This is in my lecture notes, I understand that $x=rcos\theta$ so therefore its the integral of $x^2=(rcos\theta)^2$ but why is there an r in $rdrd\theta$ in there?

Any help would be appreciated.


When we convert co-ordinates $(x,y)$ to $(r,\theta)$,

$dx\ dy = |J|\ dr \ d\theta$

where $J$ is the Jacobian of transformation.

$x = r \cos\theta \ , \ y = r \sin\theta$

Taking partial derivatives,

$x_r = \cos\theta \ , y_r = \sin\theta$

$x_{\theta} = - r \sin\theta \ , \ y_{\theta} = r \cos\theta $

$J = \begin{vmatrix} \cos\theta & \sin\theta \\ - r \sin\theta & r \cos\theta \end{vmatrix}$

$J = r(\cos\theta)(\cos\theta) + r(\sin\theta)(\sin\theta) = r(\cos^2\theta + \sin^2\theta) = r$

$$|J| = r$$ Thus, $$dx \ dy = r \ dr \ d\theta$$

  • $\begingroup$ Very clearly laid out, thank you! $\endgroup$
    – king
    May 11 '19 at 15:52
  • $\begingroup$ You're welcome. $\endgroup$
    – Ak.
    May 11 '19 at 15:52
  • $\begingroup$ good answer, but it would look better and be easier to read if you put the backslash \ in front of all the "sin" and "cos" in it instead of just two of the "cos". $\endgroup$ May 11 '19 at 17:02
  • $\begingroup$ @Paul Sinclair Thanks for your advice, I've edited it :) $\endgroup$
    – Ak.
    May 11 '19 at 17:10
  • $\begingroup$ But you didn't put in the backslashes: r\sin \theta produces this: $$r\sin\theta$$ while r sin \theta produces this: $$r sin \theta$$ Most common functions include a special operator name form like this: \sin \cos \tan \log \ln \arcsin, \min, \max, \sup, \inf, \lim, \limsup, etc. Putting them in helps the TEX engine figure out how better to format the text (more generally you can use \operatorname{myoperator} to do the same with a "myoperator" function. $\endgroup$ May 11 '19 at 17:15

Here's the underlying geometric intuition (not a rigorous argument).

In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.

In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $d\theta$ and $dr$ is $ dr \times r d\theta = r dr d\theta$.

There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html


That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $\mathrm d x\,\mathrm d y$ is replaced with $r\,\mathrm d r\,\mathrm d\theta$.

Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment $\mathrm d r$ of the radius and a small increment of the angle $\mathrm d\theta$ is approximately $\;\mathrm dr\cdot r\,\mathrm d\theta$


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