How the use of integration by parts to get the following results involves gradient and normal?

Question

How to get results that $\int_\Omega |\triangledown \phi|^2 d^nr = \int_ {\partial\Omega} \phi(\textbf{n}\cdot \triangledown)\phi dS$, where $dS$ is the element of area on the boundary and $\textbf{n}$ the outward-directed normal.

Thanks

Answer 1

$ \displaystyle \int_\Omega \nabla\Phi\cdot\nabla\Phi d^nr=\int_\Omega\nabla\cdot(\Phi\nabla\Phi)-\Phi\nabla^2\Phi d^n r=\int_{d\Omega}dS n\cdot(\Phi\nabla\Phi)-\int_\Omega\Phi\nabla^2\Phi d^nr= $

$ \displaystyle \int_{d\Omega}dS \Phi n\cdot\nabla\Phi-\int_\Omega\Phi\nabla^2\Phi d^nr$

Note that you still have to have that volume integral. The surface integral is not enough, unless some other assumptions have been made, like $\nabla^2\Phi=0$.