Answer
$4 \pi$
Work Step by Step
Apply the Divergence's Theorem: $\iint_S F\cdot dS=\iiint_E div F dV$
Here, $S$ shows that the closed surface with $E$ as the region inside that surface.
$div(F)=3(x^2+y^2+z^2)=3(r^2+z^2)$
For the cylindrical coordinates:
$0 \leq \theta \leq 2 \pi, 0 \leq r \leq 1; 0 \leq z \leq 2$
But $F=xi+yj+zk$ or, $div (F)=1+1+1=3$
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, $\iint_S F\cdot dS=\sin \phi(\sin^2 \phi \cos^2 \theta+\sin^2 \phi \sin^2 \theta+\cos^2 \phi)dA$
or, $\iint_D [\sin \phi] dA=\int_0^{\pi} \int_0^{2 \pi} [\sin \phi] d\theta d \phi=\int_0^{\pi} [\sin \phi] d\phi \int_0^{2 \pi} d\theta$
or, $[-\cos \phi]_0^{\pi}[\theta]_0^{2\pi}=(-\cos \pi +\cos 0)(2 \pi-0)$
Divergence's Theorem: $\iint_S F\cdot dS=\iiint_E div F dV$
Thus, we have $\iint_S F\cdot dS=\iiint_E div F dV=4 \pi$
