Sphere has the minimum surface area?

[persia7]

How do you prove that for a given volume, sphere has the minimum surface area?

 

[mfb]

This is a special case of the isoperimetric inequality in 3 dimensions, with that keyword you should find proofs.

 

[HallsofIvy]

It is a standard problem in "Calculus of Variations" so you might also check that.

 

[persia7]

i tried to find proofs but i don't find please show me proofs

 

[jackmell]

i tried to find proofs but i don't find please show me proofs

How about first doing it for a surface of revolution, say for y(x) between -1 and 1. Then the surface is:

$$S=2\pi \int_{-1}^1 \sqrt{1+(y')^2}dx$$

and the volume of this surface of revolution is:

$$V=\pi \int_{-1}^1 y^2 dx$$
via discs.

Now, in the language of variation, we wish to minimize S while keeping V constant. The Euler condition for this problem is:

$$\frac{\partial H}{\partial y}-\frac{d}{dx}\frac{\partial H}{\partial y'}=0$$
with
$$H=V+\lambda S$$

Now, I've never solved this particular variational problem before ok, but if it were mine, I'd first see if I can get a circle out of Euler's equation thus showing me that a sphere has the minimum surface area for a given volume.

 

[persia7]

How about first doing it for a surface of revolution, say for y(x) between -1 and 1. Then the surface is:

$$S=2\pi \int_{-1}^1 \sqrt{1+(y')^2}dx$$

and the volume of this surface of revolution is:

$$V=\pi \int_{-1}^1 y^2 dx$$
via discs.

Now, in the language of variation, we wish to minimize S while keeping V constant. The Euler condition for this problem is:

$$\frac{\partial H}{\partial y}-\frac{d}{dx}\frac{\partial H}{\partial y'}=0$$
with
$$H=V+\lambda S$$

Now, I've never solved this particular variational problem before ok, but if it were mine, I'd first see if I can get a circle out of Euler's equation thus showing me that a sphere has the minimum surface area for a given volume.

can you show me a proof in internet?

 

[Chronos]

You have your choice of proofs: http://cornellmath.wordpress.com/2008/05/16/two-cute-proofs-of-the-isoperimetric-inequality/, or, http://math.stanford.edu/~jbooher/expos/isoperimetric_promys.pdf

 

[persia7]

thanks to all

 

[jackmell]

can you show me a proof in internet?

I worked with it a bit but ran into a DE I couldn't solve or else I had errors. But that's ok guys, don't tell me how to do it. I rag on others here about that: pretty soon you'll get to a problem that no one else on Earth can help you with and then you will have to go all by yourself so might as well get some practice. :)

Oh yeah, thanks guys for those references about isoperimetric inequality. All news to me. :)

 

[HallsofIvy]

Well, what do you expect? That is a very technical, difficult proof, involving, as I said before, "Calculus of Variations". Whole courses are taught on that subject! It isn't something you can expect to go over in an afternoon!

 

[jackmell]

I really was expecting the variational problem I set up to fall right through. Spent about an hour on it yesterday. I'm missing something because a get a really messy DE but maybe my approach is not the easiest variational formulation to use. Still though if the DE I derived at is correct, then I should be able to back-substitute the equation for a circle and get equality but when I do that I do not get equality so I got a problem.

 

[jackmell]

How about first doing it for a surface of revolution, say for y(x) between -1 and 1. Then the surface is:

$$S=2\pi \int_{-1}^1 \sqrt{1+(y')^2}dx$$

Dang it! I'm missin' a y:

$$S=2\pi \int_{-1}^1 y \sqrt{1+(y')^2}dx$$


I think I need to review some things first . . .

 

[jackmell]

I get for the Euler equation:

$$1+(y')^2-yy''+\lambda y\left(1+(y')^2\right)^{3/2}=0$$
where $\lambda$ is an arbitrary parameter.

Now, I'm going to cheat a little bit because I'm not good at this: I know the solution should be a circle, so when I substitute $y(x)=\sqrt{1-x^2}$ into the left side of the DE above, in order for it to be zero, $\lambda=-2$. Therefore, we want to solve:

$$1+(y')^2-yy''-2y\left(1+(y')^2\right)^{3/2}=0$$

I started a thread in the DE forum because I needed help to solve it. It's here:

https://www.physicsforums.com/showthread.php?t=702671

Thanks to fzero and Jacquelin, we obtain:

$$\frac{dy}{dx}=\frac{\sqrt{y^2-(y^2-c_1)^2}}{y^2-c_1}$$

Now, just for the purpose of actually making progress with this problem albeit not the best way, I'm going to let $c_1=0$. Then we wish to solve:

$$\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{y}$$

with the solution being:

$$y^2+(x+c_2)^2=1$$

Letting again $c_2=0$, we get the desired extremal: $x^2+y^2=1$.

All that remains to be done is to justify my particular values of the constants $c_0, c_1,\lambda$.

If someone can do this, I think we'd have a pretty decent proof of this theorem.

Also I forgot, this is only a necessary condition for a minimum. I think we would have to look into the second variation to prove this is a minimum.