= Solution
{c}
On the unit <circle>, the positively oriented unit <tangent vector> is $\mathbf u=(-y,x,0)$, and $|\mathbf u|=1$. Hence $d\mathbf r=\mathbf u\,ds$ and
$$
\oint_C(\psi\mathbf u)\cdot d\mathbf r=\oint_C\psi\,ds.
$$
Since the <scalar field> $\psi$ has no $z$ dependence, the <curl product rule for a scalar field> gives
$$
\begin{aligned}
\nabla\times(\psi\mathbf u)
&=\psi(0,0,2)+(\psi_x,\psi_y,0)\times(-y,x,0)\\
&=(0,0,2\psi+x\psi_x+y\psi_y).
\end{aligned}
$$
On the unit upper <hemisphere> $S_1$, the outward <unit normal vector> is $(x,y,z)$, so its vertical component is $z=\cos\theta$. Applying the <Stokes theorem> to the <vector field> $\psi\mathbf u$ therefore proves
$$
\boxed{\oint_C\psi\,ds
=\iint_{S_1}\left(2\psi+x\frac{\partial\psi}{\partial x}
+y\frac{\partial\psi}{\partial y}\right)\cos\theta\,dS}.
$$
The smoothness needed here is that $\psi$ extend as a continuously differentiable <scalar field> to a neighbourhood of the projected unit <disk>. The left side is a scalar <line integral> with respect to <arc length>; the positive <boundary orientation> is used when identifying it with the vector <line integral>.
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