Confusion with usage of density function.

Question

I'm trying to solve an exercise and im starting to get very confused of my usage of the density function. Can someone explain my where my error is? I have the two coordinates $X_1,X_2$ that are uniformly distributed on the disk $D=\{(x_1,x_2)\in\mathbb{R}^2:x_1^2+x_2^2<1\}$. At first I want to calculate the density function of $(X_1,X_2)$, lets call it $f$. We know it has to be constant and the integral of $f$ over $D$ has to equal $1$. So by $$\int_Df(x_1,x_2)d(x_1,x_2)=1$$ we know that the density function has to equal $1/\pi$. But now I want to calculate the c.d.f. of $R^2=X_1^2+X_2^2$. So let $g$ be density function of $R^2$. But then I think I use the definition of density function wrong. Because if I calculate the following I get a contradiction.\begin{align*}1=\mathbb{P}=(R^2\leq1)&=\int_0^1g(r)dr\\&=\int_0^1\int_0^1f(\sqrt{r-s},\sqrt{s})dsdr\\&=1/\pi\int_0^1\int_0^11dsdr=1/\pi.\end{align*} My thought process was, that for $R^2=r$ to hold true, we can just "sum up" over all $s\in[0,1]$ with $X_1^2=r-s,X_2^2=s$. But there is obviously a logical error. Can sombody tell me where exactly my fault is and what the right approach to calculate the c.d.f. of $R^2$ is? By logical arguments I know it's $\mathbb{P}(R^2\leq r) = r$, but I'd like to calculate it with the integrals.

Answer 1

If you let $Q=R^2$, then for $0 \le r \le 1$ you have

$$\mathbb P(R^2 \le r) =\mathbb P(Q \le r) = \int_{x_2=-\sqrt{r}}^{\sqrt{r}}\int_{x_1=-\sqrt{r-x_2^2}}^{\sqrt{r-x_2^2}} \frac{1}{\pi}\, dx_1\, dx_2 =\int_{x_2=-\sqrt{r}}^{\sqrt{r}}\frac{2 \sqrt{r-{{x_2^2}}}}{\pi }\, dx_2 =r$$ as expected. It gives $1$ when $r=1$, as it should.

The corresponding pdf for $Q$ will be $1$ as this is uniform on $[0,1]$, while the pdf for R will be $\frac d{dx} x^2=2x$ also on $[0,1]$