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Material Type: Exam; Professor: Liu; Class: VECTOR ANALYSIS; Subject: Mathematics; University: University of South Carolina - Columbia; Term: Spring 2007;
Typology: Exams
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Math 550
Exam 3
April 17, 2007
C
(sin x − y
3
)dx + (x
3
where C is the unit circle.
Solution: We have
C
(sin x − y
3 )dx + (x
3
D
3 x
2
2 dxdy
2 π
0
1
0
3 r
2 rdrdθ
2 π
0
dθ ·
1
0
3 r
3
dr
= 2 π ·
π.
F = y~i − x~j. Evaluate
S
F ) · d
S where S is the surface defined
by x
2
2
2 = 1, z ≥ 0.
Solution: We have
S
F ) · d
C
F · d~r
2 π
0
sin θ(− sin θ) − cos θ cos θdθ
2 π
0
− 1 dθ
= − 2 π.
F = e
y sin z~i + e
x cos z~j + e
z (x
2
2 )
k. Use Gauss’ theorem to evaluate
∂w
F · d
where ∂W is the boundary of the cylinder x
2
2 ≤ 1 and 0 ≤ z ≤ 1.
Solution: We have
∂w
F · d
W
div(
F )dxdydz
W
e
z
(x
2
2
)dxdydz
2 π
0
1
0
1
0
e
z
r
2
rdrdzdθ
2 π
0
dθ ·
1
0
e
z
dz ·
1
0
r
3
dr
= 2 π · (e − 1) ·
π(e − 1).
−y
x
2 +y
2
i +
x
x
2 +y
2
j is well-defined in R
2 except the origin
(0, 0). Let D is a simple region in R
2
. Assume (0, 0) is not on the boundary of D.
(a) Suppose (0, 0) is not in D. Use Green’s theorem to show
∂D
F · d~r = 0.
Solution: Note that
∂x
x
x
2
2
y
2 − x
2
(x
2
2 )
2
∂y
−y
x
2
2
y
2
− x
2
(x
2
2 )
2
Apply Green’s theorem. We have
∂D
F · d~r =
D
0 dxdy = 0.
(b) Suppose (0, 0) is in D. Evaluate
∂D
F · d~r.
Solution: Suppose D contains a small ball B
center at (0, 0) with radius . Form
part (a), we have
∂(D\B)
F · d~r = 0. Thus we have
∂D
F · d~r =
∂B
F · d~r
2 π
0
− sin θ
2
(− sin θ) +
cos θ
2
( cos θ)dθ
2 π
0
dθ
= 2 π.
