Double integral with polar coordinates - radius

Question

it is me again.

$$\left(\frac{x^2}{4} + \frac{y^2}{9}\right)^2 = \frac{x^2}{4} - \frac{y^2}{9}$$

Now, obviously I introduce polar coordinates:

$$x= 2r\cos\phi$$

$$y=3r\sin\phi$$

The equation then translates to $$r = \sqrt{\cos^2\phi - \sin^2\phi}$$

Now, here is where I encounter a problem.

How do I know if the boundaries for $r$ are:

1. $ r \in \left[0, \sqrt{\cos^2\phi - \sin^2\phi}\right]$

or

2. $ r \in \left[\sqrt{\cos^2\phi - \sin^2\phi}, 1\right]$

I thought that maybe the boundaries are as in $(2)$ because the square root can be $1,$ but also lower than 1 depending on the value of the angle.

Also, please tell me if my boundaries for $\phi$ are correct. Here's my reasoning:

Since the left side of the equation is squared, it is always $\ge 0$. That means that the right side also has to be $\ge0$, so $\cos^2{\phi} \ge \sin^2{\phi}$, which is $\phi \in [-\frac{\pi}{4}, \frac{\pi}{4}]$