 Solution

Choose the upward orientation of the paraboloid cap. Its boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise as viewed from above. By Stokes theorem,On the boundary, $dz=0$ and $A\cdot dr=ydx-3xydy$. Parametrize by $x=2\cos \theta$, $y=2\sin \theta$, $0\le \theta \le 2\pi$. ThenThe second term integrates to zero over a full period, while the first contributes $-4\pi$. Equivalently, replace the cap by the upward disk at $z=1$: there $(\nabla \times A{)}_z=-3y-1$, whose integral is $-4\pi$ by symmetry.

An orientation is not stated in the original question. The displayed result uses the upward normal, also the outward normal on the curved top of the enclosed cap; the opposite orientation gives $4\pi$.