Answer
The flow rate is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 5\pi $
Work Step by Step
The net given by $y = 1 - {x^2} - {z^2}$, $y \ge 0$ can be parametrized by
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,1 - {r^2},r\sin \theta } \right)$,
where $0 \le r \le 1$, $0 \le \theta \le 2\pi $.
So,
${{\bf{T}}_r} = \left( {\cos \theta , - 2r,\sin \theta } \right)$
${{\bf{T}}_\theta } = \left( { - r\sin \theta ,0,r\cos \theta } \right)$
${\bf{N}}\left( {r,\theta } \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{ - 2r}&{\sin \theta }\\
{ - r\sin \theta }&0&{r\cos \theta }
\end{array}} \right) = - 2{r^2}\cos \theta {\bf{i}} - r{\bf{j}} - 2{r^2}\sin \theta {\bf{k}}$
Notice that the normal vector obtained above is in the negative $y$-direction. To use the positive direction, we set ${\bf{N}}\left( {r,\theta } \right) = 2{r^2}\cos \theta {\bf{i}} + r{\bf{j}} + 2{r^2}\sin \theta {\bf{k}}$.
Evaluate:
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right) = \left( {r\cos \theta - 1 + {r^2},r\sin \theta + 1 - {r^2} + 4,{r^2}{{\sin }^2}\theta } \right)$
$ = \left( {{r^2} + r\cos \theta - 1, - {r^2} + r\sin \theta + 5,{r^2}{{\sin }^2}\theta } \right)$
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right)\cdot{\bf{N}}\left( {r,\theta } \right)$
$ = 2{r^4}\left( {\cos \theta + {{\sin }^3}\theta } \right) + {r^3}\left( {2{{\cos }^2}\theta - 1} \right) + {r^2}\left( { - 2\cos \theta + \sin \theta } \right) + 5r$
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 2{r^4}\left( {\cos \theta + {{\sin }^3}\theta } \right){\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^3}\left( {2{{\cos }^2}\theta - 1} \right){\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^2}\left( { - 2\cos \theta + \sin \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 5r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \theta + {{\sin }^3}\theta } \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 2{r^4}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {2{{\cos }^2}\theta - 1} \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - 2\cos \theta + \sin \theta } \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 {r^2}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 5r{\rm{d}}r$
Now, the following integrals vanish, that is:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta = \mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\sin ^3}\theta {\rm{d}}\theta = 0$
So, the integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {2{{\cos }^2}\theta - 1} \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r + \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 5r{\rm{d}}r$
Since $2{\cos ^2}\theta - 1 = \cos 2\theta $, so
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } \cos 2\theta {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r + \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 5r{\rm{d}}r$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \frac{1}{2}\left( {\sin 2\theta } \right)|_0^{2\pi }\left( {\frac{1}{4}{r^4}} \right)|_0^1 + \left( {2\pi } \right)\left( {\frac{5}{2}{r^2}} \right)|_0^1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 0 + 5\pi $
So, the flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 5\pi $.
