Help me in this probability question Original : come si procede in generale se ho delle variabili aleatorie doppie uniformi su un triangolo o su un quadrato?come determino la densità congiunta e le marginali della x e della y? c' entrano i domini normali e se c entrano come c entrano? se ho una variabile aleatoria doppia (X;Y) uniforme sul quadrato o sul triangolo come trovo la funzione di ripartizione di una variabile aleatoria U che dipende da x e da y?ho difficoltà con il seguente esercizio:trovare la distr di U=X/X+Y se (X,Y)-Unif(0,1)*(0,1) e non ben capito perchè si distinguono due casi. dove posso trovare informazioni riguardo a questi argomenti dato che sul libro di testo sono poco trattati?

Translated :
How do I proceed in general if I have double uniform random variables on a triangle or on a square? How do I determine the joint density and marginals of x and y?
Do normal domains enter and if c enter how do c enter? If I have a double random variable (X; Y) uniform on the square or on the triangle as I find the distribution function of a random variable U which depends on x and y? 
I have difficulties with the following exercise:
Find the distribution of $U = X / (X + Y)$ if $(X, Y) \sim \operatorname{Uniform} (0,1) \times (0,1)$. Also I don't understand well why two cases are distinguished. 
Where can I find information on these topics since they are poorly treated in the textbook?
 A: $$
U = \frac X {X+Y}.
$$
The cumulative probability distribution function of the random variable $U$ gives the probability that (capital) $U$ is less any particular number (lower-case) $u$:
$$
F_U(u) = \Pr(U\le u) = \Pr\left( \frac X {X+Y} \le u \right).
$$
Since $X$ and $Y$ are positive, this fraction $X/( X+Y)$ must be between $0$ and $1.$ So we have $F_U(u) = 1$ if $u\ge1$ and $F_U(u) = 0$ if $u\le0.$ The problem is to find $F_U(u)$ when $0<u<1.$
First we can simplify the inequality $X/(X+Y) \le u$:
\begin{align}
& \frac X{X+Y} \le u \\[8pt]
& X \le u(X+Y) = uX+uY \\[8pt]
& X-uX \le uY \\[8pt]
& X(1-u) \le uY \\[8pt]
& \text{Since $0<u<1,$ we have $1-u>0$} \\
& \text{so when we divide both sides by $1-u,$} \\
& \text{the “$\le$'' remains “$\le$'' and does not} \\ & \text{become “$\ge$'' :} \\[8pt]
& X \le \frac u {1-u} Y.
\end{align}
The line $x = uY/(1-u)$ passes through the point $(0,0).$ Draw its graph and you see that it cuts the square into two parts: a triangle and a quadrilateral (except when $u=1/2,$ so that $u/(1-u)=1,$ in which case you get two isosceles right triangles.
The probability is the area of the part of the square where $x$ is smaller than it is in the other part.
A: So generally you have a joint distribution of random variables $X,Y$ which is uniform over any domain $D$ in $\mathbb{R}^2$, then
$$
A(D) = \text{Area}(D) = \iint_D \,dx\,dy
$$
and the joint probability of $(X,Y)$ being in any region $R \subseteq D$ is given by
$$
f_{XY}(R) = \frac{1}{A(D)} \iint_R \,dx\,dy
$$
Deriving the individual distributions will depend on the shape of $D$, since e.g.
$$
f_X(x) = \int_D f_{XY}(x,y) \,dy
$$
and the bounds on the integral depend on the shape of $D$...

In your other example, you are interested in the distribution of $Z = \frac{X}{X+Y}$ where $X,Y \sim \mathcal{U}(0,1)$. Here, you can compute
$$
F_Z(z)
 = \mathbb{P}[Z \le z]
 = \mathbb{P}\left[\frac{X}{X+Y} \le z\right]
 = \int_0^1 \int_0^1 \mathbb{I}_{\{\frac{x}{x+y} \le z\}} \,dx\,dy
$$
can you absorb the indicator function into the limits of the integral?
