= Solution
{c}
Take the boundary pieces in the order
$$
(2,0,0)\longrightarrow(0,2,0)\longrightarrow(0,0,4)\longrightarrow(2,0,0).
$$
On the circular arc $C_1$, use $(x,y,z)=(2\cos\theta,2\sin\theta,0)$ for $0\leq\theta\leq\pi/2$. Then $\mathbf F=(0,0,1)$ and $dz=0$, so its <line integral> is zero.
On $C_2$, use $x=0$, $z=4-y^2$, with $y$ decreasing from two to zero. Since $dx=0$ and $F_y=0$, $\mathbf F\cdot d\mathbf r=dz$, giving four. On $C_3$, use $y=0$, $z=4-x^2$, with $x$ increasing from zero to two. Again $\mathbf F\cdot d\mathbf r=dz$, giving minus four. The total is $\boxed{0+4-4=0}$, agreeing with <Stokes theorem>.
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