Given that $X$ and $Y$ are jointly distributed with pdf: $$f(x, y)=\left\{\begin{array}{cc} \frac{1}{x^{2} y^{2}}, & x \geqslant 1, y \geqslant 1 \\ 0 & \text { else } \end{array}\right\}$$
Find the joint pdf of $U=XY, V=\frac{X}{Y}$
My try: First i found the Jacobian of the transformation: $$\begin{aligned} &\Rightarrow \quad x=\sqrt{u v} \quad y=\sqrt{\frac{u}{v}} \\ &J=\left|\begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{array}\right|=\left|\begin{array}{ccc} \frac{\sqrt{v}}{2 \sqrt{u}} & \frac{1}{2 \sqrt{u v}} \\ \frac{\sqrt{u}}{2 \sqrt{v}} & \frac{-1}{2} \sqrt{u} v^{\frac{-3}{2}} \end{array}\right|=\frac{-1}{2v} \end{aligned}$$
Now we have: $$f(u, v)=|J| f\left(x\left(u, v\right), y(u, v)\right)$$ $$\begin{aligned} &f(u,v)=\frac{1}{2 v} \frac{1}{(\sqrt{u v})^{2}} \frac{1}{\left(\sqrt{\frac{u}{v}}\right)^{2}} \\ &f(u,v)=\frac{1}{2 v u^{2}} \end{aligned}$$
We have $u=xy \geq 1 $
But i am not able to find limits of $v$?
$x \geq 1$ and $y \geq 1$ become $\sqrt {uv} \geq 1$ and $\sqrt {\frac u v} \geq 1$. So the limits are $\frac 1 u \leq v \leq u$ and $u \geq 1$.