Calculus: Early Transcendentals 8th Edition

$V(E)=(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$
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. $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ Thus, we have $div F=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial z}{\partial z}=1+1+1=3$ $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_E (3)dV$ $(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_E dV$ Hence, it has been verified that $V(E)=(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$