https://www.quora.com/Can-you-do-a-surface-integral-on-a-mobius-strip
According to this link, it is possible to integrate surface area of the non-orientable Mobius strip by using density. However, I'm trying to understand explanations in the Wikipedia, I don't know from which equation I should start to calculate it. Please help me.
$\newcommand{\Reals}{\mathbf{R}}$Let $D$ be a region in the $(u, v)$-plane, such as a rectangle. To integrate a (continuous) scalar function $f$ over a (piecewise-$C^{1}$) parametrization $\Phi:D \to \Reals^{3}$, form the scalar surface element $$ dS = \|\Phi_{u} \times \Phi_{v}\|\, du\, dv, $$ and put $$ \int_{\Phi} f = \int_{D} f\, dS. $$ This formula does not depend on orientation, and it gives a meaningful answer even if, say, the image of $\Phi$ is a Möbius strip obtained by identifying boundary points of $D$.