# Gauss theorem

1. Apr 23, 2012

### matematikuvol

$$\oint_S \vec{A}\cdot d\vec{S}=\int_V div\vec{A}dv$$

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

$$\int_S \vec{A}\cdot d\vec{S}+\int_S \vec{A}\cdot d\vec{S}=\int_{V'} divAdv$$

Am I right?

Last edited: Apr 23, 2012
2. Apr 23, 2012

### matematikuvol

Here is the picture

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3. Apr 23, 2012

### HallsofIvy

Staff Emeritus
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:
$$\int_S \vec{A}\cdot\vec{dS}- \int_{S'} \vec{A}\cdot\vec{dS}= \int_V div \vec{A}dV$$