Answer
The flow rate is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \frac{{24}}{5}$
Work Step by Step
We can parametrize the region bounded by the ellipse ${\left( {\frac{x}{2}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} = 1$ in the $xy$-plane by
$G\left( {r,\theta } \right) = \left( {2r\cos \theta ,3r\sin \theta ,0} \right)$,
where $0 \le r \le 1$, $0 \le \theta \le \frac{\pi }{2}$ (since $x,y \ge 0$).
From here, we obtain
${{\bf{T}}_r} = \left( {2\cos \theta ,3\sin \theta ,0} \right)$
${{\bf{T}}_\theta } = \left( { - 2r\sin \theta ,3r\cos \theta ,0} \right)$
${\bf{N}}\left( {r,\theta } \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{2\cos \theta }&{3\sin \theta }&0\\
{ - 2r\sin \theta }&{3r\cos \theta }&0
\end{array}} \right) = 6r{\bf{k}}$
Evaluate:
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right) = \left( {2r\cos \theta ,3r\sin \theta ,12{r^3}{{\cos }^2}\theta \sin \theta } \right)$
${\bf{v}}\left( {G\left( {r,\theta } \right)} \right)\cdot{\bf{N}}\left( {r,\theta } \right) = 72{r^4}{\cos ^2}\theta \sin \theta $
We calculate the flow rate through the portion of the ellipse 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}} = 72\mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{2\pi } {r^4}{\cos ^2}\theta \sin \theta {\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = 72\mathop \smallint \limits_{r = 0}^1 {r^4}{\rm{d}}r\mathop \smallint \limits_{\theta = 0}^{\pi /2} {\cos ^2}\theta \sin \theta {\rm{d}}\theta $
Write $t = \cos \theta $. So, $dt = - \sin \theta d\theta $. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = - 72\left( {\frac{1}{5}{r^5}|_0^1} \right)\mathop \smallint \limits_{t = 1}^0 {t^2}{\rm{d}}t$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = - \frac{{72}}{5}\left( {\frac{1}{3}{t^3}} \right)|_1^0 = \frac{{72}}{{15}}$
So, the flow rate is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{v}}\cdot{\rm{d}}{\bf{S}} = \frac{{24}}{5}$.
