Area of Mobius strip

I would say your formula is correct.

Some might argue that the "correct" area of the Möbius strip is twice that value (i.e. with $\theta$ going all the way up to $4 \pi$) - the non-orientability makes things a little awkward. In particular if you use the language that a non-twisted strip has "two sides" while a Möbius strip has only one, it seems that we are counting only half the area of the Möbius strip: if you made a paper model and coloured in the area as you measured it, you would only colour half of the "side" and then abruptly stop.

For the untwisted strip, you also only colour half of the physical surface - one of the two sides - but the contiguity makes this seem more natural. I think to be consistent, if you're going to double-count the area in one case you should in both; and it's certainly the undisputed convention that we don't double-count areas of orientable surfaces. From a mathematical perspective (talking about an abstract surface with zero thickness rather than a physical object with sides) I therefore think this is the most sensible convention.