We wish to find a curve $y : C^1[-a, a]$, where $a$ is an unknown parameter, such that $y(\pm a) = 0$, and the volume of the solid of revolution generated by rotating $y$ around the $x$-axis has maximal volume, given that the surface area is fixed. In other words, we let

$$ S(y) = \int_{-a}^a2\pi y\sqrt{1 + (y')^2}dx,\quad V(y) = \int_{-a}^a\pi y^2 dx $$ Then the goal is to find $y$ and $a$ so that $V(y)$ is maximized, subject to $S(y) = A$, $y(\pm a) = 0$.

I'm expecting to end up with $y$ a semicircle, in which case $a = \frac12\sqrt{\frac A\pi}$. Solving this using Lagrange multipliers and Euler-Lagrange equations however has proven tedious.

The augmented problem I end up with is that of finding the stationary functions of the functional $$ \tilde{J}(y; a) = \int_{-a}^a\left(-\pi y^2 + 2\pi\lambda y\sqrt{1 + (y')^2}\right) dx $$ for some real number $\lambda$. If we let $f(y, y')$ be the integrand, then the Euler-Lagrange equation (which has been written in an equivalent form given that $f$ doesn't depend directly on $x$) is $$ -f + y'f_{y'} = \text{const} $$ or, letting $C$ be our constant, $$ \pi y^2 - 2\pi \lambda y\sqrt{1 + (y')^2} + \frac{2\pi\lambda y(y')^2}{\sqrt{1 + (y')^2}} = C $$ Now, one may verify (by plugging it in) that $y(x) = \sqrt{a^2 - x^2}$ is a solution to this equation, where $\lambda = a/2$ and $C = 0$, but for the life of me I cannot solve the equation in general (without the knowledge that the solution is a semicircle a priori).

If we do some rearranging, we can end up with $$ (\pi y^2 - C)\sqrt{1 + (y')^2} - 2\pi\lambda y = 0 $$ in which case introducing the parametrization $y'(x) = \tan\theta(x)$ and rearranging gives $$ y = \lambda\left(\cos\theta\pm\sqrt{\cos^2\theta+c}\right) $$ for some other constant $c$. We can then sub back in $y' = \tan\theta$ to get a separable equation in $\theta$, namely: $$ d\theta\left(-\lambda\cos\theta\pm\frac{2\cos^2\theta}{\sqrt{\cos^2\theta+c}}\right) = dt $$

According to wolframalpha, however, this does not have an analytic antiderivative in general.

How can I arrive at the desired result?


Alright I've figured it out.

The stationary functions of the augmented functional must satisfy not only the Euler-Lagrange equation, but also the additional natural boundary condition $H[y(\pm a)] = 0$, where $H[y] = -f + y'f_{y'}$ (this can be determined just by taking the Gateaux derivative of $\tilde{J}(y; a)$ and setting it equal to zero for all admissible directions).

In this case, the $H[y]$ is also just our Euler-Lagrange equation, and so $H[y(\pm a)] = 0$ tells us that $C$ equals zero, in which case our equation is reduced simply to $$ y = 2\lambda\cos\theta $$ after the parametrization $y' = \tan\theta$.


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