Möbius strip area
Jonathan Walther wrote to ask me an interesting question about the area of a Möbius strip.
I've been having a debate with a friend about how to calculate the area of a moebius strip, where the moebius strip is constructed by taking a 1" by 10" area, twisting it, and joining the ends.I have been maintaining that the area remains the same; that is, 10 square inches. My friend insists it is 20 square inches.
After discussion with my friend it became apparent that our different calculations came from our having different concepts of "area". He used a strip of paper to "illustrate" the moebius strip, and I feel this gave him erroneous intuition in this case.
My observation was that if you did make the strip from paper, you would need 20 sq. in. of paint in order to paint the whole thing. If you used 10 sq. in. of paint, you would have 10 sq. in. of surface unpainted.
However, Jonathan argues that this is a misinterpretation if the Möbius strip is seen as having zero thickness, because then points on one "side" are actually identical with the corresponding points on the "other side". He suggests that, on a zero-thickness strip, you can go only 10" before you return to your starting point. (On a strip made of paper with non-zero thickness, you must go 20" before returning to your starting point.)
Does anybody have a view to clarify this? It does just seem like a question of how to define surface area, but maybe there is a particular definition of "surface area" or "Möbius strip" which is somehow preferable.
(Normally you deal with areas in a plane, and the definition is easier. Is there something handy from multivariate calculus here?)
