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Probably simple proof (Stoke's Theorem)

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that ∫u∇v dl = - ∫v∇u dl

    Both integrals are over closed surfaces.

    2. Relevant equations

    The question is being asked in a chapter over Stoke's Theorem. However, I'm confused because I think found the solution without invoking the theorem... Which leads to...

    3. The attempt at a solution

    I used integration by parts to derive

    ∫u∇v dl = uv - ∫v∇u dl

    However since it's over a closed surface, I believe (uv) goes to zero. Is this the correct proof? Thanks.
  2. jcsd
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