# Integral of divergence over a closed surface

I am reading a paper, where an integral of a divergence over a closed surface is used without proof.

$$\oint_S [\nabla \cdot \vec{v}(\vec{r})] d\vec{s} = 0$$,

where $$\vec{v}$$ is tangential to the surface ($$\vec{v}(r)\cdot \vec{n}(\vec{r}) = 0$$)

I have looked at vector calculus identities and Green theorems and can't seem to find the expression I need. Any suggestions?

This is a direct consequence of the divergence theorem. For a vector field $$X$$ on an oriented $$n$$-dimensional Riemannian manifold $$(M, g)$$, the divergence theorem states that $$\int_M (\operatorname{div} X) dV_g = \int_{\partial M} g(X, N) dV_{\tilde g}$$ where $$N$$ is the outward-pointing normal vector at the boundary, $$dV_g$$ is the Riemannian volume form, and $$dV_{\tilde g}$$ is the induced volume form on the boundary. For a surface embedded in Euclidean space, we use the metric induced by the pullback of the inclusion $$i:S \to \mathbb{R}^k$$, i.e., $$i^*g$$ where $$g = \delta_{ij}dx^idx^j$$ is the usual Euclidean metric. Then $$dV_g = ds$$ where $$ds$$ is the area element, and $$dV_{\tilde g} = dt$$ where $$dt$$ is a length element. Since the surface in your question is closed, the boundary $$\partial S$$ is empty and the right-hand side integral is $$0$$.
If $$M$$ is not orientable, the divergence theorem still holds if you replace $$dV_g$$ and $$dV_{\tilde g}$$ with the respective densities $$d\mu_g$$ and $$d\mu_{\tilde g}$$. These densities are nothing but local volume forms on different patches of the manifold glued together with a partition of unity.
• Also, there seems to be a mistake. The LHS of your divergence theorem equation should refer to an integral of a volume, but you use the symbol $ds$ in there, which you call "the area form of the surface". This is inconsistent with the standard definition of the theorem – Aleksejs Fomins Oct 6 at 12:53
• It is not a mistake. The divergence theorem applies to manifolds of all dimensions, not just 3-dimensional manifolds. I have rewritten it in the general form to avoid confusion. Also note that "closed surface" means a compact surface without boundary, so $\partial S = \emptyset$ is part of the definition. – abhi01nat Oct 6 at 16:54