I have the following question about the Divergence theorem. I've seen in my class book the following: Divergence Theorem: Let $\Omega \subset \mathbb{R}^3$ be a symmetric elemental region and $\partial\Omega$ a closed surface with normal orientation. Let $F=(P,Q,R)$ be a differentiable field then, $$\iiint_\Omega \operatorname{div} F \,dx\, dy \,dz = \iint_{\partial\Omega} F \cdot ds.$$
On the proof given as $\Omega$ is symmetric elemental region there are $\phi_1(x,y)$ and $\phi_2(x,y)$ such that $\Omega=\{(x,y,z) \in \mathbb{R}: (x,y)\in D, \phi_1(x,y) \leq z \leq \phi_2(x,y)\}$
And the calculates $\iint_{S_1} (0,0,R) ds$
where $S_1$ is the surface with parametrization $T(x,y)=(x,y,\phi_1(x,y)).$ But to use this parametrization for calculating the integral it's necessary to differentiate $\phi_1$ and in the book, it assumes that the derivative exists and finishes the proof from there.
I have two question, first is that I don't see why I can assume that the derivative exists if I don't have the hypothesis of $\partial\Omega$ being piecewise smooth. Second, are there surfaces where the theorem does not apply?
The assumptions can be weakened somewhat, but a minimum of regularity is needed for the very formulation of the divergence theorem. How did you construct the surface element $ds$? If you check that construction you will see that it involves derivatives of the parameterization $\phi$.
The necessity of some regularity to construct the surface element is more easily seen in the one-dimensional case. Peano's curves are maps $\phi\colon [0, 1]\to [0,1]\times[0, 1]$ that are continuous and surjective. This means that you can parameterize the square, a $2$ dimensional object, with a single parameter. But certainly you cannot construct a "line element" on the square. This is no surprise, as Peano's curves are not differentiable at any point.