What is the exact, rigorous, full statement of Divergence (Gauss') Theorem in $\mathbb{R}^3$ (without being too complicated)?

Question

The wolfram page http://mathworld.wolfram.com/DivergenceTheorem.html states the formula $$ \int_{V} \nabla \cdot \mathbf{F} dS = \int_{\partial V} \mathbf{F} \cdot d\mathbf{S} $$ but it does not speak much of what kind of conditions should be imposed on $\mathbf{F}, V$ and so on.

I think it is enough for $\mathbf{F}$ to be continuously differentiable over $V$ (is it?). But what should be on $V$?

Q1) Is it enough for $V$ to have $\partial V$ as a parametrized (smooth) surface (even piecewise)?

Q2) (It may be a topological one.) But my textbook says a parametrized surface is the image of a continuously differentiable mapping $\mathbf{r} : \mathcal{R} \to \mathbb{R}^3$ where $\mathcal{R}$ is a region (i.e., open, bounded, its boundary having Jordan content 0) in $\mathbb{R}^2$. Then can a sphere have a parametrization?

Q3) What should be the exact imposition on $\mathbf{F}$ including how to specify its domain?

(I hope you'd not talk about manifolds and forms and other complex definitions...)

Answer 1

1. Yes, Divergence theorem is good for "piecewise smooth" domain, namely Lipschitz domain, for example, a regular polyhedron.

2. The unit $2$-sphere's most common parametrization: $$ \mathbf{\Phi}: [0,\pi]\times [0,2\pi] \to \mathbb{R}^3\\ (u,v)\mapsto(\sin v \cos u, \sin v \sin u, \cos v) $$ has two artificial singularities at two poles $(0,0,\pm 1)$ (in $xyz$-coordinates).

3. Normally we require $\mathbf{F}$ to be smooth on $V$ and continuous up to the boundary. However if we are allowed to go to the realm of weak derivatives, the weak divergence of certain $\mathbf{F}$ has the exact form of Divergence theorem as well. A sufficient requirement for this is that $\nabla \cdot \mathbf{F} \in L^1(V)$, $\mathbf{F}\cdot \mathbf{n}\in L^1(\partial V)$, and $\mathbf{F}\cdot \mathbf{n}$ is continuous on any surface lying within $V$.