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I'm trying to compute an integral over $\mathbb{R}^2$ using polar coordinates, as a disk with $\infty$ radius.

I think the following should work, but I got twice the expected result.

Integrate[
 HeavisideTheta[
  r - Norm[FromPolarCoordinates[{\[Rho], t}] - x0]], {\[Rho], 0, 
  Infinity}, {t, 0, 2 \[Pi]}]

however the following provides correct results:

Integrate[
 HeavisideTheta[r - Norm[{x, y} - x0]], {x, y} \[Element] 
  Disk[{0, 0}, Infinity]]

What's the difference between them?

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When my rusty brain is correct, then you forgot to include the determinant of the Jacobian matrix that you need to include in your integrand if you are integrating with a different coordinate system. In your case it is simply multiplying by $r$

$$ \iint _{\mathbf {F} (A)}f(x,y)\,dx\,dy=\iint _{A}f(r\cos \varphi ,r\sin \varphi )\,r\,dr\,d\varphi$$

This gives:

x0 = {0, 0};
r = 1;

Integrate[ρ*
  HeavisideTheta[
   r - Norm[FromPolarCoordinates[{ρ, t}] - x0]], {ρ, 0, 
  Infinity}, {t, 0, 2 π}]
(* π *)

Integrate[
 HeavisideTheta[r - Norm[{x, y} - x0]], {x, y} ∈ 
  Disk[{0, 0}, Infinity]]
(* π *)
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  • 2
    $\begingroup$ For people too lazy to remember the Jacobian determinant: CoordinateChartData["Polar", "VolumeFactor", {ρ, t}]. $\endgroup$ – J. M. will be back soon May 12 '17 at 7:04

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