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Math 53  Solutions to Final Exam Review
GSI: Santiago Canez
Disclaimer: As always, there may be some typos in this write up. If you think you
have found a mistake, please let me know! Happy studying.
1.
Compute the line integral of
F
(
x,y
) = (2
x
2
−
3
y
2
)
i
+ (2
x
+3
y
2
)
j
over the triangle with vertices (
−
2
,
0), (2
,
0), and (0
,
2) oriented counterclockwise.
Solution.
Let
C
denote the triangle. By Green’s Theorem, we have
°
C
F
·
d
r
=
°°
D
curl
F
·
k
dA
where
D
is the region enclosed by the triangle. The left side of
D
is given by
y
−
x
=2
and the right side is given by
y
+
x
= 2. Since curl
F
·
k
=2+6
y
, we have
°°
D
curl
F
·
k
dA
=
°
2
0
°
−
y
+2
y
−
2
(2 + 6
y
)
dxdy
=
°
2
0
(2
x
+6
yx
)
±
±
±
−
y
+2
y
−
2
dy
=
°
2
0
(
−
12
y
2
+ 20
y
+ 8)
dy
= 24
.
Hence
²
C
F
·
d
r
= 24.
2.
(a) DeFne the curl and divergence of a vector Feld.
(b) Determine whether or not the vector Feld
F
=
x
2
y
i
+
zye
x
j
−
2
xyz
k
is the curl
of another vector Feld.
Solution.
(a) Let
F
=
P
i
+
Q
j
+
R
k
. Then
curl
F
=(
R
y
−
Q
z
)
i
+(
P
z
−
R
x
)
j
+(
Q
x
−
P
y
)
k
and div
F
=
P
x
+
Q
y
+
R
z
.
(b) Since div curl
G
= 0 for any vector Feld
G
, if
F
was the curl of some other vector
Feld, its divergence would have to be zero. But div
F
=2
xy
+
ze
x
−
2
xy
=
ze
x
°
=0so
F
is not the curl of another vector Feld.
Date
: May 11, 2006
1
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View Full Document3.
Let
S
be the sphere of radius
ρ
.
(a) Compute the surface area of
S
.
(b) Compute the surface integral
°°
S
z dS
.
Solution.
(a) Parameterize the sphere by
r
(
φ, θ
)=
ρ
sin
φ
cos
θ
i
+
ρ
sin
φ
sin
θ
j
+
ρ
cos
φ
k
,
0
≤
φ
≤
π,
0
≤
θ
≤
2
π.
Then
r
φ
=
ρ
cos
φ
cos
θ
i
+
ρ
cos
φ
sin
θ
j
−
ρ
sin
φ
k
r
θ
=
−
ρ
sin
φ
sin
θ
i
+
ρ
sin
φ
cos
θ
j
.
We can then compute

r
φ
×
r
θ

=
ρ
2
sin
φ.
Thus the surface area of
S
is (with
D
the region where
φ
and
θ
lie)
±±
D

r
φ
×
r
θ

dA
=
±
2
π
0
±
π
0
ρ
2
sin
φ dφ dθ
=4
πρ
2
.
(b) We use the same parameterization as above. We have for
f
(
x,y,z
)=
z
±±
S
f
(
x,y,z
)
dS
=
±±
D
f
(
r
(
u,v
))

r
u
×
r
v

dA
=
±±
D
ρ
cos
φ
(
ρ
2
sin
φ
)
dA
=
ρ
3
±
2
π
0
±
π
0
sin
φ
cos
φdφdθ
=0
.
Note that this makes sense since the function
f
(
x,y,z
)=
z
is symmetric about the
xy
plane, so the integral over the bottom half of the sphere should be the negative of
the integral over the top half of the sphere and hence the integral over the whole sphere
should be 0.
4.
Let
F
(
x,y,z
)=
x
i
+
y
j
+
z
k
. Determine whether the Fux of
F
across
S
is negative,
positive, or zero for each surface
S
below.
(a) the cone
z
2
=
x
2
+
y
2
oriented outward
(b) the sphere
x
2
+
y
2
+
z
2
= 1 oriented inward
(c) the plane
y
= 2 oriented with normals having positive
j
component
Solution.
Here,
F
is just the “radial” vector ±eld, which always points radially away
from the origin.
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 Spring '07
 Hutchings
 Math

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