From an integral theorem, derive (one of Green’s Identities): $$\int_V(\psi \nabla ^2 \phi−\phi \nabla ^2 \psi) dV=\int_{\partial V}(\psi \frac {\partial \phi}{\partial n}−\phi \frac{\partial \psi}{\partial n})dS$$
Almost certainly the integral theorem to use is the divergence theorem. $\int_V \nabla \cdot {\bf F} dV=\int_{\partial V} {\bf F}\cdot d{\bf S}$. $d{\bf S}={\bf n}dS$, but I'm struggling here with what $\bf n$ is, I get that it's the outwards normal, but specifically what it is here is what I'm struggling with.
Apologies if this is a stupid question.
Thanks in advance for any help
Observe, we have the identity \begin{align} \nabla\cdot\left(\psi\nabla\phi-\phi\nabla\psi\right) = \psi\nabla^2\phi-\phi\nabla^2\psi. \end{align} Then using Gauss's Theorem, we obtain the identity \begin{align} \int_V \nabla\cdot\left(\psi\nabla\phi-\phi\nabla\psi\right)\ dV =& \int_{\partial V} (\psi\nabla\phi-\phi\nabla\psi)\cdot\mathbf{n}\ dS\\ =& \int_{\partial V} \psi\nabla\phi\cdot\mathbf{n} -\phi\nabla\psi\cdot\mathbf{n}\ dS=\int_{\partial V} \psi\frac{\partial \phi}{\partial n}- \phi \frac{\partial \psi}{\partial n}\ dS. \end{align}