= Solution
First calculate the <curl>:
$$
\nabla\times\mathbf F=(0,-2z,3y-1).
$$
By <Stokes theorem>, use the positively oriented equator $\mathbf r(\phi)=(a\cos\phi,a\sin\phi,0)$, $0\leq\phi\leq2\pi$. Its tangent is $a(-\sin\phi,\cos\phi,0)$, while the restricted field is $(0,3a^2\cos\phi\sin\phi-a\cos\phi,0)$. Thus
$$
\iint_S(\nabla\times\mathbf F)\cdot d\mathbf S
=\int_0^{2\pi}\left(3a^3\cos^2\phi\sin\phi-a^2\cos^2\phi\right)d\phi
=\boxed{-\pi a^2}.
$$
The first term integrates to zero over a full period and the second uses $\int_0^{2\pi}\cos^2\phi\,d\phi=\pi$.
One can also calculate directly on the curved <surface>. With <spherical coordinates> $0\leq\theta\leq\pi/2$, $0\leq\phi\leq2\pi$,
$$
d\mathbf S=a^2\sin\theta(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)\,d\theta\,d\phi.
$$
Substitution gives the scalar integrand $a^3\sin^2\theta\cos\theta\sin\phi-a^2\sin\theta\cos\theta$. Its azimuthal <sine> term vanishes, and the rest gives $-2\pi a^2\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta=-\pi a^2$, confirming the <orientation> and the same result.
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