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

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

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mfb

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HallsofIvy

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

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i tried to find proofs but i dont find please show me proofs

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How about first doing it for a surface of revolution, say for y(x) between -1 and 1. Then the surface is:i tried to find proofs but i dont find please show me proofs

[tex]S=2\pi \int_{-1}^1 \sqrt{1+(y')^2}dx[/tex]

and the volume of this surface of revolution is:

[tex]V=\pi \int_{-1}^1 y^2 dx[/tex]

via discs.

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

[tex]\frac{\partial H}{\partial y}-\frac{d}{dx}\frac{\partial H}{\partial y'}=0[/tex]

with

[tex]H=V+\lambda S[/tex]

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.

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can you show me a proof in internet?How about first doing it for a surface of revolution, say for y(x) between -1 and 1. Then the surface is:

[tex]S=2\pi \int_{-1}^1 \sqrt{1+(y')^2}dx[/tex]

and the volume of this surface of revolution is:

[tex]V=\pi \int_{-1}^1 y^2 dx[/tex]

via discs.

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

[tex]\frac{\partial H}{\partial y}-\frac{d}{dx}\frac{\partial H}{\partial y'}=0[/tex]

with

[tex]H=V+\lambda S[/tex]

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.

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Chronos

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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

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thanks to all

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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. :)can you show me a proof in internet?

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

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HallsofIvy

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

[tex]S=2\pi \int_{-1}^1 \sqrt{1+(y')^2}dx[/tex]

[tex]S=2\pi \int_{-1}^1 y \sqrt{1+(y')^2}dx[/tex]

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

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I get for the Euler equation:

[tex]1+(y')^2-yy''+\lambda y\left(1+(y')^2\right)^{3/2}=0[/tex]

where [itex]\lambda[/itex] 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 [itex]y(x)=\sqrt{1-x^2}[/itex] into the left side of the DE above, in order for it to be zero, [itex]\lambda=-2[/itex]. Therefore, we want to solve:

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

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:

[tex]\frac{dy}{dx}=\frac{\sqrt{y^2-(y^2-c_1)^2}}{y^2-c_1}[/tex]

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

[tex]\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{y}[/tex]

with the solution being:

[tex]y^2+(x+c_2)^2=1[/tex]

Letting again [itex]c_2=0[/itex], we get the desired extremal: [itex]x^2+y^2=1[/itex].

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

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.

[tex]1+(y')^2-yy''+\lambda y\left(1+(y')^2\right)^{3/2}=0[/tex]

where [itex]\lambda[/itex] 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 [itex]y(x)=\sqrt{1-x^2}[/itex] into the left side of the DE above, in order for it to be zero, [itex]\lambda=-2[/itex]. Therefore, we want to solve:

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

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:

[tex]\frac{dy}{dx}=\frac{\sqrt{y^2-(y^2-c_1)^2}}{y^2-c_1}[/tex]

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

[tex]\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{y}[/tex]

with the solution being:

[tex]y^2+(x+c_2)^2=1[/tex]

Letting again [itex]c_2=0[/itex], we get the desired extremal: [itex]x^2+y^2=1[/itex].

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

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.

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