# Gauss Theorem on surfaces

Given a bounded and open domain $$\Omega\subset \mathbb{R}^n$$ with $$\Gamma:=\partial \Omega \in C^\infty$$. Let $$H$$ be the mean curvature on $$\Omega$$ and $$\nu$$ the unit outer normal vector.

Furthermore we define the tangential Gradient for a smooth function $$f:\overline{\Omega}\to\mathbb{R}$$ and the tangential divergence for a smooth vector field $$v:\overline{\Omega}\to\mathbb{R}^n$$ as follows $$\nabla_\Gamma f:=\nabla_x f-(\nu\cdot\nabla_xf))\nu$$ respectively $$\operatorname{div}_\Gamma v := \operatorname{div}_x(v) - (\nu\cdot D_v)\nu,$$ where $$(D_v)_j:=\sum_{i=1}\partial_jv_i$$.

Is it now possible to show that the following theorem holds

$$\int_\Gamma f~\operatorname{div}_\Gamma v~dS = \int_\Gamma (v\cdot \nabla_\Gamma f)~dS + (n-1)\int_\Gamma f~(v\cdot\nu)H~dS.$$

• What’s your question? – Jack Lee Apr 2 at 15:10
• How to prove (or at least where to find a proof) of the theorem $\int_\Gamma f~\operatorname{div}_\Gamma v~dS = \int_\Gamma (v\cdot \nabla_\Gamma f)~dS + (n-1)\int_\Gamma f~(v\cdot\nu)H~dS$; sorry for not making this clear. I tried to clarifie my question aove. – Bara Apr 3 at 7:22