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!