$$\begin{aligned} f(x, y) &=\begin{cases}1/(x^{2} y^{2}) & \text { für } &x \geq 1, y \geq 1 \\[1ex] 0 &&\text { sonst. }\end{cases}\\[2ex] V&:=X Y\end{aligned}$$
Find the probability distribution function for $V$.
From the official solution I have that
$$\begin{aligned} F(v) &=\mathbb{P}(X Y \leq v)=\mathbb{P}(X \leq v / Y) \\ &=\int_{1}^{v} \int_{1}^{v / y} \frac{1}{x^{2} y^{2}} d x d y\end{aligned}$$
Everything is clear except the upper bound of the outer integral. Could somebody explain to me why it is $v$ and not $\infty$? I had a similar exercise with $V=\frac{X}{Y}$ where the outer upper bound was simply $\infty$ as I would have expected, why is this not the case when $V=XY$? What's the general procedure to find the bounds, so I won't run into similar problems with other definitions of $V$?
a preliminary drawing of the joint support sure can help