 Solution

Stokes theorem stateswith 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 isOn the unit sphere, the outward normal is $\hat{\mathbf{n}}=(x,y,z)$. In spherical coordinates, the surface integral over the upper hemisphere isFor the boundary calculation, parametrize $\mathbf{r}(\phi )=(\cos \phi ,\sin \phi ,0)$, with $0\le \phi \le 2\pi$. Then $\mathbf{A}=(2\cos \phi -\sin \phi ,0,0)$ andThus 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.