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!
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!