I was tutoring a Pre-Calculus student and had trouble helping him find the domain and range of the polar function $r^2=64\sin(2\theta)$. I know that its graph is the shape of a lemniscate curve that is symmetric about the origin.
Based off the shape of the graph, and evaluating the function for different values of $\theta$, I can tell that the domain of this function is $(-\infty,\infty)$. But I am suspicious of this because since $r^2$ is on the left side of the equation, then the right side cannot be negative. So my intuition tells me that the domain is really $[0,\infty)$. I'm just unsure how to verify this analytically (or if it's valid to begin with).
As for determining the range, the graphs shows that it is roughly $(-7,7)$, but an estimate is not good enough as you can imagine. I am also unsure how to find the range analytically.
One idea I had was to convert the function $r^2=64\sin(2\theta)$ to Cartesian coordinates. I know that $r^2=x^2+y^2$ and I was able to compute that $\sin(2\theta)=4x^2y^2$. This gave me $$x^2+y^2=64(4x^2y^2) \\ x^2+y^2=256x^2y^2 \\ \sqrt{x^2+y^2}=\pm16xy$$
But for this equation, I was unable to solve for $y$ in terms of $x$, and now I'm stuck. What might be a better approach for finding the domain and range of the given polar function?
Note that $r^2=64\sin(2\theta)\ge 0$, or
$$\sin2\theta \ge 0\implies 2\theta \in[0+2\pi n, \pi+2\pi n]$$ which leads to the domain $\theta \in[0+\pi n, \frac\pi2+\pi n]$. Correspondingly, the range is
$$0\le r=\sqrt{64\sin2\theta} \le 8$$