If $X$ and $Y$ are uniform$(-1,1)$, how can I find the distribution of $W=X^2+Y^2$? If $Y$ and $X$ are independent uniform (-1,1) random variables, I would like to derive the distribution of $W=X^2+Y^2$. 
At first I thought that I could use the CDF technique and a geometric argument, that is  divide the area of a circle with radius $\sqrt{w}$,   $0<w<2$ by $4$, the area of the $2$ by $2$ square that constitutes the support of $X$ and $Y$.
The problem with that approach is that the resulting pdf:
$$f_W (w)=\begin{cases} \pi /4,& 0<w<2 \\
0, & \text{elsewhere} \end{cases} $$
does not integrate to $1$. 
Could you please help me understand where I went wrong here and how to arrive at the right answer?
Thank you in advance.
 A: Your geometric approach will work. Draw a picture. If $0\lt w\le 1$, we get a circle that if fully inside the square on which the joint density "lives," and as you saw things are straightforward. 
If $1\lt w\lt 2$, we need to find the area of the part $D$ of the square that is in the disk of radius $\sqrt{w}$. Join the centre of the circle to the $8$ points where the circle meets the sides of the square. This divides our region $D$ into $8$ parts, $4$ isosceles triangles and $4$ circular sectors.
We find the combined area of these and divide by $4$, or equivalently find the area of a triangle and a circular sector, and add.
To find the area of a triangle, let us for example find the two places where $x^2+y^2=w$ meets the line $x=1$. This is at $y=\pm \sqrt{w-1}$. So the base of the triangle is $2\sqrt{w-1}$, and therefore the area is $\sqrt{w-1}$. 
The central angle of each triangle is $2\arctan(\sqrt{w-1})$, and therefore the angle of each of the $4$ circular sectors is $\frac{\pi}{2}-\arctan(\sqrt{w-1})$. For the area, divide by $2$. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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$\ds{}$
\begin{align}&\color{#66f}{\large%
\int_{-1}^{1}\int_{-1}^{1}\delta\pars{w - x^{2} - y^{2}}\,{\dd x \over 2}\,
{\dd y \over 2}}
\\[3mm]&={1 \over 4}\,2\int_{0}^{1}2\int_{0}^{1}\bracks{%
{\delta\pars{y + \root{w - x^{2}}} \over 2\verts{y}}
+{\delta\pars{y - \root{w - x^{2}}} \over 2\verts{y}}}\,\dd y\,\dd x
\\[3mm]&=\left. \half\int_{0}^{1}{\dd x \over \root{w - x^{2}}}\,
\right\vert_{w\ >\ x^{2}\,,\root{\vphantom{\Large A}w - x^{2}}\ <\ 1}
=\left. \half\int_{0}^{1}{\dd x \over \root{w - x^{2}}}\,
\right\vert_{w - 1\ <\ x^{2}\ <\ w}
\\[3mm]&=\left\lbrace\begin{array}{lcl}
\half\int_{0}^{\root{w}}{\dd x \over \root{w - x^{2}}} & \mbox{if} & 0 < w < 1
\\[2mm]
\half\int_{\root{w - 1}}^{1}{\dd x \over \root{w - x^{2}}} & \mbox{if} & 
1 \leq w < 2
\\[2mm]
0 && \mbox{otherwise}
\end{array}\right.
\\[3mm]&=\color{#66f}{\large\left\lbrace\begin{array}{lcl}
{\pi \over 4} & \color{#000}{\mbox{if}} & 0 < w < 1
\\[2mm]
\half\,\arcsin\pars{{2 \over w} - 1} & \color{#000}{\mbox{if}} & 
1 \leq w < 2
\\[2mm]
0 && \color{#000}{\mbox{otherwise}}
\end{array}\right.}
\end{align}



