +
(
) =
+ +
(
)
=
+
+
=
∫
∫
1 1
2
1
3
1
4
0
13
6
2
3
4
+ + −
=
( )
π