= Solution
The <Stokes theorem> states
$$
\int_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS
=\oint_{\partial S}\mathbf F\cdot d\mathbf r,
$$
where the boundary orientation is induced by the chosen normal through the right-hand rule.
For $\mathbf F=(y^2z,xz+2xyz,0)$,
$$
\boxed{\nabla\times\mathbf F=(-x-2xy,\ y^2,\ z),}
$$
and its divergence is $(-1-2y)+2y+1=0$, verifying $\nabla\cdot(\nabla\times\mathbf F)=0$.
Orient the paraboloid $z=3-x^2-y^2$, $1\leq x^2+y^2\leq4$, upward and the planar annulus $z=-1$ downward. Their outer boundary orientations cancel. The remaining boundary is the unit circle at $z=-1$ counterclockwise and the unit circle at $z=2$ clockwise, viewed from above. On a unit circle at height $z_0$, direct parametrization gives
$$
\oint_{\rm CCW}\mathbf F\cdot d\mathbf r=\pi z_0.
$$
Hence the boundary integral is $-\pi-2\pi=-3\pi$.
On the lower annulus, $(\nabla\times\mathbf F)\cdot(0,0,-1)=1$, giving $3\pi$. On the paraboloid, $d\mathbf S=(2x,2y,1)\,dx\,dy$, so the flux integrand is
$$
-2x^2-4x^2y+2y^3+3-r^2.
$$
The odd terms integrate to zero over the annulus, while the rest integrates to $-6\pi$. The total surface integral is $3\pi-6\pi=-3\pi$, equal to the boundary integral.
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