Answer
$0$
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.
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
This implies that $div F=\dfrac{\partial (xe^y)}{\partial x}+\dfrac{\partial (z-e^y)}{\partial y}+\dfrac{\partial (-xy)}{\partial z}$
$div F=e^y-e^y-0=0$
Now, we have
$\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV = \iiint_E (0) dV=0$
Thus, we have
$\iiint_S F \cdot dS = 0$
