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I know that when converting a double integral from Cartesian to Polar Coordinates, the Jacobian is equal to r and so we get $\iint dxdy$ $=$ $\iint rdrd\theta$. But what if I wanted to go from Polar to Cartesian Coordinates? What would the Jacobian be?

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Use $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\dfrac{y}{x}$, then obtain $$\left|\begin{array}{rr}\dfrac{\partial r}{\partial x} & \dfrac{\partial r}{\partial y} \\\dfrac{\partial \theta}{\partial x} & \dfrac{\partial \theta}{\partial y} \end{array}\right|=\dfrac{1}{\sqrt{x^2+y^2}}=\color{blue}{\dfrac1r}$$

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    $\begingroup$ (+1). OP, please note that for any valid coordinate transformation of $\mathbb{R}^2$, if the Jacobian is $J$, the Jacobian of the inverse transformation is $\frac{1}{J}$. $\endgroup$ – Mathemagical Oct 1 '17 at 5:04

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