2-dimensional uniform distribution on a given set. If $X,Y$ are uniformely  distributed on a quadrilateral with coordinates{$(0,0),(2,0),(1,1),(3,1)$}.
I want to find their marginal densities and $P(X>1|Y=\frac{1}{2}$).
I get that the joint pdf is  area$^{-1}$ multiplied by the  indicator over the domain of each variable and the double integral of this is equal to $1$.
In this shape that is given, the area is $2$ so our constant is $\frac{1}{2}$.
But I cant figure out how should I define the domain for the variables, should I make a piece wise function?
Also, is there a way to figure out the marginals just by looking at the shape of the given set?
Any help is greatly appreciated!
 A: Yes, there is a way to figure out the marginals just by looking at the shape of the set. The pdf of $X$ is a distance between lower and upper boundaries of quadrilateral, divided by the area of quadrilateral. That is
$$
f_X(x)=\frac12\times \begin{cases} x,  & 0\leq x <1,\cr 
1, & 1\leq x < 2, \cr 
1-(x-2), & 2\leq x \leq 3, \cr
0, & x\not\in[0,\,3].\end{cases}
$$
The pdf of $Y$ is a distance between left and right boundaries of quadrilateral, divided by area of quadrilateral:
$$
f_Y(y)=\frac12\times \begin{cases} 2,  & 0\leq y \leq 1,\cr 
0, & y\not\in[0,\,1].\end{cases}
$$
This is just a consequence of definition of pdf and geometric probability. Say, pdf $f_X(x)$ is a function s.t. for any interval $(a,b)$
$$
\mathbb P(X\in (a,b)) = \int_a^b f_X(x)dx. \tag{1}
$$
We can restrict ourselves with $(a,b)\subset [1,\,3]$ only since outside this interval the pdf is zero. And this is the probability that the point which is chosen randomly in quadrilateral has $x$-coordinate in $(a,b)$. This is exactly ratio of two areas: area of the part of quadrilateral over $(a,b)$ and the total area which is $2$. 
And the area of the part of quadrilateral over $(a,b)$ is the integral over $(a,b)$ from distance between upper boundary $u(x)$ and lower boundary $l(x)$. So, 
$$
\mathbb P(X\in (a,b)) = \frac{\displaystyle\int_a^b (u(x)-l(x))\, dx}{2} \tag{2}
$$ 
Comparing (1) and (2) get $f_X(x)=\frac{u(x)-l(x)}{2}$.
