Answer
The flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 4\pi $
Work Step by Step
The net given by ${x^2} + {z^2} \le 1$, $y=0$ can be parametrized by
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,0,r\sin \theta } \right)$,
where $0 \le r \le 1$, $0 \le \theta \le 2\pi $.
So,
${{\bf{T}}_r} = \left( {\cos \theta ,0,\sin \theta } \right)$
${{\bf{T}}_\theta } = \left( { - r\sin \theta ,0,r\cos \theta } \right)$
${\bf{N}}\left( {r,\theta } \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&0&{\sin \theta }\\
{ - r\sin \theta }&0&{r\cos \theta }
\end{array}} \right) = - r{\bf{j}}$
Since $r$ is always positive, the normal vector obtained above is in the negative $y$-direction. To use the positive direction, we set ${\bf{N}}\left( {r,\theta } \right) = r{\bf{j}}$.
Evaluate:
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right) = \left( {r\cos \theta ,r\sin \theta + 4,{r^2}{{\sin }^2}\theta } \right)$
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right)\cdot{\bf{N}}\left( {r,\theta } \right) = r\left( {r\sin \theta + 4} \right)$
We calculate the flow rate through the net by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\left( {G\left( {r,\theta } \right)} \right)\cdot{\bf{N}}\left( {r,\theta } \right){\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {{r^2}\sin \theta + 4r} \right){\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \left( {{r^2}\sin \theta + 4r} \right){\rm{d}}r$
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \left( {\frac{1}{3}{r^3}\sin \theta + 2{r^2}} \right)|_0^1 = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \left( {\frac{1}{3}\sin \theta + 2} \right)$
$ = \frac{1}{3}\left( { - \cos \theta } \right)|_0^{2\pi } + 4\pi = 4\pi $
So, the flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 4\pi $.
