# Is The Jacobian Needed to Find CDF for R in Polar Coordinates?

I'm attempting to use inversion sampling to generate points on a disk according to the following PDF:

$$f(r) = \dfrac{2}{\pi(1+r^2)}$$

Here, the polar angle would just be a uniform random variable in $$[0,2\pi)$$. My question is this: when I go to integrate the PDF to find the CDF, do I include the $$r$$ from the Jacobian in the integral:

$$F(R) = \displaystyle\int_0^{R}f(r)r\hspace{0.1cm}dr$$

Or do I treat $$R$$ as any other random variable and just compute the following:

$$F(R) = \displaystyle\int_0^{R}f(r)\hspace{0.1cm}dr$$

Or is there something else that I'm missing, as well? I'm relatively new to working with CDFs as my physics undergraduate education was lacking in terms of statistics, so I greatly appreciate any help with this!

No, you should not include $$r$$. As a quick check, cdfs require that $$\lim_{R \to \infty} F(R) = 1$$. Without the $$r$$, you get
$$F(R) = \int_0^R f(r) dr = \frac{2}{\pi} (\arctan(R) - \arctan(0)) = \frac{2}{\pi} \arctan(R) \to 1 \text{ as } R \to \infty,$$ which is what we expected. If you do include the $$r$$, you get
$$F(R) = \int_0^R f(r) r dr = \frac{1}{\pi} \log(1 + R^2) \to \infty \text{ as } R \to \infty,$$ which is not a cdf.
Here, the $$r$$ term that you are considering is the determinant of the Jacobian of a change-of-coordinates transformation, namely $$g(r, \theta) = (r \cos(\theta), r \sin(\theta))$$, which maps polar coordinates to Cartesian coordinates. But you aren't actually changing coordinates here, as your pdf is already given as a pdf with respect to $$r$$.