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Surface Area and Volume Relationships?

  1. Nov 27, 2005 #1
    So, I had a question for all of you, regarding the relationships between area and circumference, and surface area and volume.

    For the longest time I was confused as to how these two quantities were related. I saw that the derivative of the area of a circle was equal to its volume, but then that relationship didn't hold true for any other shape. I eventually came to the conclusion that the problem was that you had to measure your area with respect to the "radius" - the line from the center of mass of the shape to the edge. If you do that, you get back an expression which, when differentiated gives you circumference.

    Here's my work if you'd like to see; its for the simplest case of any n-sided regular polygon, but the essence of the calculations is the same for more complex cases. You just don't get the nice neat formulas at the end :P

    http://home.comcast.net/~macht/problem.jpg [Broken]

    My question for this case is, why is the derivative of the area the circumference here. I have a hunch that its due to the fact that you're defining the circumference as a function in polar coordinates and thene finding the area contained within that function, so the derivative would take you back out to the circumference function. However, I couldn't really find a solid way of proving/showing that.

    However, the second part of this is much more interesting. See, if you take the volume of a sphere and differentiate it, you get the surface area of the sphere. My initial problem was trying to work this out; the 2d version seemed to be a simpler way to approach the problem.

    However, I soon ran into a problem here. I tried to get the relationship to work for an ellipse, and failed miserably. I found the area of an ellipse with dimensions a, b, and c, to be 4/3 (abc). (Actually, it was kind of funny. I took up about two whiteboards muddling my way through an ugly, ugly triple integration with wierd square root bounds and huge expressions, and at the end it just reduced to that. Definatly one of those "Oh, duh!" moments :P).

    I tried expressing each in terms of the other, like, x, 2x, and 4x, but still to no avail. I have a feeling I'm missing something, but I can't seem to find a proper equation that defines the surface in spherical coordinates. And I know that the ellipse equations are ugly; this all started from trying to find a quicker method of calculating it.

    So I guess my question is twofold. First, can any of you confirm my hunch that the results in the 2d case extend to the 3d case? And second, can anyone give me some pointers in attempts to apply it to an ellipse?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 28, 2005 #2


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    As far a spheres go, differentiating the volume w.r.t. the radius will give you surface area; this relation even holds for n-D hyperspheres also.
  4. Nov 28, 2005 #3
    Yea, I figured that one out. The question that I had when I started this whole thing out was "why?"

    My idea is that differentiating the volume of anything, given that the volume formula is calculated by integrating over spherical coordinates the "radius" of the object (meaning finding an equation that defines the line from the centroid of the object to the surface of it), will give back surface area. This would explain it, as the formulas for circles/spheres are all actually given in polar coordinates, only r is constant with respect to theta so you never actually see the polar coordinate aspect of it.

    I know it works for the 2d case, though I'm not precisely sure why (I can give a hand-waving argument about differentiation reversing integration and integrating in polar coordinates finding the area enclosed by the line, so differentiating will give you back the line, but its far from solid [or correct, as far as I know]).
    Last edited: Nov 28, 2005
  5. Nov 28, 2005 #4
    So you are looking for a general proof that the derivative of a formula that gives volume will yield a formula for surface area (but not only in 3 dimensions), right?
  6. Nov 29, 2005 #5
    Yes, pretty much.
  7. Nov 30, 2005 #6


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    I can give you formulas for the volume and surface area of an n-d hypersphere of rudius r:

    [tex]V_{n}(r)=\frac{\pi^{\frac{1}{2}} r^{n}}{\Gamma \left( \frac{n}{2}+1\right) } [/tex]


    [tex]S_{n}(r)=\frac{d}{dr}V_{n}(r)=\frac{2\pi^{\frac{1}{2}} r^{n-1}}{\Gamma \left( \frac{n}{2}\right) } [/tex]

    where [itex]\Gamma[/itex] is the Euler gamma function.

    You also might try looking into http://planetmath.org/encyclopedia/PappussTheoremForSurfacesOfRevolution.html [Broken] (examples 2,3 and 4 on the linked page). It may help, then again, I'm not sure.

    Interesting question: thank you.
    Last edited by a moderator: May 2, 2017
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