Consider $\RR^2$ with the usual rectangular coordinates $(x,y)$. The standard basis of 1-forms is $\{dx,dy\}$. But we could also use polar coordinates $(r,\phi)$, and the corresponding basis $\{dr,d\phi\}$. How are these bases related?

We have \begin{align} x &= r\,\cos\phi \\ y &= r\,\sin\phi \end{align} which leads to \begin{align} dx &= dr\,\cos\phi - r\,\sin\phi\,d\phi \\ dy &= dr\,\sin\phi + r\,\cos\phi\,d\phi \end{align}

The inverse transformation can be obtained by inverting these expressions, or directly starting from \begin{align} r^2 &= x^2 + y^2 \\ \tan\phi &= \frac{y}{x} \end{align} which leads to \begin{align} 2r\,dr &= 2x\,dx + 2y\,dy \\ (1+\tan^2\phi)\,d\phi &= \frac{x\,dy-y\,dx}{x^2} \end{align} from which it follows that \begin{align} r\,dr &= x\,dx + y\,dy \\ r^2\,d\phi &= x\,dy-y\,dx \end{align}

What about 2-forms? There is only one independent 2-form, namely $dx\wedge dy$, and multiplying this out one obtains \begin{equation} dx\wedge dy = r\,dr\wedge d\phi \end{equation} Equivalently, multiplying the above expressions for $dr$ and $d\phi$ together yields \begin{equation} r^3\,dr\wedge d\phi = (x^2+y^2)\,dx\wedge dy \end{equation} which is of course the same thing.

It is worth noting that in the language of differential forms these two “area elements” are identically equal, whereas the geometric areas “$dx\,dy$” and “$r\,dr\,d\phi$” in fact differ by second-order terms. The differential statement “$dx\,dy = r\,dr\,d\phi$” is really an assertion about integrals, not infinitesimal areas!