Answer
$11 \pi$
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:
$0 \leq \theta \leq 2 \pi, 0 \leq r \leq 1; 0 \leq z \leq 2$
Divergence's Theorem: $\iint_S F\cdot dS=\iiint_E div F dV$
This gives:
$\int_0^{2 \pi} \int_0^{2} \int_0^1 3(r^2+z^2) r dr dz d \theta=[\theta]_0^{2 \pi} (\int_0^{2} \int_0^1 3r^3+3z^2 r dr dz)$
or, $2 \pi (\int_0^{2}[\dfrac{3r^4}{4}+\dfrac{3z^2 r^2}{2}]_0^1dz)=2\pi[\dfrac{3}{4} z+\dfrac{1}{2}z^3]_0^2$
Hence, $\iint_S F\cdot dS=2\pi[\dfrac{3}{4} (2)+\dfrac{1}{2}(2)^3]=\pi(3+8)=11 \pi$
