Area of Mobius strip

Question

I want that to give a meaning to the notion of area for Mobius strip. I know that Mobius band is nonorientable surface. How can I set up an integral to compute it?

What's your idea for the following formula?

$\boldsymbol X(t,\theta)=\left(\left(1-t\sin\dfrac\theta2\right)\cos\theta,\left(1-t\sin\dfrac\theta2\right)\sin\theta,t\cos\dfrac\theta2\right)$ on $Q=\{(t,\theta); \ -1<t<1 , 0<\theta<2\pi\}$ that $X(Q)\approx M$.

and

Area of M:=$\iint_Q |X_t\times X_{\theta}|dtd\theta$ that $Q=X^{-1}(M)$.

Answer 1

I would say your formula is correct.

Some might argue that the "correct" area of the Möbius strip is twice that value (i.e. with $\theta$ going all the way up to $4 \pi$) - the non-orientability makes things a little awkward. In particular if you use the language that a non-twisted strip has "two sides" while a Möbius strip has only one, it seems that we are counting only half the area of the Möbius strip: if you made a paper model and coloured in the area as you measured it, you would only colour half of the "side" and then abruptly stop.

For the untwisted strip, you also only colour half of the physical surface - one of the two sides - but the contiguity makes this seem more natural. I think to be consistent, if you're going to double-count the area in one case you should in both; and it's certainly the undisputed convention that we don't double-count areas of orientable surfaces. From a mathematical perspective (talking about an abstract surface with zero thickness rather than a physical object with sides) I therefore think this is the most sensible convention.

Answer 2

It's definitely a way of seeing it, from my own experience it very much depends on how you define the mobius strip in the $3$-d plane.

Personally I set the midsection of the strip in the $XY$-plane with a $Z$-value of $0$ as a perfect circle with radius $R$ in order to do my calculations easier (as most people do). I set the part where the actual strip is flat on the XY-plane at the positive X-axis and the part where it's perpendicular at the negative X-axis (Y=$0$). Keep in mind that other people may have done it differently and the parametrization is going to vary greatly because of that. The main difference of my parametrization is that I didn't assume a radius of $1$ but for a general magnitude $R$ $$\boldsymbol X(t,\theta)=\left(\left(R+t\cos\dfrac\theta2\right)\cos\theta,\left(R+t\cos\dfrac\theta2\right)\sin\theta,t\sin\dfrac\theta2\right)$$ I also chose to go around the full $4\pi$ as this is basically adding the top side of the positive $t$-value and bottom side of the negative $t$-value and although some may call it double counting the mobius strip is in fact parametrized in the $3$rd dimension, so it could be seen as a $3$-d object that just happens to be defined with $0$ volume, meaning that even if the points of both sides happen to be in the same location they are still two distinct sides that should both be counted. I also used a more general variable for the width of the band and saw that as a radius $r$ (which is more useful if expanded in higher dimensions) $$Q=\{(t,\theta); \ -r<t<r , 0<\theta<4\pi\}$$

As a matter of fact I'm doing my extended essay in the IB programme about mobius strips and therefore had to make my own definition and parametrization due to the lack of a persistent one. Much like you tried to give a notion of the area of a mobius strip at the time of your question, I am trying to give a meaning to the notion of a mobius strip in higher dimensions (not the klein's bottle, I'm talking about the true analogue that can scale with an arbitrary amountof dimensions, such as how the square can become a cube or hypercube)Your post really helped me get started in the way I should be thinking, thanks a bunch!