minimal surface of revolution when endpoints on x-axis? What is the formula for the planar curve through $(\pm a,0)$ of fixed length $l$ which has minimal-area surface of revolution when rotated about the x-axis?
I get the area of the surface to be $2\pi\int^a_{-a}y\sqrt{1+y'^2}dx$. Then the first integral of the Euler-Lagrange equation (Beltrami identity) is $y\sqrt{1+y'^2}-\frac{yy'^2}{\sqrt{1+y'^2}} = k$, which has general solution $y(x)=k\cosh(\frac{x-C}{k})$. But putting in the initial conditions $y(\pm a)=0$ gives no nontrivial solutions!
Sources such as here say the general solution is indeed $y(x)=k\cosh(\frac{x-C}{k})$, but only when the endpoints of the curve are not on the x-axis. The question I'm looking at says the solution in this case is $y(x)=k(\cosh\frac{x}{k}-\cosh\frac{A}{k})$, but this doesn't seem to satisfy the right ODE!
Many thanks for any help with this!
 A: The only minimal  surfaces of revolution about the $x$-axis are a flat plane (orthogonal to the axis) and the catenoid. The plane does intersect the axis, the catenoid simply does not. That's life. The catenoid can be translated and scaled without loosing minimality. Meanwhile, it is physically stable only for a bounded subset. If we take two equal circles of wire, and dip them in soap solution so as to form a sort of hollow tube of soap film between them, the circles (if kept co-axial) can be separated only to roughly 2/3 of the diameter of the circles. More separation and the soap film pops.   
To sum up, if you have a problem in the calculus of variations, there is little assurance that a smooth/continuous minimizer exists. If you attempt to work with this with just O.D.E. techniques the result is often nonsense.  
A: Answering my own question here...
The fact that curve length is fixed amounts to a functional constraint. So we need to use both the Euler-Lagrange equation and Lagrange multipliers. The functional to minimise is $\int^a_{-a}(2\pi y-\lambda)\sqrt{1+y'^2}dx$, where $\lambda$ is a Lagrange multiplier. For this functional, the first integral of the Euler-Lagrange equation gives a different ODE from what I got above, which has the solution $y(x)=k(\cosh\frac{x}{k}-\cosh\frac{a}{k})$ for some $k$ depending on $l$, as required.
@Will Jagy: given the endpoint constraints but not the length constraint, the curve which absolutely minimises surface area would be the x-axis, which gives a trivial surface. I can't see where you get the flat plane from? To get that as a surface of revolution, the curve would have to be a vertical line, but this doesn't satisfy the endpoint constraints!
