From $dxdy$ to $\rho d\rho d\phi$. Where am I doing wrong?

Question

A small area element in the xy plane reads $da=dxdy$. In plane polar coordinates, it reads $da=\rho d\rho d\phi$. We also know, $$x=\rho\cos\phi,~ y=\rho\sin\phi.$$ So using partial derivative formula, we are left with $$dx=\cos\phi d\rho-\rho\sin\phi d\phi,~dy=\sin\phi d\rho+\rho\cos\phi d\phi$$ so that $$dxdy=\frac{1}{2}\sin2\phi\big((d\rho)^2-\rho^2(d\phi)^2\big)+\cos2\phi(\rho d\rho d\phi)\neq \rho d\rho d\phi.$$ Where am I going wrong? If this approach is misguided I want someone to explain why.

Answer 1

$(d\rho)^2=(d\phi)^2=0$, $d\phi d\rho= -d\rho d\phi$, so the terms $\cos\phi\sin\phi (d\rho)^2, \rho^2\cos\phi\sin\phi(d\phi)^2$ are zero, while the mixed term has as coefficient $\rho(\cos^2(\phi)--\sin^2(\phi))=\rho$

Answer 2

Writing $dA=dx\,dy$ is misleading. The "product" $dx\,dy$ only makes sense in double integrals, like $$\int_0^1 \int_{a(x)}^{b(x)}f(x,y)\>dy\>dx\ .$$ In fact ${\rm d}A$ is an (unsigned) measure. With respect to the euclidean metric in the coordinates $(x,y)$ the area measure ${\rm d}A$ is just the product measure inherited from the standard measure on the axes. I write $${\rm d}A={\rm d}(x,y)\ .$$

When you introduce polar coordinates $(\rho,\phi)$ in the given $(x,y)$-plane a mapping $${\rm rect}:\quad(\rho,\phi)\mapsto (x,y):=(\rho\cos\phi,\rho\sin\phi)$$ comes into play. If you want t express the "old" euclidean area ${\rm d}A$ in terms of the auxiliary variables $\rho$, $\phi$ you need the Jacobian $$J_{\rm rect}(\rho,\phi)=\det\left[\matrix{x_\rho&x_\phi\cr y_\rho&y_\phi\cr}\right]=\ldots=\rho\ .$$ You then can say that $${\rm d}A={\rm d}(x,y)=\bigl|J_{\rm rect}(\rho,\phi)\bigr|\>{\rm d}(\rho,\phi)=\rho\>{\rm d}(\rho,\phi)\ .\tag{1}$$ The proof of $(1)$ is a long story.