3
$\begingroup$

In polar coordinate how we can get $dx\;dy=r\;dr\;d\theta$?

with these parameters:

$r=\sqrt{x^2+y^2}$
$x=r\cos\theta$
$y=r\sin\theta$

Tanks.

  • 0
    tanks guys. i just decided to remember that equation $f$or exams:D.2012-01-11

2 Answers 2

4

In general, under the change of coordinates $u=u(x,y)$ and $v=v(x,y)$, the area element changes according to the formula: (See here) $du \;dv=\left|\frac{\partial (u,v)}{\partial(x,y)}\right|dx\; dy,$ where the Jacobaian is the determinant given by $\left|\frac{\partial (u,v)}{\partial(x,y)}\right|=\left| \begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ \end{array} \right|.$

Therefore, for polar coordinates $(x,y)=(r\cos\theta,r\sin\theta)$, we have $\left|\frac{\partial (x,y)}{\partial(r,\theta)}\right|=\left| \begin{array}{cc} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\ \end{array} \right|=\left| \begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \\ \end{array} \right|=r,$ which implies that $dx\; dy=r\;dr\; d\theta.$

  • 0
    tanks paul your solution is straightforward2012-01-12
1

If a circle has radius $r$, then an arc of $\alpha$ radians has length $r\alpha$. So with an infinitesimal increment $d\theta$ of the angle, the length is the infinitesimal $r\;d\theta$. And the arc is a right angles to the radius, which changes by the infinitesimal amount $dr$. So the infinitesimal area involved is just the product $r\;dr\;d\theta$.

  • 0
    Nevermind, I get what's happening now.2014-02-19