Let $\psi=x+y+z$ then find the value of
$\iint_S \psi\nabla\psi.\hat n dS$ i know the answer is $3V$
My attempt
we know that $\iint_S \psi\nabla\psi.\hat n dS=\iiint_V (\psi \nabla^2\psi+(\nabla\psi)^2dV$ From the divergence theorem, and clearly $\nabla^2\psi=0$ so we have $\iiint_V(\nabla\psi)^2dV$ but then i dont know where to go from here. i know that $|\nabla\psi|^2=3$ but i dont understand how that fits in here, how can one just convert the $(\nabla\psi)^2$ to $\|\nabla\psi|^2$
thanks for taking time to read this, any help would be greatly appreciated, note this problem is different than the ones i am used to attempt since there are no upper or lower bounds on the integrals.