Answer
$\iint_S D_n f dS=\iiint_E \nabla^2 f dV $
Work Step by Step
We know that $D_nf=(\nabla f) \cdot n$ ...(1)
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$
From equation (1), we have
$\iint_S D_n f dS=\iiint_S (\nabla f) \cdot dS=\iint_S (\nabla F) \cdot dS $
This gives: $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV $
and $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV$
or, $\iiint_E \nabla \cdot (\nabla F) dV=\iiint_E \nabla^2 f dV $
Thus, $\iint_S D_n f dS=\iiint_E \nabla^2 f dV $
