Gauss theorem

  • #1

Main Question or Discussion Point

[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:

Answers and Replies

  • #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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