Currently, I was reading this paper "Finite element methods for surface PDEs". For the theorem 2.14 in the paper:
$ \int_{\Gamma} \nabla_{\Gamma} f \cdot \nabla_{\Gamma} g \mathrm{d} A=-\int_{\Gamma} f \Delta_{\Gamma} g \mathrm{d} A+\int_{\partial \Gamma} f \nabla_{\Gamma} g \cdot \mu \mathrm{d} A $
where $\Gamma$ is defined as a surface, $\nabla_{\Gamma}$ is the surface gradient on surface $\Gamma$, $\Delta_{\Gamma}$ is the Laplace–Beltrami operator on $\Gamma.$
My question is that, if the $\Gamma$ is a close surface, is the second term on the right hand side of the equation equals to $0$ and why?