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

  • Thread starter tom8
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  • #1
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Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
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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]
 
  • #3
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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.
 
  • #4
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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]
 
  • #5
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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]
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' ?
 
  • #6
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[itex]\psi[/itex] is only the scalar function [itex]1/|r-r'|[/itex]. Their [itex]d^3r'[/itex] is [itex]dV[/itex].
 
  • #7
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[itex]\psi[/itex] is only the scalar function [itex]1/|r-r'|[/itex]. Their [itex]d^3r'[/itex] is [itex]dV[/itex].
Now I got it. It is two steps at once, thank you so much !
 

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