vector calculus divergence theorem problem

Question

If one let $\psi(x,y,z)$ be a scalar field defined within a volume V be bounded by a simple closed surface S how would they go about proving, would it be easier to start with the rhs or the lhs? i assume i would need to use the divergence theorem?

$\iint_S\psi\nabla\psi.\bar{n}dS$=$\iiint_V(\psi\nabla^2\psi+(\nabla\psi)^2dV$

my attempt is as follows

Let $F=\psi\nabla\psi$ then $\nabla.F=\nabla.(\psi\nabla\psi)=\psi\nabla.\nabla\psi+\nabla\psi.\nabla\psi$ by the product rule.

so $\iiint_v \psi\nabla.\nabla\psi+\nabla\psi.\nabla\psi dV=\iiint_v\psi\nabla^2\psi+(\nabla\psi)^2dV$ since $\nabla.\nabla=\nabla^2$ and thus $\nabla\psi.\nabla\psi=(\nabla\psi)^2$ from identities ?

Answer 1

HINT:

From product rule differentiation, we have $$\nabla \cdot\left( \phi(\vec r)\vec A(\vec r)\right)=\phi(\vec r)\nabla \cdot \vec A(\vec r)+\vec A(\vec r)\cdot \nabla \phi(\vec r)$$

Now, let $\phi=\psi$ and $\vec A=\nabla \psi$ in the Divergence Theorem.

Answer 2

From my best guess (nabla Phi)^2 means the modulus squared. surely since nabla phi is a vector, you can't square a vector. it must mean modulus.