= Solution
Direct use of <partial derivatives> gives
$$
\boxed{\nabla\times\mathbf F=(x^2+2xy,-2xy-y^2,2).}
$$
With the upward <oriented surface element> found in part (b), the <curl> flux integrand over the <annulus> $A$ becomes
$$
2+\frac{-x^3-2x^2y+2xy^2+y^3}{\sqrt{1+x^2+y^2}}.
$$
Every term in the numerator is odd in $x$ or $y$, so its integral over the symmetric <annulus> is zero. Thus the <surface integral> is
$$
\int_S(\nabla\times\mathbf F)\cdot d\mathbf S
=2\operatorname{Area}(A)=2\pi(2^2-1^2)=6\pi.
$$
Each boundary circle has constant $z$, hence $dz=0$. Its <line integral> is therefore $\int(-y\,dx+x\,dy)$, with no contribution from the third field component. An anticlockwise radius-$r$ circle gives $2\pi r^2$. The outward boundary orientation is anticlockwise at $r=2$ and clockwise at $r=1$, so
$$
\oint_{\partial S}\mathbf F\cdot d\mathbf x=8\pi-2\pi=\boxed{6\pi}.
$$
The two computed values agree, verifying <Stokes theorem> with both boundary components. A downward choice would give $-6\pi$ on both sides.
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