= Solution
{c}
Parametrize the finite <cylindrical surface> patch by
$$
\mathbf R(t,z)=(\cos t,\sin t,z),\qquad-\pi\leq t\leq\pi,\quad0\leq z\leq\cosh t.
$$
The <surface area element> is $|\mathbf R_t\times\mathbf R_z|\,dt\,dz=dt\,dz$. The seam $t=\pm\pi$ has zero <surface area>, so it causes no overcounting. Therefore
$$
\boxed{A=\int_{-\pi}^{\pi}\int_0^{\cosh t}dz\,dt=2\sinh\pi.}
$$
Unrolling the unit <circular cylinder> preserves <surface area>, which explains the equality with the <area> under the <hyperbolic cosine> in the figure. The patch consists of the <cylindrical surface> alone; no flat base is included.
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