427L Quiz 10 Solution
Nov. 28, 2007
Consider the following
2
form in
R
3
:
ω
=
xdy
∧
dz
+
y dz
∧
dx
+
z dx
∧
dy
(
x
2
+
y
2
+
z
2
)
3
/
2
,
or upon letting
r
=
p
x
2
+
y
2
+
z
2
,
ω
= (
x/r
3
)
dy
∧
dz
+ (
y/r
3
)
dz
∧
dx
+ (
z/r
3
)
dx
∧
dy.
Problem 1
(5 pts)
.
Show that
ω
is closed, i.e., that
dω
= 0
.
Proof.
Notice that
dω
=
±
∂
∂x
x
r
3
²
dx
∧
dy
∧
dz
+
±
∂
∂y
y
r
3
²
dy
∧
dz
∧
dx
+
±
∂
∂z
z
r
3
²
dz
∧
dx
∧
dy
=
±
∂
∂x
x
r
3
+
∂
∂y
y
r
3
+
∂
∂z
z
r
3
²
dx
∧
dy
∧
dz.
But
∂
∂x
x
(
x
2
+
y
2
+
z
2
)
3
/
2
=
(
x
2
+
y
2
+
z
2
)
3
/
2

x
·
3
2
·
2
x
(
x
2
+
y
2
+
z
2
)
1
/
2
(
x
2
+
y
2
+
z
2
)
3
=
r
2

3
x
2
r
5
.
And likewise,
∂
∂y
y
r
3
=
r
2

3
y
2
r
5
,
∂
∂z
z
r
3
=
r
2

3
z
2
r
5
,
so their sum is
3
r
2

3(
x
2
+
y
2
+
z
2
)
r
5
= 0
,
and thus
dω
= 0
dx
∧
dy
∧
dz
= 0
as claimed.
Problem 2
(5 pts)
.
Show that for the sphere of radius
r
given by
S
=
{
(
x,y,z
)

x
2
+
y
2
+
z
2
=
r
2
}
, we have
ZZ
S
ω
= 4
π.
1
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View Full DocumentHint: Use the fact that for a surface with unit normal
n
= (
n
x
,n
y
,n
z
)
,
dA
=
n
x
dy
∧
dz
+
n
y
dz
∧
dx
+
n
z
dx
∧
dy,
and recall that the unit normal on
S
is given by
n
= (
x/r,y/r,z/r
)
. Thus you should
conclude that
ω
=
dA/r
2
, and remember that
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 Fall '07
 Keel
 Vector Calculus, Trigraph, SEPTA Regional Rail, dx dy, dy dz

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