= Solution
The downward drop from the starting rim to angle $\theta$ is $R\cos\theta$. With no dissipative forces, <conservation of energy> gives $mv^2/2=mgR\cos\theta$, so $v=\sqrt{2gR\cos\theta}$. Dividing the <line element> by this speed yields
$$
\boxed{T[z]=\frac1{\sqrt{2gR}}\int_{-\pi/2}^{\pi/2}\sqrt{\frac{R^2+z'^2}{\cos\theta}}\,d\theta.}
$$
The endpoint singularities are integrable for the minimizing path. The integrand is independent of $z$; its <Euler-Lagrange equation> gives a constant $K$, after absorbing the overall prefactor:
$$
\frac{z'}{\sqrt{\cos\theta}\sqrt{R^2+z'^2}}=K.
$$
Solving for $z'$ and taking the positive branch required by $z_2>z_1$ gives
$$
\boxed{\frac{dz}{d\theta}=\frac{RK\sqrt{\cos\theta}}{\sqrt{1-K^2\cos\theta}},\qquad0<K<1.}
$$
The constant is fixed implicitly by
$$
z_2-z_1=RK\int_{-\pi/2}^{\pi/2}\frac{\sqrt{\cos\theta}}{\sqrt{1-K^2\cos\theta}}\,d\theta.
$$
This integral is strictly increasing from zero to infinity as $K$ increases from zero to one, so every finite positive separation selects a unique $K$. The time integrand has second <derivative> $R^2/[\sqrt{\cos\theta}(R^2+z'^2)^{3/2}]>0$ with respect to $z'$. Its supporting-line inequality, integrated between fixed endpoints with constant first <derivative> with respect to $z'$, proves this stationary path is the unique <global minimum> within the stated single-valued class.
As $\theta\downarrow-\pi/2$, $z'\to0$. Therefore the skater should initially head \b[directly down the curved ramp, perpendicular to the cylinder axis], and gain axial motion later. The path differs from the constant-pitch shortest route because extra distance is cheaper in time near the faster, lower part of the ramp. This is the <brachistochrone on a horizontal half-cylinder>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1-half-cylinder-paths.png]
{title=Shortest and fastest routes on the unrolled half-cylinder for $z_2-z_1=2R$}
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