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Sphere has the minimum surface area?

  1. Jul 20, 2013 #1
    How do you prove that for a given volume, sphere has the minimum surface area?
  2. jcsd
  3. Jul 20, 2013 #2


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  4. Jul 20, 2013 #3


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    It is a standard problem in "Calculus of Variations" so you might also check that.
  5. Jul 21, 2013 #4
    i tried to find proofs but i dont find please show me proofs
  6. Jul 22, 2013 #5
    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]
    [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.
    Last edited: Jul 22, 2013
  7. Jul 23, 2013 #6
    can you show me a proof in internet?
  8. Jul 23, 2013 #7


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  9. Jul 23, 2013 #8
    thanks to all
  10. Jul 23, 2013 #9
    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. :)
    Last edited: Jul 23, 2013
  11. Jul 23, 2013 #10


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    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!
  12. Jul 23, 2013 #11
    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.
  13. Jul 23, 2013 #12
    Dang it! I'm missin' a y:

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

    I think I need to review some things first . . .
  14. Jul 24, 2013 #13
    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:


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


    Thanks to fzero and Jacquelin, we obtain:


    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:


    with the solution being:


    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.
    Last edited: Jul 24, 2013
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