I know the problem is traditionally solved via the isoperimetric inequality, but I was hoping to solve it by minimizing a surface of revolution subject to a volume constraint.

The surface area of a surface of revolution is:


and the volume will be:


I'd like to show that the sphere (or in this case the function $x^2+y^2=1$) minimizes the surface area functional for any fixed volume.

The combined action will then be:

$$S=\int_{-1}^{1}2\pi y\sqrt{1+\dot{y}^2}+\lambda\pi y^2 dx$$

The Euler Lagrange Equation simplifies to:

$$-\frac{2 \pi y'(x)^2}{\sqrt{y'(x)^2+1}}+2 \pi \sqrt{y'(x)^2+1}+\frac{2 \pi y(x) y'(x)^2 y''(x)}{\left(y'(x)^2+1\right)^{3/2}}-\frac{2 \pi y(x) y''(x)}{\sqrt{y'(x)^2+1}}+2 \pi \lambda y(x)=0$$

I have no clear idea how to proceed from here. I can solve for $\lambda$ using Mathematica, and get

$$\lambda=\frac{y(x) y''(x)-y'(x)^2-1}{y(x) \left(y'(x)^2+1\right)^{3/2}}$$

The ODE solution is extremely convoluted, involving about half a dozen elliptic integrals, and doesn't seem to be going anywhere. Is there something fundamentally wrong with my approach, or is there an obvious next step in the problem?

EDIT: Using the Beltrami identity instead, and reversing the problem so the volume is varied and a constant surface area is the constraint, the EL equation can be simplified to:

$$y(x)^2+\frac{2\lambda y(x)}{\sqrt{y'(x)^2+1}}=k$$

Which, while much simpler, doesn't get me closer to an answer.

If $k$ is equal to $0$ and $y(0)$ is likewise, then the ODE reduces nicely, and can be solved. One solutions is:

$$y(x)=\sqrt{4 \lambda ^2-x^2}$$

Which is the equation of a circle (which when rotated around the axis becomes a sphere). However, some assumptions are made there which are not necessarily true.

  • $\begingroup$ $S$ denotes both surface area and action. Kind of confusing. $\endgroup$ – Rodrigo de Azevedo May 25 '16 at 0:09
  • $\begingroup$ Fixed. Thanks. Accustomed to using S for almost everything. $\endgroup$ – JAustin May 25 '16 at 0:11
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    $\begingroup$ $k$ should be zero since you should assume that $y(-1) = y(1) = 0$. $\endgroup$ – user99914 May 25 '16 at 2:14

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 $2\lambda$.

$$y(x)=\pm\sqrt{4 \lambda ^2-x^2}$$

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    $\begingroup$ Radius $2 \lambda$ right? $\endgroup$ – ITA Sep 13 '16 at 15:29
  • $\begingroup$ Fixed. Thank you. For the record, I am also accepting my own answer, since the problem is solved. $\endgroup$ – JAustin Sep 14 '16 at 19:29

There are a few problems with the formulation:

  1. 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

$$S = \int_{-1}^1 \left[2\pi y \sqrt{1+y'^2} + \lambda(\pi y^2-C/2)\right]\,dx.$$

  1. 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).
  • $\begingroup$ Where is the $\mathrm{d}x$? $\endgroup$ – Rodrigo de Azevedo May 25 '16 at 1:26
  • $\begingroup$ @RodrigodeAzevedo at the end, of course. added. $\endgroup$ – user7530 May 25 '16 at 1:42
  • $\begingroup$ I had the same thoughts about boundary conditions when I began the problem, and couldn't see how to actually constrain the problem with the boundary value. I also half-deliberately ignored the magnitude of the volume because it makes the integral even more unpleasant. But hey, who doesn't love page-long elliptic integrals?. $\endgroup$ – JAustin May 25 '16 at 1:57
  • $\begingroup$ I eventually decided to reverse the problem and vary the volume subject to a constant surface area, which makes the problem marginally easier, but I'm still not getting anywhere. As I mentioned in my edit, I can get an answer if I assume $k$, $C$, and $y(0)$ are equal to zero, but then the problem isn't very meaningful. $\endgroup$ – JAustin May 25 '16 at 1:57

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 $\theta$.


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