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.

  • $\begingroup$ Why not derive $\rho d\rho d\phi$ from skratch? Like I did in this answer (using $r$ and $\theta$, but still). $\endgroup$ – Arthur Apr 24 at 12:44
  • $\begingroup$ @Arthur Do you want me to supply the steps? $\endgroup$ – mithusengupta123 Apr 24 at 12:45
  • $\begingroup$ No, I'm just asking, if you have trouble translating from rectangular to polar, why not just use polar from the start, and see that you do indeed get what you ought to get? $\endgroup$ – Arthur Apr 24 at 12:46
  • $\begingroup$ @Arthur I was trying to understand what is wrong with my approach :-( $\endgroup$ – mithusengupta123 Apr 24 at 12:50
  • $\begingroup$ That's fair enough. $\endgroup$ – Arthur Apr 24 at 12:51

$(d\rho)^2=(d\phi)^2=0$, $d\phi d\rho= -d\rho d\phi$, 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 $\rho(\cos^2(\phi)--\sin^2(\phi))=\rho$

  • $\begingroup$ Why $(d\rho)^2=(d\phi)^2=0$? Why $d\rho d\phi=-d\phi d\rho$? $\endgroup$ – mithusengupta123 Apr 24 at 13:04
  • $\begingroup$ because technically they are all differential forms so antisymmetric and the area element is the absolute value (so the correct formal way is $dxdy=|dx$^$dy|=....$ so we get stuff like $dr$^$dr$=$-dr$^$dr$, so $dr$^$dr=0$, $(dr)^2=|dr$^$dr|=0$, but intuitively $(dr)^2$ is just a one dimensional element going in the $r$ coordinate only, so its 2-dimensional area is zero, while switching $drd\phi$ means changing orientation in the $r-\phi$ plane so the minus; formally though you need differential form calculus as noted above $\endgroup$ – Conrad Apr 24 at 13:09
  • $\begingroup$ @Conrad There's still time to fix a dollar misplacement that's ruined the formatting in your comment. $\endgroup$ – J.G. Apr 24 at 13:12
  • 2
    $\begingroup$ done - annoying that one cannot see comment formatting when typing $\endgroup$ – Conrad Apr 24 at 13:13
  • 2
    $\begingroup$ If you want to avoid differential form calculus, the Jacobian approach (which is equivalent but uses implicitly the properties of differential forms through matrix determinant computation) is the one to use as noted in the comments to your question $\endgroup$ – Conrad Apr 24 at 13:24

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.

  • $\begingroup$ If anyone wants a hint (without differential forms) as to the "long story", a coordinate transformation admits a chain rule of the form $dx_i=\sum_j J_{ij}dy_j$ or equivalently $d\mathrm{x}=Jd\mathrm{y}$ for some square matrix $J$. We seek a function $f$ satisfying $d^n\mathrm{x}=f(J)d^n\mathrm{y}$ when $\mathrm{x},\,\mathrm{y}$ are $n$-dimensional. The special cases of scaling, the composition of multiple transformations etc. tell us $f(J)=|\det J|$. $\endgroup$ – J.G. Apr 24 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.