= Solution
{c}
The <partial derivatives> of this <parametrized surface> are
$$
\mathbf r_s=(t\cos t,t\sin t,2),\qquad
\mathbf r_t=s(\cos t-t\sin t,\sin t+t\cos t,0).
$$
Their <cross product> is
$$
\mathbf r_s\times\mathbf r_t
=s\bigl(-2(\sin t+t\cos t),\ 2(\cos t-t\sin t),\ t^2\bigr).
$$
The mixed terms cancel in its squared <Euclidean norm>:
$$
|\mathbf r_s\times\mathbf r_t|^2
=s^2\bigl(4(1+t^2)+t^4\bigr)=s^2(t^2+2)^2.
$$
Since $s>0$, the <surface area element> is $s(t^2+2)\,ds\,dt$. There is no repeated coverage: $z=2s$ recovers $s$, and $\sqrt{x^2+y^2}=st$ then recovers $t$. Therefore the <surface area> is
$$
\begin{aligned}
\operatorname{area}(S)
&=\int_0^{2\pi}\int_0^1s(t^2+2)\,ds\,dt\\
&=\frac12\left[\frac{t^3}{3}+2t\right]_0^{2\pi}
=\boxed{\frac{4\pi^3}{3}+2\pi}.
\end{aligned}
$$
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