Answer
$0$
Work Step by Step
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
$div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}
=\dfrac{\partial (xe^y)}{\partial x}+\dfrac{\partial (z-e^y)}{\partial y}+\dfrac{\partial (-xy)}{\partial z}=e^y-e^y-0=0$
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Thus, we have $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV = \iiint_E (0) dV=0$
Hence, we get $\iiint_S F \cdot dS = 0$
