 Solution

The Stokes theorem stateswhere the boundary orientation is induced by the chosen normal through the right-hand rule.

For $\mathbf{F}=(y^{2}z,xz+2xyz,0)$,and its divergence is $(-1-2y)+2y+1=0$, verifying $\nabla \cdot (\nabla \times \mathbf{F})=0$.

Orient the paraboloid $z=3-x^2-y^2$, $1\le x^2+y^2\le 4$, 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 givesHence 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, $d\mathbf{S}=(2x,2y,1)dxdy$, so the flux integrand isThe 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.