= Solution
For a smooth oriented <surface> $S$ with positively oriented <boundary> $C=\partial S$, <Stokes theorem> states
$$
\boxed{\iint_S(\nabla\times\mathbf F)\cdot d\mathbf S
=\oint_C\mathbf F\cdot d\mathbf r}.
$$
Here $d\mathbf S=\mathbf n\,dA$ uses the chosen <unit normal vector>, while $d\mathbf r$ is tangent to the directed <boundary>. The <boundary> <orientation> follows the <right-hand rule>: curling the fingers along $C$ makes the thumb point in the normal direction. The field must be continuously differentiable on a neighbourhood of the <surface>, with the usual piecewise-smooth allowance for its <boundary>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2-stokes-orientation.png]
{title=Outward hemisphere orientation, its counterclockwise boundary, and tangent and normal elements}
For the upper <hemisphere> with outward normal, the induced equatorial direction is counterclockwise as viewed from positive $z$.
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