I want to calculate the area of the region R between the curves
$$r = 3\sin(\theta)$$ $$r = 1 + \sin(\theta)$$
You can see the imge below, for a clear visual. I am stuck. I only got to find where the curves do intersect in terms of angles, which is when $3\sin(\theta) = 1 + \sin(\theta)$ that is $\theta = \pi/6$ and $\theta = 5\pi/6$.
Thank you!
The method I present is slow, but effective. First of all, your circle has polar equation $r = 3\sin(\theta)$ which means its diameter measures $3$. Hence the area of that circle is $\frac{9}{4}\pi$, by simple geometry.
Now, if we find the area OUTSIDE the cardioid (the curve with polar equation $r = 1+\sin(\theta)$) and substract it from the total area of the circle, we will get the area of the region $R$ you want.
Let's call $A$ the area of the region outside the cardioid, but inside the circle (it's the half moon shape region, white coloured above).
$$A = \dfrac{1}{2}\int_{\pi/6}^{5\pi/6}\left(3\sin(\theta)\right)^2\ \text{d}\theta - \dfrac{1}{2}\int_{\pi/6}^{5\pi/6}\left(1 + \sin(\theta)\right)^2\ \text{d}\theta$$
The integration is really easy, and you will get
$$A = \pi$$
Thence,
$$R = \text{Area Circle} - A$$
$$R = \dfrac{9}{4}\pi - \pi$$
$$R = \dfrac{5}{4}\pi$$