Answer
$\iint_S a \cdot n dS=0$
Work Step by Step
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, we have $S$ shows a closed surface. The region $E$ is inside that surface.
We have $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$
It must be remembered that the divergence of a constant function should be always zero.
This means that $\iint_S a \cdot n dS=\iiint_Ediv (a) dV=\iiint_E (0) dV=0 $
and $\iint_S a \cdot n dS=0$ (Verfied)
