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.
From $dxdy$ to $\rho d\rho d\phi$. Where am I doing wrong
coordinate systemsderivativespartial derivativepolar coordinates
Related Solutions
Division in partial derivatives is just a notation so $\frac{\partial r}{\partial x} \neq \frac{\partial x}{\partial r}^{-1}$.
It's not wrong. It's totally correct. Your choice of $\hat{e}_{\phi}$ (implicit in the Jacobian, as you used the function $\phi$ and not the function $R\phi$) makes its norm proportional to $R$, hence the normalising coefficient $\frac{1}{R}$.
Best Answer
$(d\rho)^2=(d\phi)^2=0$, $d\phi d\rho= -d\rho d\phi$, 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 $\rho(\cos^2(\phi)--\sin^2(\phi))=\rho$