We all know that when switching to polar coordinates we have: $$ x = r\cos\theta,\qquad y = r\sin\theta $$ and either by a geometric argument or using the Jacobian we have: $$ dx\,dy = r\,dr\,d\theta $$ But also, using the definition of differential: $$dx = x_r\,dr + x_\theta \,d\theta = \cos\theta\,dr - r\sin\theta \,d\theta$$
and $$dy = y_r\,dr + y_\theta \,d\theta= \sin\theta \,dr + r\cos\theta \,d\theta$$
giving,
\begin{align} dx\,dy & = (\cos\theta \,dr - r\sin\theta \,d\theta)(\sin\theta \,dr + r\cos\theta \,d\theta) \\ &= \sin\theta \cos\theta \,dr\,dr + r\cos^2\theta \,dr\,d\theta - r\sin^2\theta \,dr\,d\theta - r^2\sin\theta \cos\theta \,d\theta \,d\theta \tag{*} \end{align}
I do not understand how this ends up being the same as $r\,dr\,d\theta$.
I have found two separate threads with answers I find unsatisfactory:
In (1) the there is a claim that $dx$ and $dy$ are vectors. Well, so what? That is not strictly a requirement, and one still ends up with a problematic $\cos^2\theta - \sin^2\theta$ term.
In (2) the third answer explains that in the context of differential forms and wedge products that $dx\land dx = dr \land dr = 0$ and $dx\land dy = -dy \land dx$ and if one omits the $\land$ symbols and looks the other way then magically we do get $dx\,dy = r\,dr\,d\theta$.
This thread goes into more detail on the use of the wedge product:
Transforming differentials to polar coordinates
I feel uneasy with all this. Are we saying that, in fact, all the time that we have been writing things like $dx\,dy$ in undergrad we always actually meant $dx\land dy$ without knowing it? If that is true, why are we not required to change $dx\,dy$ to $-dy\,dx$ when we change the order of integration in double integrals?
Is there a way to deduce $(*) = r\,dr\,d\theta$ without resorting to weird exterior products?
