Find length of edges of Moebius strip formed by cutting along length and re-joining from a circular cylinder segment:
$( a \cos \theta, a \sin \theta, z), ( \theta, 0, 2 \pi ),( z,0,b) $ after half rotation.
Since the Moebius strip has negative Gauss curvature its edges must be longer than $ 2 \pi a $.
What is its non-isometric mapping or parametrization that includes $a$ and $b$?
The question is not about its topology but its geometry.
EDIT1:
Moebius Band from Mathworld..
a = 1;
ParametricPlot3D[{ (a + v Cos [u/2]) Cos[u], ( a + v Cos [u/2]) Sin[u],
v Sin [u/2] }, {v, -4, 3}, {u, 0, 2 Pi},
PlotStyle -> {Yellow, Opacity[0.75]},
PlotLabel -> Moebius_Band _MathWorld]
EDIT2:
The Band shown below with given parametrization has self intersections!
(I write $\phi$ instead of $\theta$.)
Imagine the Moebius strip $M$ being produced in the following way: A stick of length $2b$ moves along a circle of radius $a$ in the $(x,y)$-plane, whereby the center of the stick is always on the circle and the stick is vertical at $\phi=0$. At the same time the stick is "screwed" such that it is always in a plane with the $z$-axis and makes a half turn during its trip.
A parametrization implementing this idea is given by
$$M:\quad(\phi,t)\mapsto(a\cos\phi,a\sin\phi,0)+ t\left(\sin{\phi\over2}\cos\phi,\ \sin{\phi\over2}\sin\phi,\ \cos{\phi\over2}\right)$$ with $0\leq\phi\leq 2\pi$, $\>-b\leq t\leq b$.
The length of the boundary curve can now be computed by the standard formula; but I'm afraid that the resulting integral is not elementary.