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I need help with calculating the following integral

Calculate the area between the graphs (in polar coordinates) between $r=2$ and $r=3+2\sin(\theta)$ when $\theta\in [0,2\pi]$.

It was a multiple choice question and the answer was the area is between $20$ and $30$.

Here is what I tried:

I found where the graphs meet: when $\theta=\frac{11\pi}{6}$.

Calculated $$\int_0^{\frac{11\pi}{6}}\left(3+2\sin(\theta)-2\right)\,\text d\theta$$

But this led to the wrong answer.

I think I forgot to use the jacobian but I don't see how to add it to the integral.

What is the right way to calculate it?

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If you draw it, it is the area of "a part" of the circle. Where do they meet? Not only at $11\pi/6$, indeed. $$ 2=3+2\sin\theta\,\,\text{ iff }\,\,\theta\in\left\{\frac{7\pi}{6},\frac{11\pi}{6}\right\}. $$ Hence the area is $$ \underbrace{2^2\pi \cdot \frac{2}{3}}_{\frac{2}{3}\text{ circle}}+\int_{7\pi/6}^{11\pi/6}(3+2\sin\theta)\,\mathrm{d}\theta. $$

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