Divergence Theorem and Variations

# So v f n ds f dv 0 v v which is what we wanted show to

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Unformatted text preview: 1 ˆ 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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