Proving that a sphere has a minimal surface to volume ratio using Calculus of Variations 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:
$$A=2\pi\int_{-1}^{1}y\sqrt{1+\dot{y}^2}~dx$$
and the volume will be:
$$V=\pi\int_{-1}^{1}y^2~dx$$
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.

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