= Solution
{c}
Let $\Sigma$ be a piecewise smooth oriented <surface>, and let $\mathbf F$ be continuously differentiable on an open neighbourhood of it. <Stokes theorem> states that
$$
\boxed{\oint_{\partial\Sigma}\mathbf F\cdot d\mathbf r
=\iint_\Sigma(\nabla\times\mathbf F)\cdot\mathbf n\,dS.}
$$
The <boundary orientation> is induced by the chosen <unit normal vector> through the <right-hand rule>. Smoothness on a neighbourhood makes the spatial <partial derivatives> in the <curl> well defined.
Choose the outward <orientation of a surface> on the lower half of the <ellipsoid>. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. Use
$$
\mathbf r(t)=(2\cos t,-3\sin t,0),\qquad0\leq t\leq2\pi.
$$
The <line integral> is
$$
\begin{aligned}
\oint_{\partial\Sigma}\mathbf F\cdot d\mathbf r
&=\int_0^{2\pi}\bigl[(-6\sin t)(-2\sin t)+(6\cos t)(-3\cos t)\bigr]dt\\
&=12\pi-18\pi=\boxed{-6\pi}.
\end{aligned}
$$
For an independent <surface integral> calculation, first compute $\nabla\times\mathbf F=(2y,-2x,1)$. Parametrize the lower half by
$$
\mathbf R(u,v)=(2\sin u\cos v,3\sin u\sin v,-3\sqrt3\cos u),
\quad0\leq u\leq\frac\pi2,\quad0\leq v\leq2\pi.
$$
The outward vector <area element> is
$$
\mathbf n\,dS=(\mathbf R_v\times\mathbf R_u)\,du\,dv
=\bigl(9\sqrt3\sin^2u\cos v,6\sqrt3\sin^2u\sin v,-6\sin u\cos u\bigr)\,du\,dv.
$$
Thus
$$
\begin{aligned}
\iint_\Sigma(\nabla\times\mathbf F)\cdot\mathbf n\,dS
&=\int_0^{\pi/2}\int_0^{2\pi}
\bigl(30\sqrt3\sin^3u\sin v\cos v-6\sin u\cos u\bigr)\,dv\,du\\
&=-12\pi\int_0^{\pi/2}\sin u\cos u\,du
=\boxed{-6\pi}.
\end{aligned}
$$
The two sides agree. Choosing the inward <unit normal vector> and its corresponding counterclockwise <boundary orientation> changes both answers to $6\pi$.
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