# Math 2260 Written HW #3 Solutions y x 9

Math 2260 Written HW #3 Solutions
1. Consider the following region:
9
8
7
6
5
4
3
2
1
y  x2
y2x
1
2
3
(a) What is the volume of the solid generated by revolving this region around the x-axis?
Answer: Revolving the region around the x-axis yields the solid shown between the
two surfaces shown below:
0.0
0.5
1.0
1.5
2.0
4
2
0
-2
4
2
0
-2
-4
-4
Since we’re given both curves as graphs of functions of x, it’s going to be most convenient
to integrate in terms of x. Therefore, it’s natural to use the cross-section method with
vertical cross-sections. Such a cross section is a washer with outer radius 2x and inner
radius x2 , so the cross-sectional area, as a function of x, is
A(x) = π(2x)2 − π(x2 )2 = 4πx2 − πx4 = π(4x2 − x4 ).
1
Therefore, the volume of the solid is
Z 2
Z
A(x) dx =
0
2
π(4x2 − x4 ) dx
0
2
4 3 x5
=π x −
3
5 0
4 3 25
2 −
− (0 − 0)
=π
3
5
32 32
=π
−
3
5
64π
=
.
15
(b) What is the volume of the solid generated by revolving this region around the y-axis?
Answer: Rotating the region around the y-axis yields the solid which lies between the
two surfaces shown below:
4
-2
2 1
3
0
-1
2
1
0
-2
0
-1
1
2
It’s still desirable to integrate with respect to x, so now we’ll use the cylindrical shell method.
In this case, a cylindrical shell with radius x has height 2x − x2 , so its surface area is
A(x) = 2πx(2x − x2 ) = 2π(2x2 − x3 ).
Therefore, the volume of the solid is
Z 2
Z
A(x) dx =
0
2
2π(2x2 − x3 ) dx
0
2
2 3 x4
= 2π x −
3
4 0
2 3 24
= 2π
2 −
− (0 − 0)
3
4
16
= 2π
−4
3
8π
=
.
3
2
2. Suppose an ant travels along the floor on a path which can be closely approximated by the
3
1
curve x = y3 + 4y
from y = 1 to y = 3:
3
2
1
2
4
6
8
The ant keeps track of the distance he walks in an exercise journal. Assuming all lengths are
in centimeters, what distance should the ant put down for this journey?
Answer: To find the distance traveled by the ant, we simply integrate the arc length element
from y = 1 to y = 3:
Z 3s
Z 3p
dx
2
2
dx + dy =
1+
dy.
Distance =
dy
1
1
Now,
dx
1
= y2 − 2 ,
dy
4y
2
dx
1
1
= y4 − +
dy
2 16y 4
so
and
dx 2
1
1
1 2
4
2
1+
=y + +
= y + 2 .
dy
2 16y 4
4y
Therefore, the distance traveled by the ant is
s
s
2
2
Z 3 Z 3
dx
1
1+
dy =
y 2 + 2 dy
dy
4y
1
1
Z 3
1
=
y 2 + 2 dy
4y
1
3
3
y
1
−
=
3
4y 1
3
3
1
1 1
=
−
−
−
3
12
3 4
107
1
=
−
12
12
106
=
12
53
= .
6
3