Answer
$4 \pi$
Work Step by Step
Apply Divergence's Theorem: $\iint_S F\cdot dS=\iiint_E div F dV$
and
div $F=3(x^2+y^2+z^2)=3(r^2+z^2)$
For the cylindrical coordinates:
$0 \leq \theta \leq 2 \pi, 0 \leq r \leq 1; 0 \leq z \leq 2$
Now, we have
$ div F=1+1+1=3$
and $\iiint_E div F dV=3 \iiint_E dV=(3) \times (\dfrac{4 \pi}{3})=4 \pi$
Also,we have $F[r(\theta ,\phi)]=\sin \phi \cos \theta i+\sin \phi \sin \theta j+\cos \phi k$
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, $\iint_S F\cdot dS=[-\cos (\phi)]_0^{\pi}\times [\theta]_0^{2\pi}$
Hence, $\iint_S F\cdot dS=\iiint_E div F dV=(-\cos \pi +\cos 0)(2 \pi-0)=4 \pi$
