For the curves $r=3cos\theta$ and $r=1+\cos\theta$ i need to find the area on the intersection of two curves. Now, $3cos\theta=1+\cos\theta$ implies that $cos\theta=\frac{1}{2}$ which said that $\theta=-+\frac{\pi}{3}$. By formula i know that
$$A=\frac{1}{2}\int r^2 d\theta$$
My problem is how can set the limits looking my draw. I am not sure if the first integral need to be between $0$ and $\frac{\pi}{3}$ using the function $r=1+\cos\theta$ and my second integral would be between $\frac{\pi}{3}$ and $\pi$ but using another function $r=3\cos\theta$? Then sum the two integral and finally multiply by $2$ by symmetry in the graph? Please, can somebody give me a hint or tell me if as I said, it is correct?
In summary, is correct say that $$A=2\left(\frac{1}{2}\int_{0}^{\frac{\pi}{3}}(1+\cos\theta)^2d\theta + \frac{1}{2}\int_{\frac{\pi}{3}}^{\pi}(3\cos\theta)^2d\theta \right)?$$
Thank you
First of all, it is important to have an accurate drawing (using free software like Geogebra or Desmos) :
Then, as you would do for the difference of two areas between curves having a cartesian equation, your area is :
$$A=\left(\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(3\cos\theta)^2d\theta - \frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1+\cos\theta)^2d\theta \right)$$
Compared with your formula, the bounds are not the same, there is a minus sign and an exchange of functions : "largest swept area (by red curve) minus smallest swept area" (by green curve)".
Due to symmetry $\theta \leftrightarrow -\theta$, $A$ can be written :
$$A=2 \left(\frac{1}{2}\int_{0}^{\frac{\pi}{3}}(3\cos\theta)^2d\theta - \frac{1}{2}\int_{0}^{\frac{\pi}{3}}(1+\cos\theta)^2d\theta \right)$$
$$A=\int_{0}^{\frac{\pi}{3}}\left((3\cos\theta)^2-(1+\cos\theta)^2\right)d\theta$$
you will find $A=\pi$ (crescent area)
Now, if you want the area $B$ instead (and I understand now that it is that your question), just consider that we have the equality :
$$A+B=\pi (3/2)^2$$
(area of a disk with radius $3/2$), from which we can deduce
$$B=5\pi/4$$