Answer
$\iint_S a \cdot n dS=0$
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.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
It should be remembered that the divergence of a constant function is always zero.
Thus, we have
$\iint_S a \cdot n dS=\iiint_Ediv (a) dV=\iiint_E (0) dV=0 $
Hence, it has been verified that $\iint_S a \cdot n dS=0$
