# Integration using polar coordinates

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?

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 *)

• For people too lazy to remember the Jacobian determinant: CoordinateChartData["Polar", "VolumeFactor", {ρ, t}]. – J. M. will be back soon May 12 '17 at 7:04