Answer
$11 \pi$
Work Step by Step
Apply 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:
$0 \leq \theta \leq 2 \pi, 0 \leq r \leq 1; 0 \leq z \leq 2$
Now, we have
$\int_0^{2 \pi} \int_0^{2} \int_0^1 3(r^2+z^2) r dr dz d \theta=[\theta]_0^{2 \pi} \cdot \int_0^{2} \int_0^1 3r^3+3z^2 r dr dz $
or, $2 \pi \times \int_0^{2}[\dfrac{3r^4}{4}+\dfrac{3z^2 r^2}{2}]_0^1dz=2\pi \times [\dfrac{3}{4} z+\dfrac{1}{2}z^3]_0^2$
Thus, $\iint_S F\cdot dS=(2\pi)[\dfrac{3}{4} (2)+\dfrac{1}{2}(2)^3]=11 \pi$
