Consider the autonomous differential equation in $\mathcal{U} = \mathbb{R} \times (0, +\infty)$ given by
$$x' = \dfrac{x^2}{1+x^2y^2}, y' = 0.$$
Justify that the respective flow is complete (i.é, defined for all $t \in \mathbb{R}$). Show that given $T>0$, there exist limited open sets $X \subset \mathcal{U}$ such that $X_T = \{f^t(x,y); (x,y) \in X \hspace{0.1cm} \text{and} \hspace{0.1cm} t \in \left[0,T\right]\}$ has infinite volume measure.
Attempt: Consider the initial condition of the equation as $\gamma (0) = (x_0,y_0)$. Since $y' = 0$, I know that $y = y_0$, with $y_0 > 0$. I was able to show that the flow of this autonomous equation is
$$f^t(x_0,y_0) = \left( \frac{\left(-\left(t - \frac{1}{x_0} + y_0 x _0 \right) \overline{+} \left[\left(t - \frac{1}{x_0} + y_0 x_0 \right) + 4y_0 \right]^{\frac{1}{2}} \right)}{2y_0}, y_0 \right),$$
which is defined for every $t \in \mathbb{R}$. Hence, the respective flow is complete.
Given $T >0$, my initial idea was to consider $X = (a_1, a_2) \times (0,y_0) \subset \mathcal{U}$, with
$$a_1 = \frac{\left(-\left(t - \frac{1}{x_0} + y_0 x _0 \right) - \left[\left(t - \frac{1}{x_0} + y_0 x_0 \right) + 4y_0 \right]^{\frac{1}{2}} \right)}{2y_0}$$
$$ \hspace{0.1cm} \text{and} \hspace{0.1cm}$$
$$ a_2 = \frac{\left(-\left(t - \frac{1}{x_0} + y_0 x _0 \right) +\left[\left(t - \frac{1}{x_0} + y_0 x_0 \right) + 4y_0 \right]^{\frac{1}{2}} \right)}{2y_0}, $$
which is an open and limited set. But I'm not sure if $X_T$ is a set with infinite volume measure. Can anyone help me conclude or even help me construct a different limited open set?
Any help would be appreciated!