= Solution
Take fixed-endpoint variations $y_\varepsilon=y+\varepsilon\eta$ with $\eta(a)=\eta(b)=0$. The <first variation> is
$$
\delta I=\int_a^b(f_y\eta+f_{y'}\eta')dx=\int_a^b\left(f_y-\frac d{dx}f_{y'}\right)\eta\,dx.
$$
The boundary term vanishes by the endpoint conditions. The <fundamental lemma of the calculus of variations> gives the <Euler-Lagrange equation> $f_y-d(f_{y'})/dx=0$. Along its solutions,
$$
\frac d{dx}(f-y'f_{y'})=f_x+y'\left(f_y-\frac d{dx}f_{y'}\right)=f_x.
$$
When $f_x=0$, this proves the <Beltrami identity> $f-y'f_{y'}=\text{constant}$.
For the surface $z=F(y)$, the <line element> is $ds^2=dx^2+[1+F'(y)^2]dy^2$. The arc-length integrand for a graph $y(x)$ is $\mathcal L=\sqrt{1+[1+F'(y)^2]y'^2}$. It has no explicit $x$ dependence and satisfies $\mathcal L-y'\mathcal L_{y'}=1/\mathcal L$. Thus $[1+F'(y)^2]y'^2$ is constant. Choosing the fixed direction along a smooth nonconstant segment and integrating yields
$$
\boxed{\int^y\sqrt{1+F'(s)^2}\,ds=ax+b.}
$$
The same result follows by defining the transverse arc coordinate $u(y)=\int^y\sqrt{1+F'(s)^2}ds$, which changes the metric to $ds^2=dx^2+du^2$. Its <geodesics> are straight lines in $(x,u)$ coordinates. This proves the formula for <geodesics on a generalized cylinder> and shows it gives locally shortest paths. The additional straight lines $x=\text{constant}$ in the unrolled plane are also geodesics, but cannot be represented as graphs $y(x)$. Classical differential-geometric statements assume the surface has sufficient smoothness; the arc-length coordinate itself only requires the displayed integral to be well defined.
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