Math 241, Quiz 2 Solutions
March 3, 2014
There are four questions. Each one is worth 5 points.
1. Circle T for true or F for false.
Z πZ 1p
a) The integral
1 − r2 r dr dθ is the volume of a hemisphere of radius 1.
0
T
F
T
F
0
The integral is half the volume.
Z πZ 2
1
r cos(θ) r dr dθ evaluates to the y-coordinate of the centroid
b) The expression
2π 0 0
of the semi-circular region x2 + y 2 ≤ 4, y ≥ 0.
The expression evaluates to 0, the x-coordinate of the centroid.
c) The mass of the solid region between the paraboloids z = x2 + y 2 andpz = 8 − x2 − y 2
and inside the cylinder x2 + y 2 = 1 with density function δ(x, y, z) = x2 + y 2 can be
Z 2π Z 1 Z 8−r2
r2 dz dr dθ .
found by evaluating the integral
0
0
Z
−C
F
r2
Z
f (x, y) ds = −
d)
T
f (x, y) ds
T
F
T
F
C
Arc length integrals are not oriented.
e) If C is the oriented line starting at (−1, 1) and ending at (3, 3), then
Z
(x + y) dx − y dy = 10.
C
The line integral evaluates to 8 via the parametrization x = −1+4t, y = 1+2t, 0 ≤ t ≤ 1.
2. Set up an integral for the second moment (the moment of inertia) about the z-axis of the homogeneous (mass
density=1) solid region in the first octant bounded by the paraboloid z = 12 − x2 − 3y 2 and the planes y = 1,
x = 1, and y = 1 − x. Do not evaluate the integral.
The moment of inertia about the z-axis for solid E whose mass density is 1 is Iz =
RRR
E
(x2 + y 2 ) dV .
The paraboloid is the top of E. It lies over the triangle 0 ≤ x ≤ 1, 1 − x ≤ y ≤ 1, in the xy-plane, and
0 ≤ z ≤ 12 − x2 − 3y 2 . Consequently,
Z
1
Z
1
Z
Iz =
0
1−x
0
12−x2 −3y 2
x2 + y 2 dz dy dx.
3. Let H be the solid hemisphere x2 + y 2 + z 2 ≤ 9, z ≥ 0 and I =
ZZZ
(9 − x2 − y 2 ) dV .
H
Write the iteration of this integral in
a) Rectangular Coordinates
√
√
The base of H is the disk x2 + y 2 ≤ 9. Therefore, −3 ≤ x√ ≤ 3 and√− 9 − x2 ≤ y ≤ 9 − x2 . For each
Z 3 Z 9−x2 Z 9−x2 −y2
p
point (x, y), 0 ≤ z ≤ 9 − x2 − y 2 . Therefore, I =
9 − x2 − y 2 dz dy dx.
√
−3
− 9−x2
0
b) Cylindrical Coordinates
In cylindrical coordinates √
the disk is 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3, and for each point (r, θ), 0 ≤ z ≤
Z 2π Z 3 Z 9−r2
Therefore, I =
(9 − r2 ) r dz dr dθ.
0
0
√
9 − r2 .
0
c) Spherical Coordinates
In spherical coordinates the hemisphere is 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2, and 0 ≤ ρ ≤ 3. Since r = ρ sin(φ),
Z 2π Z π/2 Z 3
I=
(9 − ρ2 sin2 (φ)) ρ2 sin(φ) dρ dφ dθ.
0
0
0
4. Evaluate one of the integrals in problem 3 (exact answer). Show all of your work. (no calculator)
I will evaluate the integral in part
Z
2π
Z
a
3
√
Z
I=
0
Z
2π
Z
0
Z
(circle one)
9−r 2
(9 − r2 ) r dz dr dθ
3
(9 − r2 )3/2 r dr dθ
0
2π
=
0
= 2π ·
=
c
0
0
=
b
3
1
2 5/2 2
dθ
(9 − r ) · · −
5
2 0
1 5/2
·9
5
2 · 81 · 3 · π
486
=
π
5
5