how to get $dx\; dy=r\;dr\;d\theta$ 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.
 A: 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 
$$dx\; dy=r\;dr\; d\theta.$$
A: 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 $r\;dr\;d\theta$.
