= Solution
{c}
Choose the outward <unit normal vector> on the upper <hemisphere>. The <curl> of $\mathbf F=(4y,x,2z)$ is
$$
\nabla\times\mathbf F=(0,0,1-4)=(0,0,-3).
$$
In <spherical polar coordinates>, put
$$
\mathbf s(\varphi,\theta)
=a(\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi),
\qquad0\leq\varphi\leq\frac\pi2,\quad0\leq\theta\leq2\pi.
$$
Here $n_z=\cos\varphi$ and $dS=a^2\sin\varphi\,d\varphi\,d\theta$. The <surface integral> is therefore
$$
\iint_\Sigma(\nabla\times\mathbf F)\cdot\mathbf n\,dS
=-3a^2\int_0^{2\pi}\int_0^{\pi/2}\cos\varphi\sin\varphi\,d\varphi\,d\theta
=\boxed{-3\pi a^2}.
$$
The compatible <boundary orientation> follows increasing $t$ in $\mathbf r(t)=(a\cos t,a\sin t,0)$, $0\leq t\leq2\pi$. On this <circle>,
$$
\mathbf F(\mathbf r(t))=(4a\sin t,a\cos t,0),\qquad
\mathbf r'(t)=(-a\sin t,a\cos t,0).
$$
Thus the <line integral> is
$$
\oint_{\partial\Sigma}\mathbf F\cdot d\mathbf r
=a^2\int_0^{2\pi}(-4\sin^2t+\cos^2t)\,dt
=\boxed{-3\pi a^2},
$$
which agrees with the <surface integral> as required by the <Stokes theorem>.
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