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Consider the formula for the area of a circle and the formula for its circumference. If one differentiates the formula of the area with respect to $r$ (the radius), the formula for the circle's circumference pops out.

The same applies to a sphere with it's volume and surface area: differentiate the formula for volume with respect to $r$, and you obtain the formula for surface area.

I have two questions about this phenomenon:

  1. Is this an unique property of the circle and sphere?
  2. Is there mathematical reason for this?
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marked as duplicate by grand_chat, Arthur, Ethan Bolker, zipirovich, RRL Dec 27 '18 at 5:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Thank you very much. This answers question no. 2. $\endgroup$ – Bulldocarx Dec 26 '18 at 23:54
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I'll begin by answering your questions in somewhat of a reverse order (though depending on your philosophical bend, you could take this first half of my answer as an answer to both). I take the core content from Wolfram MathWorld and Wikipedia.

A foreword, I'm generalizing this to the $n$-dimensional case, to show that this holds for spheres of all dimension.

We will let $V_n$ denote the $n$-dimensional analogue of volume, which the Wolfram article calls "content." To this, there is the analogue of surface area, which we'll call hyper-surface area, and denote $S_n$.

It can be shown that $S_n$ is given by

$$S_n = \frac{2\pi^{n/2}}{\Gamma \left(\frac n2 \right)} R^{n-1}$$

and that $V_n$ is given by

$$V_n = \frac{2\pi^{n/2}}{n\Gamma \left(\frac n2 \right)} R^n$$

Okay, But Where Did The Formulas Come From?: It's only fair to wonder about where these formulas come from, instead of just taking me at my word. I'll link to some resources for the derivations; the explanations are a bit long for this post and may be above your head OP, assuming you're in an introductory calculus class as I suspect.

A derivation of the formula for volume can be found here. Dr. Peyam on YouTube did a derivation of the surface area formula, which can be found here, and includes a similar level of content (but at least more explanation). (He also touches on the volume a bit as well if you want a different explanation.)

If you're unfamiliar with the notation in the formulas above, $\Gamma(n)$ is the gamma function, and is just a generalization of the notion of factorial to non-integers. It can be given by an integral, which might be a bit beyond the scope of this discussion. The relation for integers $n$ between $\Gamma(n)$ and the factorial is

$$\Gamma(n) = (n-1)!$$

For example, $\Gamma(2) = (2-1)! = 1! = 1$. (The gamma function also has the property I see not used as well as it could be in the various links that $\Gamma(n+1) = n\Gamma(n)$. This is essentially the recursion of the factorial, i.e. $n! = n\cdot (n-1)!$.)

To convince yourself of these formulas, try a few $n$ individually: let $n=2$ to find $V_n$ (area of a circle) and $S_n$ (its circumference), for example.

In any event, to see that the derivative of content yields hyper-surface area here, note:

$$\frac{d}{dR} V_n = \frac{d}{dR} \left( \frac{2\pi^{n/2}}{n\Gamma \left(\frac n2 \right)} R^n \right) = n \cdot \frac{2\pi^{n/2}}{n\Gamma \left(\frac n2 \right)} \cdot R^{n-1} = \frac{2\pi^{n/2}}{\Gamma \left(\frac n2 \right)} \cdot R^{n-1} = S_n$$

What this hints at is that this is a property of the $n$-dimensional sphere, i.e. that a property of $n$-spheres is precisely that their "content", differentiated, yields their hyper-surface area.

This presumably answers your second question regarding "is there a mathematical reason for this fact," that being it is a property of the $n$-sphere.


As for your first question, this is noted in several different ways in the question linked as a duplicate. I favor the answer by Helmer.Aslaksen, which cites a paper which can be found here:.

The bit essential to your first question is that, no, this is not a property unique to the sphere. For any shape with area able to be written as $A(s) = cs^2$ for some constant $c$ and parameter $s$ (such as side length or radius), and $P(s) = ks$ denoting the perimeter about it, a parameterization of $x = 2cs/k$ will let us have

$$\frac{d}{dx} A(x) = P(x)$$

Whether this might hold in higher dimensions, I'm uncertain. At least in $2D$ space though, it holds for shapes such as squares and ellipses.

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