# 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.