Answer
The flow rate is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{{16}}{3}\pi $
Work Step by Step
The net given by $y = \sqrt {1 - {x^2} - {z^2}} $, $y \ge 0$ can be parametrized by
$G\left( {r,\theta } \right) = \left( {r\cos \theta ,\sqrt {1 - {r^2}} ,r\sin \theta } \right)$,
where $0 \le r \le 1$, $0 \le \theta \le 2\pi $.
So,
${{\bf{T}}_r} = \left( {\cos \theta , - \dfrac{r}{{\sqrt {1 - {r^2}} }},\sin \theta } \right)$
${{\bf{T}}_\theta } = \left( { - r\sin \theta ,0,r\cos \theta } \right)$
${\bf{N}}\left( {r,\theta } \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{ - \dfrac{r}{{\sqrt {1 - {r^2}} }}}&{\sin \theta }\\
{ - r\sin \theta }&0&{r\cos \theta }
\end{array}} \right) = - \dfrac{{{r^2}\cos \theta }}{{\sqrt {1 - {r^2}} }}{\bf{i}} - r{\bf{j}} - \dfrac{{{r^2}\sin \theta }}{{\sqrt {1 - {r^2}} }}{\bf{k}}$
Notice that the normal vector obtained above is in the negative $y$-direction. To use the positive orientation, we set ${\bf{N}}\left( {r,\theta } \right) = \dfrac{{{r^2}}}{{\sqrt {1 - {r^2}} }}\cos \theta {\bf{i}} + r{\bf{j}} + \dfrac{{{r^2}}}{{\sqrt {1 - {r^2}} }}\sin \theta {\bf{k}}$.
Evaluate:
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right) = \left( {r\cos \theta - \sqrt {1 - {r^2}} ,r\sin \theta + \sqrt {1 - {r^2}} + 4,{r^2}{{\sin }^2}\theta } \right)$
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right)\cdot{\bf{N}}\left( {r,\theta } \right)$
$ = \dfrac{{{r^4}}}{{\sqrt {1 - {r^2}} }}{\sin ^3}\theta + \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\cos ^2}\theta + {r^2}\left( {\sin \theta - \cos \theta } \right) + r\sqrt {1 - {r^2}} + 4r$
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 \dfrac{{{r^4}}}{{\sqrt {1 - {r^2}} }}{\sin ^3}\theta {\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\cos ^2}\theta {\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^2}\left( {\sin \theta - \cos \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {r\sqrt {1 - {r^2}} + 4r} \right){\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\sin ^3}\theta {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \dfrac{{{r^4}}}{{\sqrt {1 - {r^2}} }}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } {\cos ^2}\theta {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\sin \theta - \cos \theta } \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 {r^2}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {\rm{d}}\theta \left( {r\sqrt {1 - {r^2}} + 4r} \right){\rm{d}}r$
Notice that 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
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\cos ^2}\theta {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {\rm{d}}\theta \left( {r\sqrt {1 - {r^2}} + 4r} \right){\rm{d}}r$
Since $2{\cos ^2}\theta - 1 = \cos 2\theta $, so
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos 2\theta + 1} \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \left( {r\sqrt {1 - {r^2}} + 4r} \right){\rm{d}}r$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos 2\theta + 1} \right){\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\rm{d}}r$
$ + \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \left( {r\sqrt {1 - {r^2}} } \right){\rm{d}}r + 4\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 r{\rm{d}}r$
Since $\mathop \smallint \limits_{}^{} \dfrac{{{r^3}}}{{\sqrt {1 - {r^2}} }}{\rm{d}}r = - \dfrac{1}{3}\sqrt {1 - {r^2}} \left( {2 + {r^2}} \right)$ and $\mathop \smallint \limits_{}^{} r\sqrt {1 - {r^2}} {\rm{d}}r = - \dfrac{1}{3}{\left( {1 - {r^2}} \right)^{3/2}}$, so
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{1}{2}\left( {\dfrac{1}{4}\sin 2\theta |_0^{2\pi } + 2\pi } \right)\left( { - \dfrac{1}{3}\sqrt {1 - {r^2}} \left( {2 + {r^2}} \right)} \right)|_0^1$
$ - \dfrac{2}{3}\pi \left( {{{\left( {1 - {r^2}} \right)}^{3/2}}|_0^1} \right) + 8\pi \left( {\dfrac{1}{2}{r^2}} \right)|_0^1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \pi \left( {\dfrac{2}{3}} \right) + \dfrac{2}{3}\pi + 4\pi $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{{16}}{3}\pi $
So, the flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \dfrac{{16}}{3}\pi $.