Answer
$-\dfrac{1712 \pi}{15}$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_D [P \dfrac{\partial h}{\partial x}-Q+R\dfrac{\partial h}{\partial z}]dA $
where, $D$ is projection of $S$ onto xz-plane.
$\iint_S F \cdot n dS=\iint_D \sqrt {x^2+y^2}+z^3 dA=\iint_D [\sqrt {x^2+y^2}]+(x^2+y^2) \times \sqrt {x^2+y^2}dA $
or, $= \int_{1}^{3} \int_0^{2 \pi} (r^2+1) \times \sqrt {r^2} (r) d\theta dr$
or, $=[ \int_{1}^{3} (r^4+r^2) dr] \times [\int_0^{2 \pi} d\theta]$
Hence, we have $\iint_S F \cdot n dS= 2 \pi \int_{1}^{3}(r^4+r^2) dr= [\dfrac{r^5}{5}+\dfrac{r^3}{3}]_1^{3} \times (2 \pi)=-\dfrac{1712 \pi}{15}$
