area of one rose petal using iterated integral but the outer integral is of radius

Question

I have a Rose leaf, described by the equation r=a*sin(3θ), from θ=0 to $\frac{\pi}{6}$.
I need to make an iterated integral of this area.
It's easy when the outer integral is of θ ($\int_{0}^{\frac{\pi}{3}} \int_{0}^{asin(3\theta)} r dr d\theta$), but I struggled trying to swap this order. I can't determine the range of $\theta$ with fixed r.
So briefly I need boundaries {b, c} of $\int_{0}^{a} \int_{b}^{c} r d\theta dr$ for this one leaf

Answer 1

The answer is $\int_0^a \int_{(\arcsin (r/a))/3}^{\pi /3 - (\arcsin (r/a))/3} r ~d\theta ~dr$, thanks to Math Lover.