Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In school we are told that the surface area of a sphere is $4\pi$. Is it true that Archimedes found the surface area of a sphere using the Archimedes Hat-Box Theorem? Is there a simple proof for this theorem? Thank you.

Added: Does that kind of projection as mentioned in the Archimedes Hat-Box Theorem preserve the areas of any shape on the surface of the sphere?

share|cite|improve this question
I think you should loo here – Norbert Feb 7 '12 at 23:43

Yes, the mapping preserves area of any shape. You can convince yourself of this by taking by small patches on the sphere, between two constant latitude lines and two longitude lines, which I believe is what they did with the state of Colorado and the sate of Wyoming. Anyway, any nice enough shape is made up, to sufficient accuracy, by a large number of these curved rectangles, and these quite definitely are mapped in an area-preserving manner. This is the oldest example of a "symplectic" map.

I do not know that much about the history of this exact example, but I do know that a book of Archimedes called The Method was thought to be lost until about 1900, and translations are available. See EUDOXUS and METHOD and SPHERE_AND_CYLINDER finally MOOSE_AND_SQUIRREL

Alright, somebody at Wikipedia is not paying attention. The equal area MAP projection is due to Archimedes.

share|cite|improve this answer
Thank you very much! :) (btw, what does the tv show have to do with Archimedes?) – kafka Feb 8 '12 at 9:05
@kafka, I just threw that in because sphere and cylinder reminded me...The character Natasha in the cartoon never said Rocky and Bullwinkle, and she left out the word "the," it was always just moose and squirrel. – Will Jagy Feb 8 '12 at 20:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.