<s>[INST] Distribution of $\sqrt{X^2+Y^2}$ when $X,Y$ are independent $U(0,1)$ variables 
As a routine exercise, I am trying to find the distribution of $\sqrt{X^2+Y^2}$ where  $X$ and $Y$ are independent $ U(0,1)$ random variables.

The joint density of $(X,Y)$ is  $$f_{X,Y}(x,y)=\mathbf 1_{0<x,y<1}$$
Transforming to polar coordinates $(X,Y)\to(Z,\Theta)$ such that $$X=Z\cos\Theta\qquad\text{ and }\qquad Y=Z\sin\Theta$$ 
So, $z=\sqrt{x^2+y^2}$ and $0< x,y<1\implies0< z<\sqrt 2$.
When $0< z<1$, we have $0< \cos\theta<1,\,0<\sin\theta<1$ so that $0<\theta<\frac{\pi}{2}$.
When $1< z<\sqrt 2$, we have $z\cos\theta<\implies\theta>\cos^{-1}\left(\frac{1}{z}\right)$, as $\cos\theta$ is decreasing on $\theta\in\left[0,\frac{\pi}{2}\right]$; and $z\sin\theta<1\implies\theta<\sin^{-1}\left(\frac{1}{z}\right)$, as $\sin\theta$ is increasing on $\theta\in\left[0,\frac{\pi}{2}\right]$.
So, for $1< z<\sqrt 2$, we have $\cos^{-1}\left(\frac{1}{z}\right)<\theta<\sin^{-1}\left(\frac{1}{z}\right)$.
The absolute value of jacobian of transformation is $$|J|=z$$
Thus the joint density of $(Z,\Theta)$ is given by 
$$f_{Z,\Theta}(z,\theta)=z\mathbf 1_{\{z\in(0,1),\,\theta\in\left(0,\pi/2\right)\}\bigcup\{z\in(1,\sqrt2),\,\theta\in\left(\cos^{-1}\left(1/z\right),\sin^{-1}\left(1/z\right)\right)\}}$$
Integrating out $\theta$, we obtain the pdf of $Z$ as
$$f_Z(z)=\frac{\pi z}{2}\mathbf 1_{0<z<1}+\left(\frac{\pi z}{2}-2z\cos^{-1}\left(\frac{1}{z}\right)\right)\mathbf 1_{1<z<\sqrt 2}$$

Is my reasoning above correct? In any case, I would like to avoid this method and instead try to find the cdf of $Z$ directly. But I couldn't find the desired areas while evaluating $\mathrm{Pr}(Y\le \sqrt{z^2-X^2})$ geometrically.

EDIT.
I tried finding the distribution function of $Z$ as
\begin{align}
F_Z(z)&=\Pr(Z\le z)
\\&=\Pr(X^2+Y^2\le z^2)
\\&=\iint_{x^2+y^2\le z^2}\mathbf1_{0<x,y<1}\,\mathrm{d}x\,\mathrm{d}y
\end{align}
Mathematica says this should reduce to
$$F_Z(z)=\begin{cases}0 &,\text{ if }z<0\\ \frac{\pi z^2}{4} &,\text{ if } 0< z<1\\ \sqrt{z^2-1}+\frac{z^2}{2}\left(\sin^{-1}\left(\frac{1}{z}\right)-\sin^{-1}\left(\frac{\sqrt{z^2-1}}{z}\right)\right) &,\text{ if }1< z<\sqrt 2\\ 1 &,\text{ if }z>\sqrt 2 \end{cases}$$
which looks like the correct expression. Differentiating $F_Z$ for the case $1< z<\sqrt 2$ though brings up an expression which doesn't readily simplify to the pdf I already obtained.
Finally, I think I have the correct pictures for the CDF:
For $0<z<1$ :
 
And for $1<z<\sqrt 2$ :

Shaded portions are supposed to indicate the area of the region $$\left\{(x,y):0<x,y< 1\,,\,x^2+y^2\le z^2\right\}$$
The picture immediately yields 
\begin{align}
F_Z(z)&=\Pr\left(-\sqrt{z^2-X^2}\le Y\le\sqrt{z^2-X^2}\right)
\\&=\begin{cases}\frac{\pi z^2}{4} &,\text{ if } 0<z<1\\\\ \sqrt{z^2-1}+\int_{\sqrt{z^2-1}}^1 \sqrt{z^2-x^2}\,\mathrm{d}x &,\text{ if }1< z<\sqrt 2 \end{cases}
\end{align}
, as I had previously found.
 [/INST] For $0 \leq z \leq 1$, $P\left(\sqrt{X^2+Y^2} \leq z\right)$ is just the area of the quarter-circle of radius $z$ which is $\frac 14 \pi z^2$.  That is, 
$$\text{For }0 \leq z \leq 1, ~\text{area of quarter-circle} = \frac{\pi z^2}{4} = P\left(\sqrt{X^2+Y^2} \leq z\right).$$
For $1 < z \leq \sqrt{2}$, the region over which we need to integrate to find $P\left(\sqrt{X^2+Y^2} \leq z\right)$can be divided into two right triangles $\big($one of them has vertices $(0,0), (0,1)$ and $(\sqrt{z^2-1}, 1)$ while the other has vertices $(0,0), (1,0)$ and $(1, \sqrt{z^2-1})$ $\big)$ together with a sector of a circle of radius $z$ and included angle $\frac{\pi}{2}-2\arccos\left(\frac{1}{z}\right)$. The area of this region (and hence the value of $\left( P(\sqrt{X^2+Y^2} \leq z\right)$) is easily found. We have that for $1 < z \leq \sqrt{2}$,
\begin{align}\text{area of region} &= \text{area of two triangles plus area of sector}\\
&=\sqrt{z^2-1} + \frac 12 z^2\left( \frac{\pi}{2}-2\arccos \left(\frac{1}{z}\right)\right)\\
&= \frac{\pi z^2}{4} + \sqrt{z^2-1} - z^2\arccos \frac{1}{z}\\
&= \left( P(\sqrt{X^2+Y^2} \leq z\right)\end{align}
which is the result in Martijn Wetering's answer.
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