Answer
$V(E)=(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$
Work Step by Step
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$
Here, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial (x)}{\partial x}+\dfrac{\partial (y)}{\partial y}+\dfrac{\partial (z)}{\partial z}=1+1+1=3$
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
$\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_E (3)dV $
$(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_E dV$
Hence, the result has been verified such that $V(E)=(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$
