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In the common divergence theorem, shall the boundary (surface) not be smooth everywhere? Is there a version of this theorem where the boundary is nowhere differentiable?

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This kind of things belongs to the realm of geometric measure theory, more than multivariable calculus (whatever those artificial classifications actually mean, of course). I'm by no means an expert; however, I'm told that the book Geometric measure theory by Federer is a classic. – Giuseppe Negro Jan 20 '13 at 1:59

I do not think you can in the ordinary sense, because the divergence theorem states $$\int dFdv=\int F\cdot nds$$

Now if $S$ is not differentiable, then the integration of the differential form $F\cdot nds$ is not really well defined. You can still integrate $F\cdot nds$ as a measurable function on $S$, where $ds$ become a certain unit surface element. For example there are "exotic spheres" constructed from plumbing which does not admit a differentiable structure, and I assume a suitable modification of standard "area form" in polar coordinate might work. Then the theorem would carry through. But to write down such an integral explicitly would be very difficult.

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Here is a link that you will find helpful:


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I am assuming that is $S$ is defined piecewise $S=S_1\cup S_2$ where $S_1$ and $S_2$ are differentiable surfaces and not differentiable along $S_1\cap\S_2$ which is a line, the usual theorem still holds. – pluton Jan 20 '13 at 1:13

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