= Solution
{c}
Choose the outward <orientation> on the upper half of the <ellipsoid>, giving counterclockwise <boundary orientation> on $C$ as viewed from $z>0$. Parametrize the boundary <circle> by
$$
\mathbf c(t)=(\cos t,\sin t,0),\qquad0\leq t\leq2\pi.
$$
Then $\mathbf u(\mathbf c(t))=(-\sin t,\cos t,0)=\mathbf c'(t)$, so the <line integral> is
$$
\oint_C\mathbf u\cdot d\mathbf r
=\int_0^{2\pi}(\sin^2t+\cos^2t)\,dt=2\pi.
$$
For a direct <surface integral>, use the <parametrized surface>
$$
\mathbf s(\varphi,t)=(\sin\varphi\cos t,\sin\varphi\sin t,\alpha\cos\varphi),
\qquad0\leq\varphi\leq\frac\pi2,\quad0\leq t\leq2\pi.
$$
Its outward <vector surface element> is supplied by the <cross product>
$$
\mathbf s_\varphi\times\mathbf s_t
=(\alpha\sin^2\varphi\cos t,\alpha\sin^2\varphi\sin t,
\sin\varphi\cos\varphi).
$$
Using the <curl> from <Question 3/a>,
$$
\iint_{S_\alpha}(\nabla\times\mathbf u)\cdot\mathbf n\,dS
=\int_0^{2\pi}\int_0^{\pi/2}2\sin\varphi\cos\varphi\,d\varphi\,dt
=\boxed{2\pi}.
$$
The two integrals agree for every $\alpha>0$, verifying the <Stokes theorem>. Only the vertical component of the <vector surface element> enters, which explains the absence of $\alpha$ in the answer.
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