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 01:59
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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.
Bombyx mori
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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 01:13