The joint distribution of $X,Y$ is given by $f(x,y)=\frac{1}{4}(1+x^3y^3)$

Question

The joint distribution of $X,Y$ is given by $f(x,y)=\frac{1}{4}(1+x^3 y^3)$ Find the joint distribution of $X^2,Y^2$.

I observed that $X,Y \sim \text{Uniform}(-1,1)$ Also, $X,Y$ are not independent as $f_{Y|X}(y|x)=\frac{1}{2} \neq f(y)$ The usual approach would be transforming $(X,Y) \rightarrow (U,V) $ where $U=X^2, V=Y^2$ The transformation is not one-one as $X,Y$ can take negative values. It becomes complicated after that. Help!

Answer 1

Let $u, v \in (0,1),$

\begin{align} P(X^2 \le u, Y^2 \le v) &=P(-\sqrt{u} \le X \le \sqrt{u}, -\sqrt{v} \le Y \le \sqrt{v})\\ &= \int_{-\sqrt{u}}^\sqrt{u} \int_{-\sqrt{v}}^\sqrt{v} f(x,y) \,\, dydx\\ &= \frac14\int_{-\sqrt{u}}^\sqrt{u} \int_{-\sqrt{v}}^\sqrt{v}(1+x^3y^3) \,\, dydx\\ \end{align}

Hopefully you can evaluate the integral.