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?


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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