Flux through a Mobius strip A friend of mine asked me what is the flux of the electric field (or any vector field like 
$$
\vec r=(x,y,z)\mapsto \frac{\vec r}{|r|^3}
$$ where $|r|=(x^2+y^2+z^2)^{1/2}$) through a Mobius strip. It seems to me there are no way to compute it in the "standard" way because the strip is not orientable, but if I think about the fact that such a strip can indeed be built (for example using a thin metal layer), I also think that an answer must be mathematically expressible.
Searching on wikipedia I found that
http://en.wikipedia.org/wiki/Mobius_resistor
A Möbius resistor is an electrical component made up of two conductive surfaces separated by a dielectric material, twisted 180° and connected to form a Möbius strip. It provides a resistor which has no residual self-inductance, meaning that it can resist the flow of electricity without causing magnetic interference at the same time.
How can I relate the highlighted phrase to some known differential geometry (physics, analysis?) theorem?
Thanks a lot!
 A: David Speyer already explained it very well. But if you want more details of the explicit calculations, you can look here:
http://www.math.uwaterloo.ca/~karigian/teaching/multivariable-calculus/moebius.pdf
I made this handout many years ago for a multivariable calculus class. Enjoy.
A: If you cut a Möbius band the short way, to produce a rectangle, and orient that rectangle, then the flux through that rectangle is well-defined. The same procedure on a cylinder produces a constant function, since the flux does not depend on the cut. Because the orientation on the rectangle disagrees with itself across the cut for a Möbius band, the flux is not constant. It is a continuous function which might be viewed as spin-$1/2$, or antiperiodic with antiperiod $2\pi$: The flux reverses sign when you continuously change the cut and orientation around the Möbius band since that reverses the orientation of the rectangle.
A: You probably know this, but if you want to define numerical value of flux through a surface, you need to specify a "plus" direction and a "minus" direction, so we can say whether a small chunk of flow through the surface contributes positively or negatively to the total.  Since a Möbius strip has no consistent choice of normal direction, this can't be done.  The physicality of a Möbius strip doesn't help you define flux, but it does drive home the point that you can't paint such an object with two different colors with no color discontinuities.
It is possible to consider a notion of gross flow, where you integrate the absolute value of flux.  I don't know what people do with it, but it doesn't seem to be stable against high-frequency perturbations.
The claims about the Möbius resistor seem to be overselling the merits a bit.  This probably depends on your chosen application, but I'm pretty sure you can minimize residual self-inductance with counterrotating coils of resistive wire.
A: The Mobius resistor concept seems like bunk/hokum because there is only one side to the resistor: both leads connecting to such a resistor would be connected to the same side and would theoretically have $0$ (zero) resistance for circuit-modeling purposes.  
The three answers prior to my comment here by Douglas Zare, David Speyer, and Scott Carnahan should all be sufficient to convince you that since you can't orient this surface, you cannot define a positive/negative orientation for the direction of flow through this surface and thus cannot define a flux through a mobius strip.
A: As you say, there is no standard definition of the flux through a nonorientable surface. I will try to convince you that you shouldn't want to define this. 
There are two standard facts about flux. The first is that, if $V$ is a three dimensional volume, with boundary $\partial V$, and $F$ is a vector field, then $\int_{\partial V} F \cdot n = \int_V \nabla \cdot F$. But a Mobius band cannot be embedded in a closed surface which contains a volume $V$, so this formula doesn't apply. Physically speaking, what I am saying is that you would never care about Gauss's law for a Mobius band, because you can't make sense of the notion of the band being part of a surface which encloses a charge. 
The other key mathematical fact is that, if $S$ is a surface with boundary $\partial S$ and $F$ a vector field, then $\int_{\partial S} F = \int_{S} \nabla \times F$. I will show that there is no hope of a formula like this for a Mobius strip. Specifically, I will show that there is a vector field $F$ which has $\nabla \times F=0$ everywhere on the Mobius strip $M$, yet $\int_{\partial M} F \neq 0$. 
Let the $z$ axis pass up through the center of the Mobius strip, with $\partial M$ winding twice around this axis and staying outside a cylinder of radius $\epsilon$ around the $z$-axis. Take the field at point $(x,y,z)$ to be $(y/(x^2+y^2), -x/(x^2+y^2), 0)$, for any $(x,y,z)$ with $x^2+y^2 \geq \epsilon^2$, and interpolate however you like within that vertical cylinder. (Physically, this is the $B$-field from a current along the $z$-axis. Or, if it isn't, then replace my formula by that one.) Then $\nabla \times F=0$ outside the vertical cylinder, and in particular everywhere on the Mobius strip. But $\int_{\partial M} F = 4 \pi$.
I previously had a physical example here, but I think I got some of the details wrong, so it is gone now.
I don't know about the Mobius resistor question. You might get a better answer at physics.stackexchange.com.
