= Solution
Away from the origin, write $\mathbf F=r^{-\beta-1}\mathbf r$. The <curl product rule for a scalar field> and the <gradient> of a radial function give
$$
\nabla\times\mathbf F
=\nabla(r^{-\beta-1})\times\mathbf r
+r^{-\beta-1}\nabla\times\mathbf r
=-(\beta+1)r^{-\beta-3}\mathbf r\times\mathbf r=0.
$$
Consequently the <surface integral> of <curl> over the upper hemisphere is zero. Its outward <unit normal vector> induces a counterclockwise equator when viewed from above. Parametrize that boundary by
$$
\mathbf r(\phi)=r_2(\cos\phi,\sin\phi,0),\qquad
d\mathbf r=r_2(-\sin\phi,\cos\phi,0)\,d\phi.
$$
The radial <vector field> is perpendicular to this tangent, so
$$
\boxed{\oint_{\partial S}\mathbf F\cdot d\mathbf r=0
=\iint_S(\nabla\times\mathbf F)\cdot\hat{\mathbf n}\,dS.}
$$
This verifies <Stokes theorem> directly without passing through the singular origin.
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