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Spherical proof

  1. Mar 6, 2008 #1
    This is really more for a friend than for me. Basically hes doing a test for the navy officers position, and one of the questions on the review says to derive the proof that shows the sphere is the most efficient form of surface area to volume. Neither of us really know where to start. Any ideas?

  2. jcsd
  3. Mar 6, 2008 #2


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    I think any proof of something like is going to be really technical. You really have to think about how to define 'surface' and 'volume'. It's an example of an isoperimetric inequality. On the other hand, it pretty easy to show a sphere beats any simple family of shapes, like cylinders or prisms. Can you quote the exact question?
  4. Mar 7, 2008 #3


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    1. How would you or us doing the problem show anything about your friends qualifications for a Navy officer's rank?

    2. What does "most efficient form of surface area to volume" mean? Least surface area for a fixed volume?

    3. A rigorous proof would involve "Calculus of Variations" which is a rather specialized field of mathematics. Seems a peculiar question to put on a Navy test! Is your friend planning to serve in nuclear submarines? I second Dick- exactly what does the question say?
  5. Mar 7, 2008 #4
    yes, hes going into the nuclear engineering department. Its not that I want anyone to solve it for us, just to get a train of thought going. He already has a biomedical engineering degree. He probably could have done this last year right after he graduated, but hes admittedly a bit rusty. I mean we did sit down for a while trying it. The original question was basically what I said. "Derive the proof showing that the sphere is the most efficient ratio of surfact area to volume". So yes, less surface for a given volume. Looking at it, surface area can increase minimally, and effect volume in much higher magnitude.
  6. Mar 7, 2008 #5


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    I don't think any proof of that would be accessible to, or necessary for, someone with an engineering degree of any kind. If you could ask a more recent graduate what a proof of this would look like, I'd love to hear it.
  7. Mar 7, 2008 #6
    The best you're going to get is anecdotal evidence of the sphere's superiority (e.g. comparing it to other shapes, referencing bubbles minimizing surface tension, etc). I'm a recent EE grad and I've never seen something like this from an engineering class. HOWEVER, viewing this as having potential analogues to physics, I think it might be possible to concoct something out of the divergence theorem http://en.wikipedia.org/wiki/Divergence_theorem and taking F to be the magnetic flux density [itex]\vec{D}[/itex], which turns the generalized divergence theorem into Gauss' Law. It would by no means be a proof though.
    Last edited: Mar 7, 2008
  8. Mar 11, 2008 #7
    I'm actually the friend he was talking about, but I ran across this thread totally at random. Odd.

    The exact question is "What geometric surface encloses the maximum volume with the minimum surface area? How would you prove it?"

    My first thought was to try to define some general expression for both surface area and volume and minimize their ratio. Or you could do something anecdotally like jhicks said... The best hand-waiving argument I've come up with is since surface area is length squared, and volume is length cubed, their ratio SA/VOL could be minimized by letter VOL increase unbounded while an increase in SA would basically have to mean the number of sides n increases also. SA will never overcome VOL based on their powers. As n increases, it approaches a continuous surface, and if you assume it to be symmetrical, that would be a sphere... But again, it's vague and requires some willingness to accept the end conclusion. I want to know how to prove it explicitly.
  9. Mar 11, 2008 #8


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    Physical proof …

    Hi klaarn! Welcome to PF! :smile:

    The question doesn't specify that the proof has to be mathematical.

    If a physical (experimental) proof will do, I'd say take a blob of mercury and see what shape it forms.

    Since the surface tension depends on the area, it should try to form the smallest area! :smile:

    Oh, and I'd deform it by hitting it a few times wth a hammer, to make sure that it returns to the same shape (in other words, prove that this is a stable equilibrium position).

    Will that impress the Navy? :smile:
  10. Mar 11, 2008 #9


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    The only reasonable thing I can think of that would even approach a suggestive argument (but is nowhere near a rigorous proof) would be to note that since surface area and volume are both unchanged by rotations, we would expect the shape that extremizes their ratio to share this symmetry.

    Specifically, if it didn't have this symmetry, we could rotate it to get a new solution (whether two shapes related by a rotation are different is arguable, but if you set up the problem, say, in a fixed coordinate system, the actual parametric equations for the two surfaces will be different, so you could reasonably call them different solutions), but it seems natural that there should be some unique minimal surface (if you're really shifty, you can use the fact that they implicitly assumed a unique solution in their question).

    The only shape invariant under rotations is the sphere, so this is all it could be. Of course, the same argument might show the sphere is the surface of maximal area for a given volume, which isn't true, so clearly there's something missing here.
  11. Mar 12, 2008 #10
    the mercury idea sounds interesting... i'm not too familiar with the properties of mercury surface tension or how or why it would try to minimize it's surface area. wouldn't its behavior depend on whether or not the droplet was suspended in air, water, or some other fluid?

    there has to be some rigorous mathematical proof though... i've even tried looking through books of proofs in the library, but found very little. i don't want to go bother any of my old profs... but it may come to that.
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