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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.

Michael Hardy
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r.zarei
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    If you want to know the intuition behind this, this answer and this question could be very useful. – Martin Sleziak Jan 11 '12 at 19:18
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    The area of a circular sector with outer radius $r_o$, inner radius $r_i$ and "spread" $\Delta\theta$ is $${\Delta\theta\over2}({r_o^2-r_i^2}) = {\Delta\theta\over2}({r_o +r_i }) ({r_o -r_i })={\Delta\theta } \cdot r_{avg}\Delta r\approx r\Delta\theta\Delta r. $$ When setting up a double integral, $r,dr,d\theta$ becomes your area element. – David Mitra Jan 11 '12 at 19:19
  • tanks guys. i just decided to remember that equation for exams:D. – r.zarei Jan 11 '12 at 19:27

2 Answers2

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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 $$dx\; dy=r\;dr\; d\theta.$$

Michael Hardy
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Paul
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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 $r\;dr\;d\theta$.

Michael Hardy
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