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How a vol. integral becomes a vol. integral plus surface integral

  1. Sep 9, 2012 #1
    Hello,

    Please see this pdf at some universities website:

    http://physics.ucsc.edu/~peter/110A/helmholtz.pdf

    In line 14 the author claims using integration by parts...I do not understand
    who could the integration by parts be used here.

    I understand the general case where we have der(a*b)=a der(b) + b der(a)

    but I am not able to see how this happened here..

    Thanks!
     
  2. jcsd
  3. Sep 9, 2012 #2
    You use the fact that, for example, for scalar [itex]\psi[/itex] and vector [itex]C[/itex],

    [tex]\int \nabla (\psi C) dV = \oint \psi C \cdot dA[/tex]

    This is a generalization of the fundamental theorem of calculus, and it's used even in the 1D case:

    [tex]\int_a^b u v' \; dx + \int_a^b u' v \; dx = \int_a^b (uv)' \; dx = uv|_a^b[/tex]
     
  4. Sep 9, 2012 #3
    I understand the divergence theorem, but in my example I have two terms on the right, one of which is a surface integral and one is vol. integral. Thus this cannot be div. theorem.
     
  5. Sep 9, 2012 #4
    It's the divergence theorem and integration by parts (or the product rule).

    [tex]\int \nabla(\psi C) \; dV = \int C \cdot \nabla \psi \; dV + \int \psi \nabla \cdot C \; dV = \oint \psi C \cdot dA[/tex]
     
  6. Sep 9, 2012 #5
    Perfect, can you tell me what do psi corresponds to in my example?

    Does correspond only to the rational (1/|r-r'|) or tothe rational and to d3r' ?
     
  7. Sep 9, 2012 #6
    [itex]\psi[/itex] is only the scalar function [itex]1/|r-r'|[/itex]. Their [itex]d^3r'[/itex] is [itex]dV[/itex].
     
  8. Sep 9, 2012 #7
    Now I got it. It is two steps at once, thank you so much !
     
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