## Calculus 8th Edition

$\iint_S a \cdot n dS=0$
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$ Remember 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 $\iint_S a \cdot n dS=0$