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The task is to rewrite the following differential form in polar coordinates:

$$w = \sqrt{x^2 + y^2} \, dx \land dy $$

I did it by a direct substitution:

$$\begin{cases}x = r \cos \varphi \\ y = r \sin \varphi \end{cases}, \;\;\;\;\;\;\;\;\; \begin{cases} dx = \cos \varphi \, dr - r \sin \varphi \, d \varphi \\d y = \sin \varphi \, dr + r \cos \varphi \, d \varphi \end{cases} $$

$$\begin{multline} w = r (cos \varphi \, dr - r \sin \varphi \, d \varphi) \land (\sin \varphi \, dr + r \cos \varphi \, d \varphi) = \\ r^2 (\cos^2 \varphi \, dr \land d \varphi - \sin^2 \varphi \, d \varphi \land dr) = r^2 \, dr \land d \varphi \end{multline}$$

Is there any other algebraical way to do it? That is I don't mean any drawing (geometry), just tossing symbols around, like pulling some terms under $d$.

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Look for the Jacobian. –  Sigur Jul 4 '12 at 18:53

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The way you did it is entirely correct and appropriate. Of course, if you have to do another transformation like this, you will remember that $dx\wedge dy = r dr\wedge d\varphi$. And yes, as @Sigur points out, this latter identity can be obtained by taking the determinant of the Jacobian matrix $$ \begin{pmatrix} \partial x/\partial r & \partial x/\partial \varphi \\ \partial y/\partial r & \partial y/\partial \varphi \end{pmatrix} = \begin{pmatrix} \cos\varphi & -r\sin\varphi \\ \sin\varphi & r\cos \varphi \end{pmatrix} $$ However, this changes nothing conceptially, only notation is different. Each row of the Jacobian matrix encodes a 1-form ($dx$ and $dy$), and the determinant is their wedge product. There is less writing to do since $dr$ and $d\varphi$ do not appear explicitly. But presumably, the purpose of the exercise was to use the differential form notation.

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