# RV's X,Y uniformly distributed in the 2D shape — More information?

In my study of probability I've come across exercise questions talking about random variables uniformly distributed in disks and triangles.

I'd like to know if there are "standard" ways of approaching questions like "Find the density of the random variable $X$", "Find the joint density of the random vector $(X,Y)$" and "Find the joint distribution function of the random vector $(X,Y)$" when $X$ and $Y$ are random variables uniformly distributed over a miscellaneous two dimensional shape.

Note: My definitions of disk and triangle are $D := \{ (x,y) : x^2 + y^2 \leq r^2 \}$ and $T := \{ (x,y) : x \leq 1, 0 \leq y \leq x \}$ but I have seen others.

So far the only concrete thought I have come across is that the joint density of a random vector $(X,Y)$ is $\frac{1}{A}$ where $A$ is the area of the shape in question. This holds true since the integral of the joint density over it's domain is defined to be $1$, and integrating $1$ over that same domain is equivalent to finding it's area.

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The vector $(X,Y)$ is uniformly distributed on the domain $S$ of finite positive area $A$ if and only if the density $f$ of $(X,Y)$ is $f(x,y)=A^{-1}\mathbf 1_S(x,y)$ (and not just $A^{-1}$). In particular, the density $f_X$ of $X$ is such that $$f_X(x)=A^{-1}\int_{\mathbb R}\mathbf 1_S(x,y)\mathrm dy,$$ and the joint distribution function $F$ of $(X,Y)$ is such that $$F(x,y)=P[X\leqslant x,Y\leqslant y]=\iint_{u\leqslant x,v\leqslant y}f(u,v)\mathrm du\mathrm dv=A^{-1}\iint_{u\leqslant x,v\leqslant y}\mathbf 1_S(u,v)\mathrm du\mathrm dv.$$