Find the area between the cardoid $r=a(1+\cos \theta)$ and the circle $r=\frac {3a}{2}$

Question

Find the area between the cardoid $r=a(1+\cos \theta)$ and the circle $r=\frac {3a}{2}$

My Attempt:

Here, the cardioid is $r=a(1+\cos \theta)$ and the circle is $r=\frac {3a}{2}$. Solving these two equations: $$\frac {3a}{2}=a(1+\cos \theta)$$ $$\cos \theta=\frac {1}{2}$$ $$\theta=\frac {\pi}{3}$$ Thus, the point of intersection is $(\frac {3a}{2},\frac {\pi}{3})$. How to proceed further?

Answer 1

Consider $f(\theta) = a(1+\cos \theta) - \frac{3a}{2}$

It is easy to observe that $f(\theta) = f(-\theta)$, hence the graph is symmetric about x-axis

Further, for $f(\theta) < 0$, the cardoid is inside the circle, and will be the function to integrate, and vice versa for $f(\theta ) > 0$

After this setup, what we get as the area is

$$A = 2\int_0^{\frac{\pi}{3}}\frac{1}{2}\left(\frac{3a}{2}\right)^2d\theta + 2\int_{\frac{\pi}{3}}^\pi\frac{1}{2}a^2(1+\cos \theta)^2d\theta$$

The factor of two comes from symmetry about x-axis