= Solution
The ellipse bounding the surface is $u=1$. With the upward normal, its positively oriented boundary is traversed by increasing $v$ from zero to $2\pi$:
$$
\mathbf r(v)=(a\cosh1\cos v,a\sinh1\sin v,c),\qquad d\mathbf r=(-a\cosh1\sin v,a\sinh1\cos v,0)dv.
$$
The given components are Cartesian components, so their ordinary <dot product> with this tangent is
$$
\mathbf F\cdot d\mathbf r=-2a\sinh1\cosh1(\sin^2v+\cos^2v)dv.
$$
Therefore <Stokes theorem> gives
$$
\boxed{\int_S(\nabla\times\mathbf F)\cdot d\mathbf S=-4\pi a\sinh1\cosh1=-2\pi a\sinh2.}
$$
A direct check uses $F_x=2y/a$ and $F_y=-2x/a$, hence $(\nabla\times\mathbf F)_z=-4/a$, and multiplies by the ellipse area $\pi a^2\sinh1\cosh1$. The third component cannot affect either calculation because the surface and its boundary lie at fixed $z$. Equivalently one may apply Stokes to the smooth planar field $(2y/a,-2x/a,0)$, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.
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