For future reference, we define
$$
\vec{r}(t,s) \equiv \left( x(t,s), y(t,s), z(t,s) \right)
$$
By Stokes' Theorem, the given integral is
$$
\iint_S \nabla \times F\ dS = \oint_{\partial S} F \cdot d\ell
$$
where $\partial S$ is the boundary of $S$. This boundary can be broken up into four separate parametric curves:
- $\vec{r}(t, 0)$, for $t$ running from 0 to 2π;
- $\vec{r}(2\pi, s)$, for $s$ running from 0 to π;
- $\vec{r}(t,\pi)$, for $t$ running from 2π to 0; and
- $\vec{r}(0,s)$, for $s$ running from π to 0.
The first integral is (setting $s = 0$ here)
\begin{align*}
I_1 &= \int_0^{2\pi} \left( F \cdot \frac{\partial \vec{r}}{\partial t}\right) dt \\
&= \int_0^{2 \pi} \left(z \frac{\partial x}{\partial t} + y \frac{\partial z}{\partial t} \right) dt \\
&= \int_0^{2 \pi} \left[ \left(9 + \frac{2}{3}\sin t \right)(0) + \left( 9 + 2 \cos t \right)\left( \frac{2}{3} \cos t \right) \right] \, dt \\
&= \frac{4}{3} \int_0^{2 \pi} \cos^2 t = \frac{4\pi}{3}.
\end{align*}
Meanwhile, the third integral is (setting $s = \pi$ here)
\begin{align*}
I_3 &= \int^0_{2\pi} \left( F \cdot \frac{\partial \vec{r}}{\partial t}\right) dt \\
&= \int^0_{2\pi} \left(z \frac{\partial x}{\partial t} + y \frac{\partial z}{\partial t} \right) dt \\
&= \int^0_{2\pi} \left[ \left(9 + \frac{2}{3}\sin t \right)(0) + \left( 9 - 2 \cos t \right)\left( \frac{2}{3} \cos t \right) \right] \, dt \\
&= -\frac{4}{3} \int^0_{2\pi} \cos^2 t = \frac{4\pi}{3}.
\end{align*}
Using similar techniques, one can show that the second and fourth integrals vanish. Thus, the total integral is
$$
\boxed{ \iint_S \nabla \times F\ dS = \frac{8\pi}{3}.}
$$
But wait! rebo79's answer seems to show that this is a closed surface, so why is this giving us a non-zero answer? The answer seems to be that the parametrization of the surface leads to an inconsistent orientation along the boundary curves. We can see this by calculating the surface normals
$$
\hat{n} = \frac{\partial \vec{r}}{\partial t} \times \frac{\partial \vec{r}}{\partial s}
$$
and plotting them at several points:
We can see that the surface orientation is not consistent, and thus we cannot apply Gauss's Law to get the integral over $S$: the normal of $S$, as parametrized, is not always the outwards-pointing normal to the region "bounded" by $S$. The problematic points seem to be when $t = \pi/2$ and $t = 3 \pi/2$; at these points, the surface intersects itself. It can also be shown that curves 1 & 3 are the same ellipse in the $yz$-plane traversed in the same direction, so their contributions to the boundary integral reinforce rather than cancel. Curves 2 & 4 are similarly the same arc traversed in the same direction, and so their contributions would reinforce rather than cancelling; but it happens that both contributions just happen to be zero for this particular choice of $F$.
Finally, note that if the parametrization of the curve had been $t \in [-\pi/2, \pi/2]$ and $s \in [0, 2 \pi]$, this problem would not have arisen. In this case, two of the "boundary curves" would have been closed loops running along line segments parallel to the $z$-axis (and so their integrals would have vanished), while the other two would have been curves running along a "line of latitude" in opposite directions (and so their integrals would have cancelled.) I suspect that there is either some kind of transcription error involved, or an instructor decided to tweak things around without thinking too carefully about the consequences; because this problem is quite devious as it stands.
