447
Answers
701–800
Answers and Explanations
Note that the lines x = 0 and x = 1 correspond to the limits of integration and that the
shell height is
1
2
0
2 2
π
e
x
−
−
, or simply 1
2
2 2
π
e
x
−
. To find the volume using cylindrical
shells, use the following integral:
2
1
2
2
2
2
2 2
0
1
2
0
1
π
π
π
π
x
e
dx
xe
dx
x
x
−
−
=
∫
∫
Now use the substitution u
x
= − 1
2
2 so that du = –x
dx, or –du = x
dx. Find the new limits
of integration by noting that if x = 1, then u = − 1
2
, and if x = 0, then u = 0. Using these
new values gives you the following:
−
=
=
( )
=
−
(
)
=
−
−
−
−
−
∫
∫
2
2
2
2
2 1
2 1 1
0
1 2
1 2
0
1 2
0
1 2
π
π
π
π
π
π
e du
edu
e
e
e
u
u
u
710.
2 2
7
20
2 2
11 2
π
+ −
ln
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.
Notice that on the interval [1, 2], you have x
x
≥ ln , so the shell height is x
x
− ln .
Therefore, the integral to find the volume is
2
2
1
2
3 2
1
2
π
π
x x
x dx
x
x x dx
−
(
) =
−
(
)
∫
∫
ln
ln
To evaluate x xdx
ln
∫
, use integration by parts with u = ln x so that du x
dx
= 1 , and use
dv = x
dx so that v x
=
2
2
. These substitutions give you
x xdx x
x
x
x
dx
x
x
x dx
x
x x
C
ln
ln
ln
ln
∫
∫
∫
=
−
=
−
=
−
+
2
2
2
2
2
2
2
1
2
2
2
4
Answers
701–800
Answers and Explanations
Note that the lines x = 0 and x = 1 correspond to the limits of integration and that the
shell height is
1
2
0
2 2
π
e
x
−
−
, or simply 1
2
2 2
π
e
x
−
. To find the volume using cylindrical
shells, use the following integral:
2
1
2
2
2
2
2 2
0
1
2
0
1
π
π
π
π
x
e
dx
xe
dx
x
x
−
−
=
∫
∫
Now use the substitution u
x
= − 1
2
2 so that du = –x
dx, or –du = x
dx. Find the new limits
of integration by noting that if x = 1, then u = − 1
2
, and if x = 0, then u = 0. Using these
new values gives you the following:
−
=
=
( )
=
−
(
)
=
−
−
−
−
−
∫
∫
2
2
2
2
2 1
2 1 1
0
1 2
1 2
0
1 2
0
1 2
π
π
π
π
π
π
e du
edu
e
e
e
u
u
u
710.
2 2
7
20
2 2
11 2
π
+ −
ln
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.
Notice that on the interval [1, 2], you have x
x
≥ ln , so the shell height is x
x
− ln .
Therefore, the integral to find the volume is
2
2
1
2
3 2
1
2
π
π
x x
x dx
x
x x dx
−
(
) =
−
(
)
∫
∫
ln
ln
To evaluate x xdx
ln
∫
, use integration by parts with u = ln x so that du x
dx
= 1 , and use
dv = x
dx so that v x
=
2
2
. These substitutions give you
x xdx x
x
x
x
dx
x
x
x dx
x
x x
C
ln
ln
ln
ln
∫
∫
∫
=
−
=
−
=
−
+
2
2
2
2
2
2
2
1
2
2
2
4