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In particular, choosing a = ˆ, ˆ and k, we see that all three components of the vector
ı
ˆ
f n dS − V ∇ f dV are zero. So
∂V
ˆ
f n dS − ∇f dV = 0
V ∂V which is what we wanted show.
To prove the third formula, assuming the ﬁrst, apply the ﬁrst with F replaced by
F × a, where a is any constant vector.
ˆ
F × a · n dS = ∇ · (F × a) dV
V ∂V = (∇ × F) · a − F · ∇ × a dV
V = (∇ × F) · a dV
V To get the second line, we used vector identity # 9. To get the third line, we just used
that a is a constant, so that it is annihilated by all derivatives. For all vectors
ˆˆ
ˆ
F×a·n=n·F×a=n×F·a
so
ˆ
n × F dS = a · a· ∇ × F dV
V ∂V ˆ
n × F dS − =⇒ a · ∇ × F dV =0 V
∂V ˆ
In particular, choosing a = ˆ, ˆ and k, we see that all three components of the vector
ı
ˆ
n × F dS − V ∇ × F dV are zero. So
∂V
ˆ
n × F dS − ∇ × F dV = 0
V ∂V which is what we wanted show. March 3, 2013 Divergence Theorem and Variations 2...
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 Fall '13
 JoelFeldman
 Calculus

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