How to find the limits of integration to get the area for a loop of a lemniscate? I know how to integrate the squared radius to get the equation that'll give me the area, like such for a lemniscate with $r^2=8\sin(2\theta)$ :
$$1/2\int 8sin(2\theta) = 4 \int \sin(2\theta) = 4 * \frac{-\cos(2\theta)}{2} = -2\cos(2\theta)$$
But how do you find the limits which you'll plug into that equation for the area of the lemniscate loop?

I actually found the answer myself just after posting this question, so I'll post it for anyone with the same question.
The limits occur when $r=0$.
The question becomes: what angle(s), when multiplied by two, has a sin of zero?
$$8\sin(2\pi)=0$$
$$\frac{arcsin(0)}{2} = 0, \frac{\pi}{2}. $$
In this case, the limits are from zero to pi/2.
