Answer
$\iint_S D_n f dS=\iiint_E \nabla^2 f dV $
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, $S$ is a closed surface and $E$ is the region inside that surface.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Since, we have $D_nf=(\nabla f) \cdot n$
This implies that $\iint_S D_n f dS=\iiint_S (\nabla f) \cdot dS=\iint_S (\nabla F) \cdot dS $
Also, $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV $
or, $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV=\iiint_E \nabla \cdot (\nabla F) dV=\iiint_E \nabla^2 f dV $
Hence, it has been verified that $\iint_S D_n f dS=\iiint_E \nabla^2 f dV $
