Math 103  Exam 3
Instructions:
•
You are not permitted to use a calculator or any other materials on this exam.
•
Show all your work on the test. You may use the back of the pages if needed.
•
Simplify all answers. Give exact answers (i.e. fractions, radicals, etc.) unless instructed
otherwise.
•
Sign the Honor Pledge after you finish the exam.
I have neither given nor received any unauthorized help on this test, and I
have conducted myself within the guidelines of the university honor code.
Also I will not discuss the contents of this exam with any other person until
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Question:
1
2
3
4
5
6
7
Total
Points:
7
8
7
8
7
7
7
51
Score:
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1. (7 points) Find the volume of the solid with uniform density that lies inside the sphere
ρ
=
a
and above the cone
φ
=
π
4
.
Solution:
Since the density is uniform, then we just need to integrate the constant
δ
=
c
. Here it will be easier to use spherical coordinates to integrate over
T
which
region indicated in the problem.
V
=
ZZZ
T
c dV
=
c
Z
2
π
0
Z
π
4
0
Z
a
0
ρ
2
sin
φ dρ dφ dθ
=
c
Z
2
π
0
Z
π
4
0
1
3
ρ
3
sin
φ

a
0
dφ dθ
=
c
Z
2
π
0
Z
π
4
0
1
3
a
3
sin
φ dφ dθ
=
c
Z
2
π
0

1
3
a
3
cos
φ

π
4
0
dθ
=
c
Z
2
π
0

√
2
6
a
3
+
1
3
a
3
dθ
=
ca
3
(
2

√
2
3
)
π
2. (8 points) Using integration show that the surface area of a sphere of radius
R
is 4
πR
2
.
Solution:
Recall that that
dS
=

∂
r
∂u
×
∂
r
∂v

du dv
, so we first need a parametrization
of the sphere. Using symmetry we need only consider the top half of the sphere which
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