To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$
$f(x,y) = \begin{cases} \dfrac{1}{4}(1+xy), & |x|<1,|y|<1 \\ 0, & \text{otherwise} \end{cases}$
The given solution starts with trying to prove that
\begin{align} $P(X^2\leq x \cap Y^2\leq y) & = P(|X|\leq \sqrt{x} \cap |Y|\leq \sqrt{y}) \\ & = \int\limits_{-\sqrt{x}}^{\sqrt{x}}\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(u,v)\,du\,dv \\ & = \int\limits_{-\sqrt{x}}^{\sqrt{x}}\int\limits_{-\sqrt{y}}^{\sqrt{y}}\dfrac{1}{4}\,du\,dv \end{align}
I did not understand how we got to $f(u,v)=\dfrac{1}{4}$ rather that $\dfrac{1}{4}(1+xy)$
I tried with $X^2=u$ and $Y^2=v$, does not work out.
Please help.
The answer is rather simple. Just verify by a simple calculation that
$\frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv = 0$
and you're done.
Edit: here all steps: $\int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} f(u,v) \, du dv = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \int\limits_{-\sqrt{x}}^{-\sqrt{x}} \left[ \frac{1}{2} uv^2 \right] _{-\sqrt{y}}^{-\sqrt{y}} \, du = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \int\limits_{-\sqrt{x}}^{-\sqrt{x}} 0 \, du = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv$