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Let $b > a > 0$. Fixing $y(a) = y(b) = 0$ and the total length $L = \int_a^b \sqrt{1+y'^2} dx$, I want to find the curve $y(x)$ that maximizes the volume of the (roughly toroidal) volume between the $x-z$ plane and the surface of revolution of $y(x)$ about the $y$ axis; that is, maximize $\int_a^b x y\, dx$. Can this be solved analytically?

This question is similar to Calculus of variations: find $y(a/2)$ if $y(x)$ maximizes the volume of rotation, but I want to rotate the curve about the other axis. (I also didn't see a way to enforce $y(b) = 0$ in that approach.) Following the solution approach there blindly and using WolframAlpha on the final differential equation results in a monstrous solution that doesn't show any signs of passing through $0$ at $x=b$.

The only sanity check I can think of is that when $a, b \gg L$, the problem is very close to the standard isoperimetric inequality, so the resulting curve should be close to a circular arc passing through $a$ and $b$ and having total length $L$. I am interested in the regime when $a$, $b$, and $L$ are all comparable.

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  • $\begingroup$ Can you clarify what you're asking for in your first paragraph? I think something was mistyped. $\endgroup$ – lordoftheshadows Sep 29 '16 at 19:38
  • $\begingroup$ I want to find the curve $y=y(x)$ that maximizes the value of $\int_a^b xy\, dx$ subject to fixing the length $\int_a^b \sqrt{1+y'^2} dx = L$ and the endpoints $y(a)=y(b)=0$. (This is equivalent to maximizing the volume of the shape bounded by the $xz$ plane and the surface formed by rotating the curve $y=y(x)$ about the $y$ axis.) $\endgroup$ – Elena Sep 29 '16 at 22:04

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