Ηοw to find the area of this region I have two functions $$r=2$$
$$r= 3+2sin\theta$$
and I want to find the area of the yellow region in the picture below.
The limits of the integral solving the equation must be $\theta=-\pi/6,7\pi/6$.
Can someone explain how to do this?

 A: I do not think you can express the area using only the limits of integration that you specify.
However, if
$\ \ A=$ the area of the region outside the circle and inside the cardioid (the pink region below),
$\ \ B=$ the area of the region inside the cardioid (pink and brown below)
and
$\ \ C=$ the area of the yellow region,
then $C=B-A$.
You can evaluate both $B$ and $A$ in polar coordinates.
The Cardioid is generated once, in the counterclockwise direction, as $\theta$ takes values from $0$ to $2\pi$. For any particular value of $\theta$ $r$ takes values from $0$ to $3+2\sin \theta$. So
$$
B=\int_0^{2\pi}\int_0^{3+2\sin\theta} r dr\,d\theta
$$
Note that the region $A$ is described by the set 
$$
\{ (r,\theta) : -\pi/6\le\theta\le 7\pi/6,\ 2\le r\le 3+2\sin\theta \};
$$
in particular the points of intersection of the circle with the cardioid are given when $\theta=-\pi/6$ and $\theta=7\pi/6$. (The maroon line segment below represents the typical $r$ range over the ray with angle $\theta$.) 
We have:
$$
A=\int_{-\pi/6}^{7\pi/6}\int_2^{3+2\sin\theta} r dr\,d\theta
$$

Alternatively (and more simply), you can take the area of the circle, $4\pi$,
and subtract the area of the green region below:
$$
\int_{7\pi/6}^{11\pi/6}\int_{3+2\sin\theta}^2 r dr\,d\theta
$$
 
A: Hint : 


*

*Search the intersection of the two curves in polar coordinates (you should find your angles)

*Draw the two lines joining the intersection points to the origine

*Compute the surface at the upper part (first equation) and lower part (second equation) delimited by these two lines in polar coordinates (you get two 'pie charts)'. 

A: I suspect you will be using polar coordinates to solve and use an iterated integral. There are two different boundary curves: the circle and the whatchamacallit (lemniscate? one of those). You need to find out where the circle and the whatchamacallit intersect. So do that.
You think $-\pi/6$ is one of those spots. Let's see: $\sin \pi/6 = 1/2$, so $3 + 2 \sin \pi/6 = 2$. Great! That's where these bounds come in.
So you now set up your integral. For the arc from $-\pi/6$ to $7 \pi / 6$, you integrate the radius up to the whatchamacallit. For the rest, just $r$ from $0$ to $2$.
EDITED to respond to OP in comments
The OP states: Well, looking at the picture a possible solution might be $2\int_0^2\int_{-\pi/6}^0 2\sin\theta + \int_0^\pi r$
This is a very intriguing guess. I want to consider only the second term for a moment, $\int_0^\pi r \mathrm{d} \theta$. This evaluates to $\pi r$. Why is that $r$ there? We know the final answer should be a number, as there is no uncertainty here. So perhaps you should explain how you got at this (incomplete) answer.
