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Divergence theorem that I learnt in analysis:

Let $U \rightarrow \mathbb R^n$ be an open, bounded, regular set and let $f : \bar{U} \rightarrow \mathbb R^n$ be such that $f$ is bounded and continuous in $U$ and there exist the partial derivatives of $f$ in $\mathbb R^n$ at all $x \in U$ and they are continuous and bounded. Then $$\int_U \operatorname{Div} f(x) dx = \int_{\partial U} f(x) \cdot \nu d\mathcal H^{n-1},$$

where $\nu$ is normal vector.

If I go back to calc 3, I see many instances where I "exploit" divergence theorem on non-smooth surfaces (such as polygons bounded by the first octant and plane $\{x+y+z = 1\}$). How can I justify divergence theorem on such non-smooth surfaces?

Let $V$ be the non-smooth open set where divergence theorem is applied. Should I take a sequence of regular set $\{U_n\}_{n \in \mathbb N}$ with $U_n \subset V$ such that $\mu(V - U_n) \rightarrow 0$ and take apply divergence theorem on $U_n$. I am assuming the contribution of vector field on $(V - U_n)$ would be small with $\mu(V - U_n) < \epsilon$. However, I am bogged down by the possibility that something strange can happen near the sharp corner of the boundary of $V$. Can anyone justify how I can ignore the "sharp" corner points?

Thanks in advance.

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As Spivak develops the generalized Stokes's theorem in Calculus on Manifolds, integration is really about differential forms on smooth singular chains. That is, we're not integrating over manifolds per se, but instead integrating over formal integer linear combinations of images of cubes or simplices under smooth maps.

The polygonal/polyhedral regions in multivariable calculus fit immedietaly into this framework.

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