= Solution
{c}
Choose the <scalar field>
$$
f(x,y,z)=\frac{x^2+y^2+z^2}{2},\qquad\nabla f=\mathbf r.
$$
The allowed <vector-valued Stokes identity> gives
$$
\oint_{\partial\Sigma}f\,d\mathbf r
=\iint_\Sigma\mathbf n\times\mathbf r\,dS
=-\iint_\Sigma\mathbf r\times d\mathbf S.
$$
Every boundary point lies on the unit <circle> in the $xy$-plane, so $f=1/2$ throughout the boundary. Since the <line integral> of $d\mathbf r$ around a closed <curve> is the zero displacement,
$$
\boxed{\iint_\Sigma\mathbf r\times d\mathbf S
=-\frac12\oint_{\partial\Sigma}d\mathbf r=\mathbf0}.
$$
Only the boundary value of this <scalar field> mattered: the interior of the <surface> need not be part of a <sphere>. Either consistent choice of <orientation> gives the same zero answer.
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