<s>[INST] Joint probability distribution (over unit circle) A couple of two continuous random variables $(X,Y)$ is distributed uniformly over the closed unity circle (so $-1\leq x \leq 1$ , $y$ analog). $U$ is defined as the distance from $O$ to the point $(X,Y)$. Calculate U:
My calculations so far: 
$$
f_{XY}(x,y) = \frac{1}{\pi} ; (x,y) \in S
$$ 
Where S is the unit circle. 
$$
\begin{align}
F_{U}(u) &= Pr((x,y): \sqrt{X^2+Y^2}\leq u)\\
&= \int_{S}\int f_{X}(x)f_{Y}(y)dxdy\\
&= \int_{-u}^{u}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{4}{\pi^2}dydx\\
&= \frac{8}{\pi^2}\left(u\sqrt{1-u^2}+\arcsin{u}\right)
\end{align}
$$
When $0<u\leq1$. However the answer is $F_{U}(u) = u^2$ when $0<u\leq1$. 

Where am I going the wrong way in my argumentation? 

 [/INST] Answer: On the second line of your derivation.
Let's do it correctly:
$$
\begin{eqnarray}
   F_U\left(u\right) &=& \Pr\left(\sqrt{X^2+Y^2} \leqslant u \right) \\
   &=& \int_S f_{X,Y}\left(x,y\right) I\left(\sqrt{X^2+Y^2} \leqslant u\right) \mathrm{d}x\mathrm{d}y \\
   &=& \left.\int_{-1}^{1} \left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}   \frac{1}{\pi} I\left(\sqrt{X^2+Y^2} \leqslant u\right) \mathrm{d} y \right)\mathrm{d} x \right|_{\text{assuming } 0<u \leqslant u} \\
   &=& \int_{-u}^{u} \left(\int_{-\sqrt{u^2-x^2}}^{\sqrt{u^2-x^2}}   \frac{1}{\pi} \mathrm{d} y \right)\mathrm{d} x \\
   &=& \left.\int_{-u}^{u} \frac{2}{\pi} \sqrt{u^2-x^2} \mathrm{d}x \right|_{x = z \cdot u} \\
  &=& u^2 \int_{-1}^{1} \frac{2}{\pi} \sqrt{1-z^2} \mathrm{d}z = u^2
\end{eqnarray}
$$
As a side note, it would have been easier to work in polar coordinates here.
</s>