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Since Möbius strips cannot be oriented, the calculation of surface area is meaningless; perhaps that is the reason why I have never seen such problem. However, the following φ is a parameter representation of the Mobius strip and it should be possible to mechanically fit the usual surface area formula to the phi above (See the problem, below).

$\varphi:\left(r,t\right)\in D \mapsto(\varphi_1\left(r,t\right),\varphi_2\left(r,t\right),\varphi_3\left(r,t\right)\in\mathbb{R^3}$

  • $\varphi_1\left(r,t\right)=\cos{\left(2t\right)}\left(r\cos{\left(t\right)}+2\right)$
  • $\varphi_2\left(r,t\right)=\sin{\left(2t\right)}\left(r\cos{\left(t\right)}+2\right)$
  • $\varphi_3\left(r,t\right)=r\sin{\left(t\right)}$
    D:={(r,t)|-1≦r≦1，0≦t≦π}

So that, the problem we need to address can be formulated as follows;

The Problem

Calculate the the following double-integral; $$\int_{r=-1}^{r=1}\int_{t=0}^{t=\pi}f\left(r,t\right)\ \ drdt$$ Here,
$f:D\rightarrow\mathbb{R}$, D:={(r,t)|-1≦r≦1，0≦t≦π} be defined as
$$f(r,t) = \left\| \frac{\partial \varphi}{\partial r} \times \frac{\partial \varphi}{\partial t} \right\|$$

As far as I have calculated, $$\frac{\partial\varphi}{\partial r}=\left(\begin{matrix}cos\left(2t\right)cos\left(t\right)\\sin\left(2t\right)cos\left(t\right)\\sin\left(t\right)\\\end{matrix}\right)$$ $$\frac{\partial\varphi}{\partial t}=r\left(\begin{matrix}-2sin(2t)cos(t)-sin(t)cos(2t)\\2cos(2t)cos(t)-sin(2t)sin(t)\\cos(t)\\\end{matrix}\right)+4\left(\begin{matrix}-sin(2t)\\cos(2t)\\0\\\end{matrix}\right) =\left(\begin{matrix}r(-2sin(2t)cos(t)-sin(t)cos(2t))-4sin(2t)\\r(2cos(2t)cos(t)-sin(2t)sin(t))+4cos(2t)\\rcos(t)\\\end{matrix}\right)$$

So, $$\left(\frac{\partial\varphi}{\partial r}\times\frac{\partial\varphi}{\partial t}\right)=\left(\begin{matrix}sin(2t)cos(t)rcos(t)\ \\sin(t)\left(r(-2sin(2t)cos(t)-sin(t)cos(2t))-4sin(2t)\right)\\cos(2t)cos(t)(r(2cos(2t)cos(t)-sin(2t)sin(t))+4cos(2t))\\\end{matrix}\right) -\left(\begin{matrix}sin(t)\ \ \left(r(2cos(2t)cos(t)-sin(2t)sin(t)\ +4cos(2t)\right)\\cos(2t)cos(t)rcos(t)\\sin(2t)cos(t)\left(r(-2sin(2t)cos(t)-sin(t)cos(2t))-4sin(2t)\right)\\\end{matrix}\right) =\left(\begin{matrix}rsin\left(2t\right)-cos\left(2t\right)sin\left(2t\right)-4sin\left(t\right)cos\left(2t\right)\\-rsin^2\left(2t\right)-rcos\left(2t\right)-4sin\left(t\right)sin\left(2t\right)\\2rcos^2\left(t\right)+4cos\left(t\right)\\\end{matrix}\right)$$

So, the $f$ may be $$f\left(r,t\right)=\sqrt{r^2+4\left(rcos\left(t\right)+2\right)^2}.$$ So the double-integral, we'd like to caluculate is $$\int_{r=-1}^{r=1}\int_{t=0}^{t=\pi}\sqrt{r^2+4\left(rcos\left(t\right)+2\right)^2}\ \ drdt$$

So, anyway What I would like to ask of you all are

  • (1)Confirm that there are no calculation errors in the calculations of the $f$
  • (2)Graphing of the correctly calculated $f$
  • (3)Calculate the double integral of $f$
  • (4)If possible, a discussion of why the calculation of the area is inconvenient even though this integral can probably be done (perhaps different areas for the same Möbius wheel depending on how the integral interval is taken?)

See also, Derive Cartesian cubic Möbius strip from parametric

Blue Various
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    The way one would do an area calculation in this case is to find a suitable fundamental domain, calculate the first fundamental form, and then evaluate the corresponding integral involving the square root of the determinant of the first fundamental form. The kind of extrinsic calculation you are proposing is too tedious and therefore unreliable in practical calculation. – Mikhail Katz Dec 06 '23 at 16:07
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    Surface area is perfectly meaningful without orientability; it’s flux or the integral of a $2$-form that requires an orientation. I’m not sure what embedding of the Möbius strip — if any — gives a nice integral here. – Ted Shifrin Dec 06 '23 at 16:14
  • @Mikhail Katz 　What happens in the case of the Mobius ring? – Blue Various Dec 06 '23 at 16:16
  • What is the Möbius ring? You mean the Klein bottle? – Ted Shifrin Dec 06 '23 at 17:53
  • @Ted Shifrin Oops, the Mobius ring is a Mobius strip, or Mobius band typo. Sorry. – Blue Various Dec 23 '23 at 18:59

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