Part II: The Answers
448
Answers
701–800
Evaluating the integral 2
3 2
1
2
π
x
x x dx
−
(
)
∫
ln
gives you
2
2 2
5
2
4
2 2
5
3 2
1
2
5 2
2
2
1
2
5 2
π
π
π
x
x x dx
x
x
x x
x
−
(
)
=
−
−
=
∫
ln
ln
− −
+
=
−
+
−
−
x
x x
2
2
1
2
5 2
2
2
5 2
2
2
4
2
2
5
2
2
2
2 2
4
2
5
1
1
2
1
ln
ln
ln
π
+ +
=
−
+ − −
=
+ −
1
4
2 2
5
2 2 1 2
5
1
4
2 2
7
20
2 2
7 2
11 2
π
π
ln
ln
711.
π
2
– 2π
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
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.
Here, the limits of integration correspond to the lines y = 0 and y = π
2
, and the shell
height is cos y. To find the volume using cylindrical shells, use the integral
2 0
2
π
π y
ydy
cos
∫
To evaluate the integral, use integration by parts with u = y so that du = dy, and let
dv = cos y
dy so that v = sin y:
2
2
2
0
2
0
2
0
2
0
2
0
2
π
π
π
π
π
π
π
π
y
ydy
y
y
ydy
y
y
y
cos
sin
s in
sin
c os
∫
∫
=
−
(
)
=
+
(
)
= =
−
(
) + −
(
)
=
−
( )
=
−
2
2
2
0
0
2
0
2 2
1
2
2
π π
π
π
π π
π
π
sin
sin
cos
cos
448
Answers
701–800
Evaluating the integral 2
3 2
1
2
π
x
x x dx
−
(
)
∫
ln
gives you
2
2 2
5
2
4
2 2
5
3 2
1
2
5 2
2
2
1
2
5 2
π
π
π
x
x x dx
x
x
x x
x
−
(
)
=
−
−
=
∫
ln
ln
− −
+
=
−
+
−
−
x
x x
2
2
1
2
5 2
2
2
5 2
2
2
4
2
2
5
2
2
2
2 2
4
2
5
1
1
2
1
ln
ln
ln
π
+ +
=
−
+ − −
=
+ −
1
4
2 2
5
2 2 1 2
5
1
4
2 2
7
20
2 2
7 2
11 2
π
π
ln
ln
711.
π
2
– 2π
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
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.
Here, the limits of integration correspond to the lines y = 0 and y = π
2
, and the shell
height is cos y. To find the volume using cylindrical shells, use the integral
2 0
2
π
π y
ydy
cos
∫
To evaluate the integral, use integration by parts with u = y so that du = dy, and let
dv = cos y
dy so that v = sin y:
2
2
2
0
2
0
2
0
2
0
2
0
2
π
π
π
π
π
π
π
π
y
ydy
y
y
ydy
y
y
y
cos
sin
s in
sin
c os
∫
∫
=
−
(
)
=
+
(
)
= =
−
(
) + −
(
)
=
−
( )
=
−
2
2
2
0
0
2
0
2 2
1
2
2
π π
π
π
π π
π
π
sin
sin
cos
cos