Let $\Omega \subseteq \mathbb{R}^n$ a bounded domain with boundary $\partial \Omega \in C^1$. Then if $w \in C^1(\Omega)$ it holds $$\int_\Omega \text{div} \ w \ dV = \int_{\partial \Omega} \ w \cdot \nu \ d\sigma$$
This is a classical formulation of the divergence theorem.
My question is: if $w \in C^1_c(\Omega)$ is the hypothesis $\partial \Omega \in C^1$ still crucial?
I think the answer is no since $w = 0$ near the boundary but I'm not very sure how to prove this formally.
EDIT
I think the following is trivial but an equivalent formulation of my question is:
Let $\Omega \subseteq \mathbb{R}^n$ a bounded domain and $w \in C^1_c(\Omega)$. Does it hold $\int_\Omega \text{div} \ w \ dV = 0$ ?
This is equivalent since $w|_{\partial \Omega} = 0$ and the right integral in the divergence theorem becomes exactly $0$
This could be an elegant solution:
Since $\Omega$ is bounded, there exists a ball of radius $R$ big enough such that $\Omega \subseteq B_R(0)$. Then we can extend $w$ to all the ball so that $w = 0$ out of its support.
Now we can apply the usual divergence theorem and we can conclude.