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Gauss theorem

  1. Apr 23, 2012 #1
    [tex]\oint_S \vec{A}\cdot d\vec{S}=\int_V div\vec{A}dv[/tex]

    Suppose region where [tex]\vec{A}(\vec{r})[/tex] is diferentiable everywhere except in region which is given in the picture. Around this region is surface [tex]S'[/tex]. In this case Gauss theorem leads us to

    [tex]\int_S \vec{A}\cdot d\vec{S}+\int_S \vec{A}\cdot d\vec{S}=\int_{V'} divAdv[/tex]

    Am I right?
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2
    Here is the picture
     

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  4. Apr 23, 2012 #3

    HallsofIvy

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    Well, what you wrote is not correct because you have "S" on both integrals and I suspect you want "S' " on one. But even then it not correct- you need to subtract not add:
    [tex]\int_S \vec{A}\cdot\vec{dS}- \int_{S'} \vec{A}\cdot\vec{dS}= \int_V div \vec{A}dV[/tex]
     
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