How to derive the surface area of a sphere?

1. Jul 14, 2013

mesa

So, I have been trying to figure out how to derive the equation for the surface area of a sphere. All attempts have resulted in colossal failure and as such are not even worth posting on the forum.

I know Archimedes was the first to come up with the formula but I have not been able to find something on the web that goes into a detailed proof that is understandable. There are of course several sites that detail a circumscribed sphere in a cylinder of height equal to twice the radius of the sphere and how it has the same surface area (not including end caps) but how was that connection made?

I am interested in any solutions (*EDIT* - no calculus) not just that of Archimedes.

Last edited: Jul 14, 2013
2. Jul 14, 2013

WannabeNewton

Are you ok with a calculus proof? Consider a spherical surface and take any ring cross section of the sphere that has infinitesimal height along the sphere. Let $R$ be the radius of the sphere then the radius of this ring will be $R\sin\theta$ where $\theta$ is the angle made with respect to the vertical ($\theta \in [0,\pi]$ as you can see). Also note that the infinitesimal height of the ring will be given by $Rd\theta$ so the surface area of the ring will be $dA = 2\pi R^{2}\sin\theta d\theta$ (circumference times height). The entire sphere is built up from all such rings so the surface area of the sphere will be $A = 2\pi R^{2}\int _{0}^{\pi}\sin\theta d\theta = 2\pi R^{2}cos\theta|_{\pi}^{0} = 4\pi R^{2}$.

3. Jul 14, 2013

DeIdeal

I don't know how Archimedes proved it, but luckily we have more advanced mathematical tools available today, and you did say you are OK with any proofs:

Let S be the surface of a sphere of radius r.

$$A =\iint_S \mathrm{d}A=\intop_0^{2\pi}\intop_0^{\pi} r^2\sin\theta\mathrm{d}\varphi\mathrm{d}\theta=r^2\intop_0^\pi \sin\theta\intop_0^{2\pi}\mathrm{d}\varphi\mathrm{d}\theta$$
$$=r^2\cdot 2\pi\intop_0^\pi \sin\theta \mathrm{d}\theta=r^2\cdot 2\pi\cdot(\cos 0-\cos\pi)=4\pi r^2$$

If Archimedes did indeed come up with the formula, there's bound to be a geometrical argument to it, but this is the simplest, although maybe not the most intuitive proof, especially so if you're not familiar with calculus.

EDIT: Argh, what's up with me being ninja'd constantly? Can't I get a notification when somebody posts into a thread before I do

4. Jul 14, 2013

mesa

Calculus does a fine job of getting a solution but we are in the general math section (although my post did say "any" solutions so I apologize for the confusion).

Any ideas?

5. Jul 14, 2013

WannabeNewton

6. Jul 14, 2013

DeIdeal

Hm. This might seem a bit circular, but if you accept the volume of a sphere as known, another non-rigorous way of showing it is as follows:

Consider a thin spherical shell of thickness a and outer radius r, so that the volume of the shell is $V_S=V_r-V_a=\frac{4}{3}\pi r^3 - \frac{4}{3}\pi(r-a)^3$. Expanding the second term and cancelling one term it becomes $V_S=4\pi a(r^2-ra+\frac{1}{3}a^2)$. The thickness of the shell is a, so its area is $A=V_S/a=4\pi(r^2-ra+\frac{1}{3}a^2)$. When we want to find out the area of a sphere instead of the shell, a is considered to be very small so that $A\rightarrow 4\pi(r^2+0)=4\pi r^2$.

EDIT: But again, this is a very heuristic proof. In addition to the previous links, http://en.wikipedia.org/wiki/Cavalieri's_principle]this[/PLAIN] [Broken] seems to have some information about it, but like it says on the page, "Today Cavalieri's principle is seen as an early step towards integral calculus", so at least in some way that's still what's going on.

Last edited by a moderator: May 6, 2017
7. Jul 14, 2013

mesa

8. Jul 14, 2013

WannabeNewton

9. Jul 14, 2013

mesa

Sorry WannabeNewton, I did not mean to be taken literally.

Google yielded me essentially nothing over the last couple hours and you came back with two solutions in a few minutes hence the 'better version of Google'.

10. Jul 14, 2013

WannabeNewton

Ooooh ok lol; I should apologize because it was I who didn't interpret your sentence correctly. My first instinct is to usually check math.stackexchange is all; you can keep that in mind if you want as it might come in handy later who knows :)

11. Jul 14, 2013

mesa

Well it is still very clever. Thanks for the post!

Last edited by a moderator: May 6, 2017
12. Jul 14, 2013

mesa

This is the first time seeing that site, might be a late night tonight, ha ha ha!

13. Jul 14, 2013

WannabeNewton

I like the purely geometric proofs too, more so than the calculus proofs, but they are not as simple/short in nature (but more minimalist in a sense). It wouldn't hurt to get used to the calculus based proofs though because IIRC you are also doing electrodynamics so the "building up infinitesimals" argument will show up over and over.

14. Jul 14, 2013

mesa

I love them both too. As a parent I realized we shouldn't pick favorites but geometry is tough to beat although Calculus can be so damn powerful...

15. Jul 15, 2013

coolul007

In the book: "The Works of Archimedes", Chapter: "On the Sphere and Cylinder I" Proposition 33 is a proof of contradiction based on the previous propositions dealing with circumscribed and inscribed polyhedra surfaces. Archimedes did his Pi approximations the same way with polygons and the circle. A sort of geometry Calculus.

16. Jul 17, 2013

mesa

Unfortunately neither I nor my local library has that book. Can you send me a copy of his proof or perhaps your interpretation of it?

17. Jul 17, 2013

coolul007

I will try to scan the chapter, it requires previous methods. I'm out of town so it will be about a week or so.

18. Jul 17, 2013

gerben

19. Jul 17, 2013

mesa

Wonderful, I look forward to it.

20. Jul 17, 2013

mesa

Hah, I missed your post while responding to coolul007, thanks for the link!

coolul007, you are 'off the hook' as they say!

21. Jul 18, 2013

lurflurf

By the Pappus centroid theorem we have.

A=length half circle * circumference of circle with centroid radius
$$= (\pi r )(2 \pi \bar{r})=(\pi r )(2 \pi \cdot 2r / \pi)=4 \pi r^2$$

The idea is that due to the rotational symmetry we can mush the sphere in such a way that the surface area is unchanged.