= Solution
{c}
Use the outward <orientation> on the unit <hemisphere> and counterclockwise <boundary orientation> on the unit <circle>. With $\mathbf r(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector <line integral> is
$$
\begin{aligned}
\oint_{\partial\Sigma}f\,d\mathbf r
&=\int_0^{2\pi}(-\cos t\sin t,\cos^2t,0)\,dt\\
&=\boxed{(0,\pi,0)}.
\end{aligned}
$$
For the <surface integral>, $\nabla f=\mathbf i$ and the outward <unit normal vector> is $\mathbf n=(x,y,z)$, since the radius is one. The <cross product> is
$$
\mathbf n\times\nabla f=(0,z,-y).
$$
The <integral> of $y$ is zero by <reflection symmetry>. Using the same <spherical polar coordinates> as in <a/ii>, with $a=1$, the other component is
$$
\iint_\Sigma z\,dS
=\int_0^{2\pi}\int_0^{\pi/2}\cos\varphi\sin\varphi\,d\varphi\,d\theta
=\pi.
$$
Hence
$$
\boxed{\iint_\Sigma(\mathbf n\times\nabla f)\,dS=(0,\pi,0)},
$$
verifying the <vector-valued Stokes identity> directly, component by component. The wedge symbol in the paper denotes the <cross product> here.
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