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Divergence Theorem: Multiplied by Scalar Field

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    whupg.png



    2. Relevant equations

    Definitely related to the divergence theorem (we're working on it):

    wEw1l.png


    3. The attempt at a solution

    I'm a bit confused about multiplying a scalar field f into those integrals on the RHS, and I'm not sure if they can be taken out or not. If they can be, I evaluated the RHS out to be 0 (zero), which doesn't make sense with my evaluation of the LHS, which is just grad f dotted into F.

    On the other hand, if it CAN'T be taken out of the integral, I'm at a loss as of how this relates to the divergence theorem..
    I'm not sure what I'm missing here :( Help would be very much appreciated!
     
  2. jcsd
  3. Oct 28, 2012 #2

    LCKurtz

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    Start by using the divergence theorem on the first term on the right$$
    \iint_{\partial R}(f\vec F)\cdot \hat n\, dA = \iiint_R\nabla \cdot (f\vec F)\, dV$$Work out that ##\nabla \cdot (f\vec F)## in the integrand and go from there.
     
  4. Oct 28, 2012 #3

    Dick

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    First show,
    [tex]\nabla \cdot (fF)=\nabla f \cdot F+f \nabla \cdot F[/tex]
    If you write it out in components it's just the product rule.
     
  5. Oct 28, 2012 #4

    lurflurf

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    This is the product rule
    [tex]\nabla \cdot (\psi\mathbf{A}) = \mathbf{A} \cdot\nabla\psi + \psi\nabla \cdot \mathbf{A}[/tex]
    wrapped up with the divergence theorem.
     
  6. Oct 28, 2012 #5
    For the integrand, I'm getting:
    [itex]\partial (f A) / \partial x + \partial (f B) / \partial y + \partial (f C) / \partial z [/itex], where [itex]\vec F = (A, B, C)[/itex]

    Am I on the right track?
     
  7. Oct 28, 2012 #6
    Double Post, but:

    Ahh so if I use the Divergence Thm as LCKurtz suggested on the first term on the RHS, I get the product rule in the form:

    [tex]\nabla f \cdot F = \nabla \cdot (fF) - f \nabla \cdot F[/tex]

    Except with the terms as integrands. I'm not sure if this is sufficient to prove the validity of the equation though? I'm sorry guys, I feel like you guys are putting the answer right in my face but I'm just not getting it :(
     
  8. Oct 28, 2012 #7

    LCKurtz

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    Assuming ##\vec F = \langle A,B,C\rangle##, Yes. Keep going...
     
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