= Solution
The closed boundary includes the curved unit hemisphere and the disk at $y=0$. On the disk, $F=(0,y,0)=0$, so that contribution vanishes. On the curved part the outward normal is $n=(x,y,z)$, giving $F\times n=(yz,0,-xy)$.
Use <spherical coordinates> with the polar axis along positive $y$:
$$
x=\sin\theta\cos\phi,\quad y=\cos\theta,\quad z=\sin\theta\sin\phi,
\quad 0\le\theta\le\frac\pi2,\quad0\le\phi\le2\pi.
$$
The area element is $\sin\theta\,d\theta\,d\phi$. Consequently
$$
\iint_SF\times dS
=\int_0^{\pi/2}\cos\theta\sin^2\theta\,d\theta
\int_0^{2\pi}(\sin\phi,0,-\cos\phi)\,d\phi
=\boxed{(0,0,0).}
$$
This directly performs the surface integral and includes the planar cap, rather than applying the closed-surface theorem to only an open hemisphere.
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