= Solution
Choose the outward <normal vector> on the side of the <elliptic cylinder>. Parametrize it by
$$
\mathbf r(\theta,z)=(\cos\theta,\tfrac12\sin\theta,z),\qquad
-1\le z\le1.
$$
Since $\mathbf r_\theta\times\mathbf r_z=(\tfrac12\cos\theta,\sin\theta,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical <curve> with opposite directions. Their <line integrals> cancel for a single-valued <vector field>. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021-stokes-cut.png]
{title=An elliptic cylindrical side cut open into a parameter rectangle, with compatible outward boundary directions}
To apply <Stokes theorem> to a <flux> rather than a <curl> integral, first find a <vector potential>:
$$
\mathbf A=(0,0,xyz^2),\qquad
\nabla\times\mathbf A=(xz^2,-yz^2,0)=\mathbf F.
$$
This <vector potential> is smooth everywhere. On either horizontal rim, $d\mathbf r$ is horizontal whereas $\mathbf A$ is vertical, giving $\mathbf A\cdot d\mathbf r=0$. For a cut at $\theta_0=\pi/4$, $xy=1/4$, so the upward cut contribution is $\int_{-1}^1z^2/4\,dz=1/6$ and the downward one is $-1/6$. Consequently
$$
\boxed{\iint_S\mathbf F\cdot d\mathbf S
=\oint_{\partial S}\mathbf A\cdot d\mathbf r=0.}
$$
A direct check uses the <oriented surface element> above: $\mathbf F\cdot(\mathbf r_\theta\times\mathbf r_z)=\tfrac12 z^2\cos2\theta$, whose <integral> over a full turn vanishes. Reversing the chosen <orientation> still gives zero.
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