Answer
$4 \pi$
Work Step by Step
Divergence's Theorem states that $\iint_S F\cdot dS=\iiint_E div F dV$
Here, $S$ is a closed surface and $E$ is the region inside that surface.
Thus, div $F=3(x^2+y^2+z^2)=3(r^2+z^2)$
In the cylindrical coordinates, we have:
$0 \leq \theta \leq 2 \pi, 0 \leq r \leq 1; 0 \leq z \leq 2$
Here, we have $F=xi+yj+zk \implies div F=1+1+1=3$
and $\iiint_E div F dV=3 \iiint_E dV=(3)(\dfrac{4 \pi}{3})=4 \pi$
Also, $F(r(\theta \phi))=\sin \phi \cos \theta i+\sin \phi \sin \theta j+\cos \phi k$
Thus, $\iint_S F\cdot dS=\sin \phi(\sin^2 \phi \cos^2 \theta+\sin^2 \phi \sin^2 \theta+\cos^2 \phi)dA=\iint_D \sin \phi dA$
or, $\int_0^{\pi} \int_0^{2 \pi} \sin \phi d\theta d \phi=\int_0^{\pi} \sin \phi d\phi \int_0^{2 \pi} d\theta$
or, $\iint_S F\cdot dS=[-\cos \phi]_0^{\pi}[\theta]_0^{2\pi}=(-\cos \pi +\cos 0)(2 \pi-0)=4 \pi$
Hence, $\iint_S F\cdot dS=\iiint_E div F dV=4 \pi$
