Tensor calculus proof How can I show that
$$\int_{\Sigma}(φ ψ_{,i} - ψ φ_{,i})d^{2}x \ = \ \int_{\Omega}(φ ψ_{,ii} - ψ φ_{,ii})d^{3}x $$, where $ψ,φ$ are scalar functions of the coordinates. Then $i, ii$ designate differentiation of first and second order, and  $Σ, Ω$ are the surface and volyme respectively?
 A: The Divergence theorem on $\mathbb{R^3}$ (Gauss Theorem) for the vector field $\mathbf{F}=\varphi\nabla\psi$ is
$$
\int_\Sigma\varphi\nabla\psi\cdot \mathrm{d}\Sigma=\int_\Omega\nabla\cdot(\varphi\nabla\psi)\mathrm{d}\Omega\tag{1}
$$
With $\nabla\cdot(\varphi\nabla\psi)=\nabla\varphi\cdot \nabla\psi+\varphi\nabla^2\psi$ and using $\frac{\partial \psi}{\partial \nu}=\nabla \psi\cdot\hat{\nu}$ for the surface normal we can rewrite this as
$$
\int_\Sigma\varphi\frac{\partial \psi}{\partial \nu}\mathrm{d}\Sigma=\int_\Omega\nabla\varphi\cdot\nabla\psi\,\mathrm{d}\Omega+\int_\Omega\varphi\nabla^2\psi\,\mathrm{d}\Omega\tag{2}
$$
This is a version of Greens first identity
Now we swap the scalar functions $\varphi$ and $\psi$ in $(2)$ and subtract from $(2)$ and end up with
$$
\int_\Sigma\left(\varphi\frac{\partial \psi}{\partial \nu}-\psi\frac{\partial \varphi}{\partial \nu}\right)\mathrm{d}\Sigma=\int_\Omega\left(\varphi\nabla^2\psi-\psi\nabla^2\varphi\right)\mathrm{d}\Omega
\tag{3}$$
This is a version of Greens second identity.
