Applying surface integral on the Mobius strip I'm trying to apply the surface integral on the Mobius strip. I know the Mobius strip's surface area can be easily calculated by getting the area of a piece of paper before it got twisted but this is my challenge through mathematics and I really want to complete this.
I first tried to apply the surface integral for scalar-valued one. According to this link, in the 2nd part of the anonymous' answer it is said that I need to first split up the Mobius strip so that it becomes orientable. Is this step actually required?   
Because someone suggested me from my previous question that I don't actually need to split it as     
1) I need to do the integration from $0$ to $4\pi$, thinking like painting,   and
2) that gives to intervals with distinct opposite normal vectors
I agree with this idea, but still I'm confused.  
Is it then more appropriate to interpret the mobius strip as a vector form and apply the surface integral of vector fields? 
 A: In order to compute the surface area of any surface, orientable or not, we need an essentially bijective parametric representation of that surface.  Nonorientability of this surface is of no harm if you just want to know its area. 
But the flux of some extraneous vector field ${\bf v}$ through such a surface makes no sense, because it is impossible to define in a coherent way whether  the flux of ${\bf v}$ through a surface element at some point ${\bf x}\in S$ should be counted positive or negative.
Now for the area: Assume that your Moebius band $S\subset{\mathbb R}^3$ of width $2b$ and soul radius $R$ is presented in the form
$$S:\quad (\phi,t)\mapsto{\bf x}(\phi,t):=\left(R+t\cos{\phi\over2}\right)(\cos\phi,\sin\phi,0)+t\sin{\phi\over2}(0,0,1)$$
with parameter domain $B:=[{-}\pi,\pi]\times[-b,b]$. In this way the complete set $S$ is produced in a one-one way, except for the following: There is a (vertical) suture segment corresponding to $\phi=\pm\pi$ and $-b\leq t\leq b$ which is covered twice "with opposite direction". 
In order to compute the area of $S$ we need not bother about the surface normal; we just compute the integral
$${\rm area}(S)=\int_B\>\bigl|{\bf x}_\phi\times{\bf x}_t\bigr|\>{\rm d}(\phi,t)\ .\tag{1}$$
I didn't compute this integral. Its value is about $2\pi R\cdot2b$. If you want to paint the Moebius band on both sides you need paint for twice this area, as you would if you  paint a rectangle of given area on both sides.
