Proving that a sphere has a minimal surface to volume ratio using Calculus of Variations
I spent a few more hours on the problem, and eventually found the solution. No major breakthroughs, but a lot of algebra.
First, I reversed the problem, instead maximizing the volume subject to a surface area constraint. Not strictly necessary, but it simplifies the computation. I also gave the surface area a magnitude $k$, per user7530's comment. The new action is:
$$S=\int_{-\lambda/2}^{\lambda/2}\pi y^2 + 2\lambda\pi y\sqrt{1+\dot{y}^2}-\lambda k~dx$$
The integration boundaries were also changed for convenience.
Then, because the Lagrangian does not depend explicitly on the variable $x$, I was able to apply the Beltrami identity instead of the traditional EL equation.
$$\frac{d}{dx}\left(\pi y(x)^2 + 2\lambda\pi y(x) \sqrt{1+y'(x)^2}-\lambda k - \frac{2\pi\lambda y'(x)^2y(x)}{\sqrt{1+y'(x)^2}}\right) = 0$$
Integrating both sides, we find that the functional inside the derivative is equal to a constant. Because $y$ is equal to $0$ at $-\lambda/2$ and $\lambda/2$, and each term in the derivative depends on $y(x)$ except for the $\lambda k$ term, the arbitrary additive constant must be $-\lambda k$, which cancels with the $-\lambda k$ on the left side.
The DE becomes:
$$\pi y(x)^2 + 2\lambda\pi y(x) \sqrt{1+y'(x)^2} - \frac{2\pi\lambda y'(x)^2y(x)}{\sqrt{1+y'(x)^2}} = 0$$
Multiplying the second term by $\frac{\sqrt{1+y'(x)^2}}{\sqrt{1+y'(x)^2}}$, that in turn simplifies to:
$$y(x)^2 + \frac{2\lambda y(x)}{\sqrt{1+y'(x)^2}}=0$$
This can be solved for $y(x)$, and with the constant chosen to equal zero, the solution is the equation of a circle of radius $2\lambda$.
$$y(x)=\pm\sqrt{4 \lambda ^2-x^2}$$
There are a few problems with the formulation:
- The volume constraint: right now you are constraining the total volume to be zero (check by differentiating $S$ with respect to $\lambda$). If you want the total volume to be equal to some fixed value $C$, you need instead the action
$$S = \int_{-1}^1 \left[2\pi y \sqrt{1+y'^2} + \lambda(\pi y^2-C/2)\right]\,dx.$$
- The boundary conditions: your problem restricts your surface of revolution to touch the bounding points $(\pm 1,0,0)$. Of course, depending on the value of $C$ the sphere of volume $C$ will not usually touch these points. You can still minimize surface area, given these boundary conditions, but realize you are solving a harder version of the isoperimetric problem. (Which I also cannot solve at the moment).
I have to solve a similar problem, here are my thoughts.
We could express the problem in spherical coordinates with the domain $\Omega \in \mathbb{R}^3$ delimited by $\partial \Omega$, the Surface $A = \vert \partial \Omega \vert$ and the Volume $V = \vert \Omega \vert = constant$:
$$min\{A[R]\} = \int_{\partial\Omega} R^2(\varphi,\theta) {\rm sin}(\theta){\rm d}\varphi {\rm d}\theta$$ $$constant = V = \int_{\Omega} r^2 {\rm sin}(\theta) {\rm d} r {\rm d} \varphi {\rm d} \theta = \int_{\partial\Omega} \left( \int \limits_{0}^{R(\varphi,\theta)} r^2 {\rm d} r \right) {\rm sin}(\theta) {\rm d}\varphi {\rm d}\theta = \int_{\partial\Omega} \frac{R^3(\varphi,\theta)}{3} {\rm sin}(\theta) {\rm d}\varphi {\rm d}\theta $$
Now we have two integrals over the same domain $\partial\Omega$ and we can use formulas for problems under isoperimetric conditions.
With $F := R^2(\varphi,\theta){\rm sin}(\theta)$ and $G := \frac{R^3(\varphi,\theta)}{3} {\rm sin}(\theta)$ and $F^* = F + \lambda G$, we can use the formula $$ 0 = \underbrace{ \frac{{\rm d}}{{\rm d} \theta} \frac{ \partial F^* }{ \partial \frac{ \partial R }{ \partial \theta}} }_{=0} + \underbrace{ \frac{{\rm d}}{{\rm d} \varphi} \frac{ \partial F^* }{ \partial \frac{ \partial R }{ \partial \varphi}} }_{=0} - \frac{ \partial F^* }{ \partial R} $$
Therein is $$ F^* = \underbrace{ R^2(\varphi,\theta) {\rm sin}(\theta) }_{=F} + \lambda \cdot \underbrace{ \frac{R^3(\varphi,\theta)}{3} {\rm sin}(\theta) }_{=G} $$
So it follows $$ 0 = \frac{ \partial F^* }{ \partial R} = 2 R (\varphi,\theta){\rm sin}(\theta) + \lambda \cdot R^2(\varphi,\theta) {\rm sin}(\theta) $$
Solving the equation for $R$: $$ R(\varphi,\theta) = \frac{2}{\lambda} $$
So we have shown, that $R(\varphi,\theta) = constant$, because $\lambda$ is constant in $\varphi$ and $\theta$.