## Calculus: Early Transcendentals 8th Edition

$\dfrac{4\pi}{3}$
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV$ Here, $S$ shows a closed surface and $E$ is the region inside that surface. $div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}=\dfrac{\partial (2)}{\partial x}+\dfrac{\partial (2)}{\partial y}+\dfrac{\partial z}{\partial z}=0+0+1=1$ The integral $\iiint_E div F dV$ is defined as the volume of the region $E$ that lies inside a sphere having radius $1$. The volume of the region E is: $\iiint_E dV=\dfrac{4\pi(1)^3}{3}=\dfrac{4\pi}{3}$ Hence, we have: $\iint_S F \cdot n dS=\iiint_Ediv \overrightarrow{F}dV =\dfrac{4\pi}{3}$