Math 13 Fall 2010: Exam 3
Tuesday April 27, 2010
Name:
Instructions: There are 4 questions on this exam each of which is scored out of 8 points for a total
of 32 points. You may not use any outside materials(eg. notes or calculators). You have 50 minutes to
complete this exam. Remember to fully justify your answers.
Score:
1 Problem 1. Evaluate
Z
0
1
Z √1−y2
0
1
dxdy
1 + x2 + y 2
Proof. Changing to polar we have
Z 1 Z √1−y2
Z π/2 Z 1
Z π/2
¯
r
1 ¯¯
1
dxdy
=
drdθ
=
ln 1 + r2 ¯|10 dθ
2
2
2
1+x +y
2
0
0 1+r
0
0
0
Z π/2
ln 2
π ln 2
=
dθ =
.
2
4
0
2 Problem 2. Consider the region R that lies outside the cone z 2 = 3(x2 + y 2 ) and inside the sphere
x2 + y 2 + z 2 = 4 in the first octant. Set up the integral for the volume of R in rectangular, cylindrical,
and spherical coordinates.
Proof. The curve of intersection is given by 4(x2 + y 2 ) = 4 and √
hence x2 + y 2 = 1. We also need to know
the cone equation is spherical coordinates, which is tan φ = 1/ 3 and so φ = π6 . So we have
Z
1
√
1−x2
Z
Z √3(x2 +y2 )
V =
Z
2
dzdydx +
0
Z
0
π/2
Z
1
Z
0
√
r 3
V =
Z
π/2
Z
0
2
Z
rdzdrdθ +
0
Z
π/2
Z
0
0
π/2 Z 2
V =
0
π/6
Z
√
4−x2
√
1−x2
√
4−r 2
Z √4−x2 −y2
dzdydx
0
rdzdrdθ
0
1
0
ρ2 sin φdρdφdθ.
0
or we can subtract the area inside the cone from the sphere. We have
Z 1 Z √1−x2 Z √4−x2 −y2
4
V = π−
√ 2 2 dzdydx
3
0
0
3(x +y )
Z π/2 Z 1 Z √4−r2
4
rdzdrdθ
V = π−
√
3
0
0
r 3
Z π/2 Z π/6 Z 2
4
V = π−
ρ2 sin φdρdφdθ.
3
0
0
0
3 Problem 3. Consider the region R above the upper sheet of the hyperboloid x2 + y 2 = z 2 − 1 and below
the plane z = 3 with density δ(x, y, z) = 2z.
1. Find the mass of R.
2. Set-up but do not evaluate the integral for the moment of inertia about the z-axis for R.
Proof. We have that the vertex of the upper sheet is given by z = 1 at (0, 0, 1) and at z = 3 we have
x2 + y 2 = 8.
1. For the mass (using cylindrical coordinates) we have
Z
2π
Z
√
8
M=
Z
0
0
2π
Z
=
Z
0
Z
√
√
8
Z
3
2π
0
Z
√
8
2rzdzdrdθ =
1+r 2
0
9r − r(1 + r2 )drdθ
0
8r − r3 drdθ
0
=
2π
Z
√
4r2 − r4 /4]0 8 dθ =
2π
32 − 16dθ
0
= 32π.
2. For the moment of inertia we have
Z
2π
√
Z
8
Iz =
0
Z
3
√
1+r 2
0
4
2r3 zdzdrdθ. Problem 4. Let R be the parallelogram bounded by the lines x+y = 1, x+y = 2, 2x−3y = 2, 2x−3y = 4.
Evaluate
ZZ
5xdA.
R
Proof. Set u = x + y and v = 2x − 3y and so we have x = 15 (3u + v) and y = 51 (2u − v). The jacobian
of the transformation is given by
¯
µ3
¶¯
1
¯
¯ 1
5
5
¯
¯
J = ¯det 2
1 ¯= .
−
5
5
5
So we have
ZZ
Z
2
Z
5xdA =
R
1
5
Z
=
1
2
2
4
1
(3u + v)dvdu
5
6u + 6du
1
1
(9 + 6) = 3.
5
5