The random variables $X_1$ and $X_2$ are independent and equidistributed with density function: $f(x) = 4x^3$, for $0 ≤ x ≤ 1$ and $0$, otherwise.
Set $Y_1 = X_1\sqrt{X_2}$ and $Y_2 = X_2\sqrt{X_1}$. Find the joint PDF of $Y_1;Y_2$
I found the marginals for $Y_1$ and $Y_2$, but how can I find the joint?
$U=\sqrt{X_2}=>X_2=U^2=>f_u(u)=f_{x_2}(u^2)2u=\int_0^12u4(u^2)^3dx_2=8u^7$ the same for $\sqrt{X_1}$ then I used independence and found the marginals of $Y_1, Y_2$.
It is a quite tedious job but I give you some hints: Let $X_1,X_2$ be the two independent random variables with densities: $f(x)=4x^3$, $0<x<1$. Because they are independent, their joint density is just the product: $$f_{(X_1,X_2)}(x_1,x_2)=16x_1^3x_2^3,\quad (x_1,x_2)\in[0,1]^2.$$
Define the funcion $g:\mathbb{R}^2\rightarrow\mathbb{R}^2$ by $g(x_1,x_2)=(x_1\sqrt{x_2}, x_2\sqrt{x_1})$. You want to find the joint density of the vector $(Y_1,Y_2)=g(X_1,X_2)$.
In general the joint density of a tranformation of a (two-dimensional) random vector is given by: $$f{(Y_1,Y_2)}(y_1,y_2) = f_{(X_1,X_2)}(g^{-1}(y_1,y_2))|Jg^{-1}(y_1,y_2)|,$$ where $Jg^{-1}(y_1,y_2)$ is the Jacobian matrix of the inverse tranformation.
The first step is then to find the inverse $g^{-1}$ which happens to be: $$g^{-1}(y_1,y_2)=\left( \frac{y_1^{4/3}}{y_2^{2/3}}, \frac{y_2^{4/3}}{y_1^{2/3}}\right).$$
The Jacobian of $g^{-1}$ is: $$Jg^{-1}(y_1,y_2)=\begin{pmatrix} \frac{4}{3}\frac{y_1^\frac{1}{3}}{y_2\frac{2}{3}}& -\frac{2}{3}\frac{y_1^\frac{4}{3}}{y_2\frac{5}{3}} \\ -\frac{2}{3}\frac{y_2^\frac{4}{3}}{y_1\frac{5}{3}}&\frac{4}{3}\frac{y_2^\frac{1}{3}}{y_1\frac{2}{3}} \end{pmatrix}$$ which has determinant: $|Jg^{-1}(y_1,y_2)|=\frac{4}{3}\frac{1}{(y_1y_2)^{1/3}}$.
The joint density of $(Y_1,Y_2)$ is thus $$f{(Y_1,Y_2)}(y_1,y_2) =16\left(\frac{y_1^{4/3}}{y_2^{2/3}}\right)^3\left(\frac{y_2^{4/3}}{y_1^{2/3}}\right)^3 \frac{4}{3}\frac{1}{(y_1y_2)^{1/3}},$$ and you may simplify things.
The most demanding part in my opinion is to find the exact domain of $(y_1,y_2)$ when $(x_1,x_2)\in[0,1]^2$. Observe that they are interrelated and that $y_1$ and $y_2$ can be negative although $x_1,x_2$ are positive.
Check for eventual mistakes too! but you see the procedure. I hope it helped! Good luck!