 Solution

The radius of the surface of revolution is $|y(x)|$, so its area and the curve's arc length areFor the proposed curve, $y<0$ on the interior. On this branch $A=-2\pi S$ with $S=\int y\sqrt{1+y^{\prime 2}}dx$. Extremizing $S$ subject to $L=l$ is therefore equivalent to making the actual area stationary. Introduce a Lagrange multiplier $\lambda$ and the augmented integrand $F=(y+\lambda )\sqrt{1+y^{\prime 2}}$. The Beltrami identity givesFor the fixed-length stationary surface of revolution, takeIndeed $y^{\prime }=\sinh (kx)$ and $\sqrt{1+y^{\prime 2}}=\cosh (kx)$, so the Beltrami equation holds identically. It also follows directly that $F_y=\cosh (kx)$ and $d(F_{y^{\prime }})/dx=\cosh (kx)$, verifying the Euler-Lagrange equation. The endpoints vanish, and the length condition becomesFor endpoint-fixed variations preserving this length, the augmented first variation is zero; subtracting the zero length variation proves that the area's first variation is zero. Its value, if desired, is $A=\pi [\sinh (2ka)-2ka]/k^2$.

To prove uniqueness, put $s=ka>0$. The equation is $\sinh s/s=l/(2a)>1$. Its derivative is $(s\cosh s-\sinh s)/s^2$. The numerator is zero at $s=0$ and has derivative $s\sinh s>0$, so the ratio increases strictly from one to infinity. The intermediate value theorem gives exactly one positive $k$. The stated upper bound on $l$ is not needed for this uniqueness argument. Stationarity has been proved, without making an unsupported global-minimum claim.