Answer
$8 \pi$
Work Step by Step
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
Here, we have $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV =\int_0^{2 \pi}\int_0^2\int_0^{4-r^2} (3r \cos \theta+1) r dz dr d\theta $
$=\int_0^{2 \pi}\int_0^2\int_0^{4-r^2} 3r^2 \cos \theta+ r dz dr d\theta $
$=\int_0^{2 \pi}\int_0^2 3(4-r^2)r^2 \cos \theta+(4-r^2) r dr d\theta $
$=\int_0^{2 \pi}\int_0^2 3(4r^2-r^4) \cos \theta+(4r-r^3) dr d\theta $
$=\int_0^{2 \pi}[3(\dfrac{4r^3}{3}-\dfrac{r^5}{5}) \cos \theta+(2r^2-\dfrac{r^4}{4})]_0^2 d\theta $
$=\int_0^{2 \pi} 3(\dfrac{32}{3}-\dfrac{32}{5}) \cos \theta+(8-\dfrac{16}{4}) d\theta $
$=[3(\dfrac{32}{3}-\dfrac{32}{5}) \sin \theta+4 \theta]_0^{2 \pi} $
$=8 \pi$
