Area inside cardiod $r=2-2 cos (θ) $ and circle $r=-6cosθ$

Question

I found the points of intersection $(3,2π/3)$ and $(3, 4π/3)$ but now I'm stuck and don't know how to continue. I don't know how to choose the range of numbers to integrate.

The answer is 5π if it helps.

Answer 1

The region in question is:

I recommend we flip the region, thus it is equivalent to find the area between the cardioid $r = 2 + 2\cos\theta$ and the circle $r = 6\cos \theta$, that is,

You can verify that the points of intersection occurs at $(3, \pi/3)$ and $(3,-pi/3)$.

So the area enclosed by the two polar curves is given by

$$\int_{\pi/3}^{\pi/2} \frac{(6\cos\theta)^2}{2} d\theta + \int_{-\pi/3}^{-\pi/2} \frac{(6\cos\theta)^2}{2} d\theta+ \int_{-\pi/3}^{\pi/3} \frac{(2 + 2\cos\theta)^2}{2} d\theta$$

The above integrals can be simplified by using symmetry about the $\theta$ axis, thus, we get

$$2\int_{\pi/3}^{\pi/2} \frac{(6\cos\theta)^2}{2} d\theta + 2\int_{0}^{\pi/3} \frac{(2 + 2\cos\theta)^2}{2} d\theta$$

If you go through the integration, you will indeed find that the area enclosed is $5 \pi$, but I leave that to you as it is only a matter of computation.