= Solution
The radius of the <surface of revolution> is $|y(x)|$, so its area and the curve's <arc length> are
$$
\boxed{A[y]=2\pi\int_{-a}^{a}|y|\sqrt{1+y'^2}\,dx,\qquad
L[y]=\int_{-a}^{a}\sqrt{1+y'^2}\,dx.}
$$
For the proposed curve, $y<0$ on the interior. On this branch $A=-2\pi S$ with $S=\int y\sqrt{1+y'^2}\,dx$. Extremizing $S$ subject to $L=l$ is therefore equivalent to making the actual area stationary. Introduce a <Lagrange multiplier> $\lambda$ and the augmented integrand $F=(y+\lambda)\sqrt{1+y'^2}$. The <Beltrami identity> gives
$$
F-y'F_{y'}=\frac{y+\lambda}{\sqrt{1+y'^2}}=C.
$$
For the <fixed-length stationary surface of revolution>, take
$$
y=\frac{\cosh(kx)-\cosh(ka)}k,\qquad
\lambda=\frac{\cosh(ka)}k,\quad C=\frac1k.
$$
Indeed $y'=\sinh(kx)$ and $\sqrt{1+y'^2}=\cosh(kx)$, so the Beltrami equation holds identically. It also follows directly that $F_y=\cosh(kx)$ and $d(F_{y'})/dx=\cosh(kx)$, verifying the <Euler-Lagrange equation>. The endpoints vanish, and the length condition becomes
$$
\boxed{l=\frac{2\sinh(ka)}k.}
$$
For 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 \b[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.
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