# Reference for the divergence theorem for embedded $C^1$-submanifolds of $\mathbb R^d$ with boundary

I'm aware of Gauss's theorem (aka the divergence theorem) for compact subsets $$K$$ of $$\mathbb R^d$$ with "$$C^1$$-boundary"$$^1$$.

I know that there are several generalizations of this theorem, but since I'm not familiar with general differential geometry, I'd like to find a reference which particularly considers the case of the manifold boundary $$\partial M$$ of a "$$d$$-dimensional embedded $$C^1$$submanifold of $$\mathbb R^d$$ with boundary"$$^2$$.

Is there a nice reference which presents and proofs this case in a rigorous way?

Please note that my main background is abstract measure/probability theory.

$$^1$$ i.e. for each $$p\in\partial K$$, there is an open neighborhood $$U$$ of $$p$$ and a $$\psi\in C^1(U)$$ with $$K\cap U=\{\psi\le0\}$$ and $$\psi'(x)\ne0$$ for all $$x\in U$$. $$\partial K$$ is denoting the topological boundary of $$K$$ and it is a $$(d-1)$$-dimensional embedded $$C^1$$-submanifold of $$\mathbb R^d$$.

$$^2$$ i.e. each point of $$M$$ is locally $$C^1$$-diffeomorphic to $$\mathbb H^d:=\mathbb R^{d-1}\times[0,\infty)$$ and the manifold boundary $$\partial M$$ is a $$(d-1)$$-dimensional embedded $$C^1$$-submanifold of $$\mathbb R^d$$ (without boundary).

• Unfortunately, Do Carmo's book is not a good reference for this result. The book uses the term "differentiable" to mean $C^{\infty}$, and assumes this of all manifolds and forms in Stokes's theorem. You can see this when he proves that $d^2=0$ for differentiable forms, which is clearly not true for $C^1$ forms. The problem is to find a reference for the $C^1$ case. I think we need to look for something else, because I think do Carmo's proof only works for $C^1$ differential forms on $C^2$ manifolds with boundary. – Ben McKay Jul 19 at 16:28
• @BenMcKay: regarding smoothness of the manifold one can work in a $C^\infty$ atlas that is compatible with the given $C^1$ atlas. Smoothness of the form is also not a problem because one can approximate $C^1$ forms by $C^\infty$ forms, so the $C^\infty$ Stokes theorem implies the $C^1$ version. – Igor Belegradek Jul 19 at 18:25