= Solution
{c}
Use the <spherical coordinates> parametrization
$$
\mathbf R(\phi,\theta)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),
\qquad \alpha\leq\phi\leq\beta,\quad0\leq\theta\leq2\pi.
$$
Then $\mathbf R_\phi\times\mathbf R_\theta=\sin\phi\,\mathbf R$, so the outward <unit normal vector> is $\mathbf n=\mathbf R=\sin\phi\,\mathbf e_r+\cos\phi\,\mathbf e_z$ and the <surface area element> is $\sin\phi\,d\phi\,d\theta$. Since $\sin\phi>0$ throughout this band, the <vector field> is smooth near it. The <surface integral> of the <curl> is
$$
\begin{aligned}
\iint_\Sigma(\nabla\times\mathbf F)\cdot\mathbf n\,dS
&=2\pi\int_\alpha^\beta\bigl[f(\cos\phi)\cos\phi-f'(\cos\phi)\sin^2\phi\bigr],d\phi\\
&=2\pi\int_\alpha^\beta\frac{d}{d\phi}\bigl[\sin\phi f(\cos\phi)\bigr],d\phi\\
&=\boxed{2\pi\bigl[\sin\beta f(\cos\beta)-\sin\alpha f(\cos\alpha)\bigr].}
\end{aligned}
$$
For the induced <boundary orientation>, the lower circle $\phi=\beta$ is traversed with increasing $\theta$, and the upper circle $\phi=\alpha$ with decreasing $\theta$. One way to check the signs is that the outward <conormal> is $\mathbf e_\phi$ at the lower circle and $-\mathbf e_\phi$ at the upper circle; the positive <unit tangent vector> is $\mathbf n\times\text{conormal}$.
Along a circle of fixed $\phi$, $d\mathbf r=\sin\phi\,\mathbf e_\theta\,d\theta$. Therefore the <circulation> is
$$
\oint_{\partial\Sigma}\mathbf F\cdot d\mathbf r
=\int_0^{2\pi} f(\cos\beta)\sin\beta\,d\theta
-\int_0^{2\pi}f(\cos\alpha)\sin\alpha\,d\theta,
$$
which equals the boxed <surface integral> and verifies <Stokes theorem> with the required <orientation>.
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