Connection between the circumference/area of circles, and between the volume/surface area of spheres? [duplicate]

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.

1. Is this an unique property of the circle and sphere?
2. Is there mathematical reason for this?

marked as duplicate by grand_chat, Arthur, Ethan Bolker, zipirovich, RRLDec 27 '18 at 5:19

• Thank you very much. This answers question no. 2. – Bulldocarx Dec 26 '18 at 23:54

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.