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Lowest surface to volume ratio for an uncovered vessel

  1. Jul 2, 2012 #1

    It is well known that a sphere has the lowest surface to volume ratio. However, a related question is: What is the shape that gives the lowest surface to volume ratio if you do not include the "top" in the surface. That is, what is the maximal volume of an uncovered vessel of a fixed surface area?

    For example, a cylinder whose height is equal to its radius, has a volume Pi R^2 h = Pi R^3 and a surface (without cover) of Pi R^2 + 2*Pi *R*h = 3 Pi R^2. If we fix the volume to unity, the surface in this case is 3.Pi^(1/3).

    In comparison, a half sphere of unity volume has a smaller surface -- (18 Pi)^(1/3).

    However, it's clear this is not the best shape - a sphere cut a bit above half its volume beats the half sphere.

    So, what is the shape that gives the lowest surface to volume ratio if you do not include the "top" in the surface ?

  2. jcsd
  3. Jul 2, 2012 #2


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    Hey eee000 and welcome to the forums.

    I'm a little confused about what you mean by "top".

    With a sphere, there is no discontinuous boundary like you have with say a cube so in these kinds of situations (where the surface is completely smooth), it's really hard to define a "top".

    Based on this, do we need the shape to have something corresponding to a "top" where the top is an area that is "smooth" respect to the rest of the body?
  4. Jul 2, 2012 #3


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    That depends strongly on how you define "top" for a general surface.
  5. Jul 2, 2012 #4
    Thank you,
    The top I have in mind is defined by gravity. I want to have a vessel full of water that do not spill out. That is, if we define the z axis to be the direction of gravity, I think of a surface that may have holes in it, but the volume that we count is only integrated up to lowest z coordinate in the holes.
  6. Jul 2, 2012 #5


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    There are well-known formulae for the volume and surface of a spherical cap: http://en.wikipedia.org/wiki/Spherical_cap

    (removed wrong stuff).
    Last edited: Jul 2, 2012
  7. Jul 2, 2012 #6
    Thank you for the spherical cap formulae, but I think your algebra is wrong. Fixing the sphere radius R, and parameterizing the cap by h, I find the volume to be V=Pi h^2 (R - h/3) and the surface S=2 Pi R h. The lowest ratio is obtained when h=3R/2, and then V=9Pi/8 R^3, and S=3Pi R^2. Setting the volume to unity (R=(8/9Pi)^(1/3)), I find S=(64 Pi/3)^(1/3), better than the half sphere with (18 Pi)^(1/3) and better than the full sphere with (36 Pi)^(1/3).
  8. Jul 2, 2012 #7


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    You're right, I made a silly mistake.

    What I should've done is simply expressed V/A as a function of h, and that's [itex]\frac{V}{A} = \frac{1}{6}(3h - \frac{h^2}{r})[/itex]. Taking the derivative and setting it to zero yields your result of [itex]h = \frac{3r}{2}[/itex].

    Intuitively, shouldn't this (partial sphere to a height of 1.5r) be the open vessel with the lowest A/V ratio?
  9. Jul 2, 2012 #8
    I'm sorry,I made a mistake too...
    Since the surface to volume ratio scales as 1/V^(1/3) we need to do the optimization for fixed volume. This yields that the optimal shape is exactly the half sphere.

    Obviously, my statement "(64 Pi/3)^(1/3), better than the half sphere with (18 Pi)^(1/3)" is wrong, as 64/3 > 18 ...
  10. Jul 2, 2012 #9


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    I didn't see that, but I wasn't following that anyway, because I think your method of visualising the problem is unnecessarily confusing. Why not just let r = 1 and figure out the value of h that minimises the A/V ratio (or maximises the V/A ratio, as I did) for the open partial sphere? Now I'm sure that h = 1.5 (for r = 1) is the right answer.
  11. Jul 2, 2012 #10
    sure it is, but I think the correct formulation of the problem is that of a fixed volume, especially if one is interested in he general question - what is the best shape of all. Since the A/V ratio generally scales with 1/V^(1/3), comparison must be made for equal volumes.
  12. Jul 2, 2012 #11


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    OK, I see where you're coming from. To state the problem rigorously, "For an open partial sphere of unit volume, what value of h/r will yield minimal surface area?"

    Can we agree on that statement?

    The way to go about it then is to put h = kr (where 0 < k <= 2), set V(h,r) = 1, then work out r in terms of k alone.

    Then plug that expression for r into A(h,r) to get A in terms of k alone.

    Finally, minimise A(k) over the domain (0,2].

    I have to turn in now - I'm already up too late with a bad cold and I have to work tomorrow. So please go ahead (and if you've done it, please post the basic steps, if you can). If the problem remains unsolved tomorrow, and I'm free, I'll work on it.
  13. Jul 2, 2012 #12
    For the partial sphere, the solution is easy - the half sphere is optimal.
    The open problem (to the extent of my minimal knowledge) is:

    "For an open manifold of unit volume(*) , what shape will yield minimal surface(*) area?"

    where Volume is the volume contained up to the first hole, and surface includes the whole manifold but the holes
  14. Jul 2, 2012 #13


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    But have you proved it? Just asking.

    EDIT: Proved it myself.

    [itex]A(k) = (2){(9\pi)}^{\frac{1}{3}}.k^{-\frac{1}{3}}.{(3-k)}^{-\frac{2}{3}}[/itex]

    and this has a minimum at k = 1 (when h = r), and the minimum is [itex]{(18\pi)}^{\frac{1}{3}}[/itex], yielding a half-sphere of unit volume.

    I cannot provide any insight on the more general problem.
    Last edited: Jul 2, 2012
  15. Jul 3, 2012 #14


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    Suppose you have an optimal vessel and it holds more than a half-sphere of the same surface area of vessel. Necessarily the upper rim will lie in one horizontal plane. Create a cap for it by reflection in that plane. You now have a closed vessel of twice the volume and twice the area, and it would seem that it holds more fluid than a closed sphere of the same surface area.
  16. Jul 3, 2012 #15
    Thank you haruspex! That's what I was looking for.
    I was thinking in more local terms, along the lines of Steiner's arguments, but your one-liner is much better.
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