Answer
$\iint_S (f \nabla g-g \nabla f) \cdot n ds=\iiint_E (f \nabla^2g-g \nabla^2f) dV$
Work Step by Step
We know that $D_nf=(\nabla f) \cdot n$
$\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV$
But $F=\nabla g$
Now, we have $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV=\iiint_E f(\nabla \cdot ( \nabla g) +\nabla f \cdot (\nabla g) dV$
$=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$
$=[\iiint_E (f \nabla^2g+\nabla f \cdot (\nabla g) )dV]-[\iiint_E g ( \nabla^2f)+\nabla g \cdot (\nabla f) )dV$
$=\iiint_E f \nabla^2g-g ( \nabla^2f) dV$
Thus, we have $\iint_S (f \nabla g-g \nabla f) \cdot n ds=\iiint_E (f \nabla^2g-g \nabla^2f) dV$ (Verified)
