Find the probability density function on a unit disk Let $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq1\}$ be a unit disk and $(X,Y)$ be uniformly distributed on D. Find the probability density function of W for

*

*$W=X+Y$

*$W=Y/X$
I think I need to find a cdf and then differentiate it to find the pdf. However, I have never tried finding a cdf for a circle.
Should I split the circle into four parts and then find the pdf for each of them so that each part os either strictly increasing or stringly decreasing?
 A: The joint probability density function is $f_{\small X,Y}(x,y)= \dfrac{1}{\pi}\mathbf 1_{x^2+y^2\leqslant 1}$.
Your solution requires integrating wrt $x$ and $y$ then deriving wrt the composite variable so... use the fundamental theorem of calculus to avoid some unnecessary calculations.

Thus
$$\begin{align}f_{\small X+Y}(w) &= \frac{\mathrm d ~~}{\mathrm d w}\iint_{x^2+y^2\leq 1 , x+y\leqslant w}f_{\small X,Y}(x,y)\mathrm d x\mathrm d y\\&=\mathbf 1_{-\surd 2\leqslant w\lt \surd 2}\int_{x^2+(w-x)^2\leq 1}\left\lvert\dfrac{\partial (w-x)}{\partial w}\right\rvert f_{\small X,Y}(x,w-x)\,\mathrm d x\\&=\mathbf 1_{-\surd 2\leqslant w\lt\surd 2}\int_{(w-\sqrt{2-w^2})/2}^{(w+\sqrt{2-w^2})/2}\frac 1\pi\,\mathrm d x\\&~~\vdots\end{align}$$

Likewise
$$\begin{align}f_{\small Y/X}(w) &= \frac{\mathrm d ~~}{\mathrm d w}\iint_{x^2+y^2\leq 1 , y/x\leqslant w}f_{\small X,Y}(x,y)\mathrm d x\mathrm d y\\&=\mathbf 1_{-1\leqslant w\lt 1}\int_{x^2+(xw)^2\leq 1}\left\lvert\dfrac{\partial xw}{\partial w}\right\rvert f_{\small X,Y}(x,xw)\,\mathrm d x\\&=\mathbf 1_{-1\leqslant w\lt 1}\int_{x^2\leq 1/(1+w^2)} \lvert x\rvert\cdot\frac 1\pi\,\mathrm d x\\&~~\vdots\end{align}$$
