Math 2260 Written HW #2 Solutions
1. Find the volume of the solid given by revolving the shaded region around the x-axis:
1
Π
0
2
Π
2
In the picture, the red curve is the curve y =
p
cos(x) and the blue line is the line y = 1.
Answer: Rotating this region around the x-axis yields the solid which lies between the two
surfaces shown below:
Now, each cross section is a washer with outer radius 1 and inner radius
cross-sectional area is
2
√
A(x) = π(1)2 − π cos x = π − π cos x = π(1 − cos x).
1
√
cos x, so the
Therefore, the volume of the solid is
Z π/2
Z π/2
A(x) dx =
π(1 − cos x) dx
−π/2
−π/2
π/2
= π [x − sin x]−π/2
h π
π
i
=π
− sin(π/2) − − − sin(−π/2)
2
2
= π(π − 2)
= π 2 − 2π,
which is approximately 3.59.
2. Use integration to prove that the volume of a cone with height h and base radius r is 31 πr2 h.
(Hint: think about generalizing the example given in class on Monday. The first thing to do
is to realize the cone as a solid of revolution.)
Answer: To make a cone of height h and radius r which is a solid of revolution, the first
problem is to figure out the equation of the line that goes through both the origin and the
point (h, r):
r
h
Since this is a line of slope
r−0
h−0
=
r
h
which has y-intercept zero, the equation of the line is
y=
r
x.
h
Now, rotating the shaded region around the x-axis yields the desired cone of height h and
radius r.
r
r
h
2
Rh
Therefore, the volume of the cone will be 0 A(x) dx where A(x) is the area of the crosssection, which is
r 2
r2
A(x) = π
x = π 2 x2 .
h
h
Hence, the volume of the cone is
Z
h
h
Z
0
0
r2 2
x dx
h2
3 h
x
3
3 0 h
−0
3
π
A(x) dx =
=π
r2
h2
=π
r2
h2
1
= πr2 h,
3
as expected.
3