Let the joint probability density function of $X,Y$ is given as:
$f(x,y)=\frac{1}{4}[1+x^3y^3]$ where $-1\leq x \leq 1$ and $-1\leq y \leq 1$. It is required to obtain the joint probability density function of $(X^2,Y^2)$.
My approach
I tried to solve this problem using CDF approach:
$F(X^2\leq x, Y^2\leq y) = P(-\sqrt{x}\leq X \leq \sqrt{x},-\sqrt{y}\leq Y \leq \sqrt{y} )$
The above can be easily evaluated using the double integral. It turns out:
$\int_{-\sqrt{x}}^{\sqrt{x}}\int_{-\sqrt{y}}^{\sqrt{y}}\frac{1}{4}(1+x^3y^3)dydx$
After solving the above integral, the following result is obtained which is nothing but the joint CDF of $X^2$ and $Y^2$:
$F(X^2,Y^2)=\sqrt{yx}$ where $0\leq x \leq 1$ and $0\leq y \leq 1$. Now, differentiating this CDF two times with respect to both $x$ and $y$, we get:
$f(x,y)=\frac{1}{4\sqrt{xy}}$, where $0 \leq x \leq 1$ and $0 \leq y \leq 1$.
What is wrong with this approach that I just used since my manual used the transformation approach and come up with a different answer. Please ignore my choice of variable's symbols.
Thanks.