I'm trying to prove that the space region that contains maximum volume for a given fixed surface area is a sphere. I've done this for the analogous 2d isoperimetric problem using calculus of variations.

Here's what I have:

Let our region of space be V and the enclosing surface $\displaystyle \partial V$. We want to minimize the volume $\displaystyle I = \iiint\limits_{V}\,dV$

subject to the constraint $\displaystyle S = \iint\limits_{\partial V} \, dS$.

We parametrize $\displaystyle \partial V$ with parameters u and v. Then $\displaystyle dS=\left | \mathbf{r}_u \times \mathbf{r}_v \right | du dv$, where $\displaystyle \mathbf{r}_u = \left (\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} \right ) = \left (x_u , y_u , z_u \right )$ and $\displaystyle \mathbf{r}_v = \left (\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} \right ) = \left (x_v , y_v , z_v \right )$.

Let us define $\displaystyle \mathbf{F}(x,y,z) = \frac{x}{3}\mathbf{i} + \frac{y}{3}\mathbf{j} + \frac{z}{3}\mathbf{k}$. Then $\displaystyle \nabla \cdot

\mathbf{F}(x,y,z) = 1$, and $\displaystyle \iiint\limits_{V}\nabla \cdot \mathbf{F}(x,y,z)\,dV = \iiint\limits_{V}\,dV = I$.

Thus, by the divergence theorem,

$\displaystyle I = \iiint\limits_{V} \nabla \cdot \mathbf{F}(x,y,z) \,dV $

$\displaystyle = \iint\limits_{\partial V} \mathbf{F} \cdot \mathbf{\hat n} \, dS$,

where $\displaystyle \mathbf{\hat n}$ is the outward unit normal vector. For our parametrization of $\displaystyle \partial V$, we have $\displaystyle \mathbf{\hat n} dS = \mathbf{r}_u \times \mathbf{r}_v du dv$ and thus

$\displaystyle I = \iint\limits_{\partial V} \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, du \, dv$, and thus we want to find the extremum of:

$\displaystyle M = \iint\limits_{\partial V} \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) + \lambda \left | \mathbf{r}_u \times \mathbf{r}_v \right | \, du \, dV$

With $\displaystyle \mathbf{r}_u \times \mathbf{r}_v = (y_u z_v - z_u y_v) \mathbf{i} + (z_u x_v - x_u z_v) \mathbf{j} + (x_u y_v - y_u x_v) \mathbf{k}$

And so $\displaystyle \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) = \tfrac{1}{3} (y_u z_v - z_u y_v) x + (z_u x_v - x_u z_v) y + (x_u y_v - y_u x_v) z)$

and $\displaystyle | \mathbf{r}_u \times \mathbf{r}_v | = \sqrt{(y_u z_v - z_u y_v)^2 + (z_u x_v - x_u z_v)^2 + (x_u y_v - y_u x_v)^2}$

And we put these into the formula for M to get a double integral overdudvof a function $\displaystyle L(x,y,z,x_u,y_u,z_u,x_v,y_v,z_v)$.

Were there only one independent variable, I could find the Euler-Lagrange equations for x, y and z from L; or more specifically, I'd use the lack of explicit dependence on the independent variable to apply the Beltrami Identity to find lower order differential equations for x, y, z. However, this has two independent variables. How do I set up and solve the Euler-Lagrange equations for this sort of problem? Is there a Beltrami Identity equivalent for multiple independent variables when the integrand has no explicit dependence on both of the independent variables?

--Kevin C.