Section 13.8 wrapup:
Let
F
=
xz
i
+
yz
j
+
xy
k
as in Example 2, and let
S
be the
sphere of radius 1 centered at (1,2,3).
Find
∫∫
S
curl
F
⋅
d
S
.
..?..
Surface integrals over closed surfaces: If
S
is a closed
surface,
(*)
∫∫
S
curl
F
⋅
d
S
= 0.
Proof #1: If
S
is a closed surface, its boundary is the empty
set, so
∫∫
S
curl
F
⋅
d
S
=
∫
∂
S
F
⋅
d
r
= 0.
Proof #2: Let
C
be a piecewisesmooth simple closed curve
that divides
S
into two pieces
S
+
and
S
–
, so that
C
is
positively oriented with respect to
S
+
and negatively
oriented with respect to
S
–
.
[Draw picture.]
Then
∫∫
S
curl
F
⋅
d
S
=
∫∫
S
+
curl
F
⋅
d
S
+
∫∫
S
–
curl
F
⋅
d
S
=
∫
C
F
⋅
d
r
+
∫
–
C
F
⋅
d
r
=
∫
C
F
⋅
d
r
–
∫
C
F
⋅
d
r
= 0.
Hybrid proof: Draw a curve
C
, but let it shrink down to a
point.
Suppose we have to evaluate
∫
C
F
⋅
d
r
where
C
is the unit
circle in the
x
,
y
plane, and we suspect that Stokes’ Theorem
would be helpful.
What are some surfaces
S
for which
∂
S
=
C
?
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 Fall '11
 Staff
 Integrals, Vector Calculus, Stokes' theorem, curl F dS

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