= Solution
<Stokes theorem> states
$$
\boxed{\oint_C\mathbf A\cdot d\mathbf r=\iint_S(\nabla\times\mathbf A)\cdot\hat{\mathbf n}\,dS},
$$
with boundary orientation induced by the chosen <surface> normal and with sufficient smoothness of the field and <surface>. Choose the outward normal on the upper unit hemisphere; its equatorial boundary runs counterclockwise as viewed from $+z$.
For the stated field, the <curl> is
$$
\nabla\times\mathbf A=\bigl(-2yz-(-2yz),\ 0-0,\ 0-(-1)\bigr)=\hat{\mathbf z}.
$$
On the unit sphere, the outward normal is $\hat{\mathbf n}=(x,y,z)$. In <spherical coordinates>, the <surface> integral over the upper hemisphere is
$$
\int_0^{2\pi}\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\,d\varphi=\pi.
$$
For the boundary calculation, parametrize $\mathbf r(\varphi)=(\cos\varphi,\sin\varphi,0)$, with $0\le\varphi\le2\pi$. Then $\mathbf A=(2\cos\varphi-\sin\varphi,0,0)$ and
$$
\oint_C\mathbf A\cdot d\mathbf r=\int_0^{2\pi}\left[-2\cos\varphi\sin\varphi+\sin^2\varphi\right]d\varphi=\pi.
$$
Thus both sides equal $\boxed{\pi}$. Reversing the hemisphere orientation reverses both signs. The hemisphere's flux calculation can also be read as the projected area of its equatorial unit disk.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-conservative-paths-and-stokes-surface.png]
{title=Two equal-endpoint paths and the oriented hemisphere used in Stokes' theorem}
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