Answer
$\iint_S D_n f dS=\iiint_E \nabla^2 f dV $
Work Step by Step
We know that $D_nf=(\nabla f) \cdot n$
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ shows a closed surface. The region $E$ is inside that surface.
We have $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$
Now, we have $\iint_S D_n f dS=\iiint_S (\nabla f) \cdot dS=\iint_S (\nabla F) \cdot dS $
This can be re-arranged as: $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV $
$\implies \iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV$
or, $\implies \iiint_E \nabla \cdot (\nabla F) dV=\iiint_E \nabla^2 f dV $
Thus, we have $\iint_S D_n f dS=\iiint_E \nabla^2 f dV $ (Verified)
