<s>[INST] How to integrate this area? I need to find the following area between the shapes made by $r=3+2\sin{t}$ and $r=2$:

My problem is if I need to integrate with any function or not. First, to check the angle I just did $2=r=3+2\sin{t} \Rightarrow \sin{t} = -\frac{1}{2} \Rightarrow$ the angle goes from $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$. For the radius, it's from $2$ to $3+2 \sin{t}$. My problem is in the integrals.
It it $\int^{\frac{11\pi}{6}}_{\frac{7\pi}{6}} \int^{3+2 \sin{\theta}}_{2} (1) r drd\theta$? Or $\int^{\frac{11\pi}{6}}_{\frac{7\pi}{6}} \int^{3+2 \sin{\theta}}_{2} (3+2 \sin{\theta} )r drd\theta$?
 [/INST] Consider the following diagram. Here $dθ$ is the infinitesimally small angle such that we the variation of $r$ with $θ$ can be neglected.

Clearly, we need to find the summation (integral) of the orange areas :

Now consider the following diagrams:


Our Orange area = Blue Area - Green Area
Also, the figures can be approximated as sectors of a circle as the variation of $r$ with $θ$ is negligible.
So, the Orange area,
$$dA=\frac{1}{2}\left(r_{blue}\right)^{2}d\theta\ -\ \frac{1}{2}\left(r_{green}\right)^{2}d\theta$$
$$\int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}dA=\int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}\frac{1}{2}\left(3+2\sin\theta\right)^{2}d\theta\ -\ \int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}\frac{1}{2}\left(2\right)^{2}d\theta$$
Which comes out to be approximately as $24.1870451584$
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