Answer: The circle is centred at the pole with radius \(a\). The cardioid \(r=a(1-\cos\theta)\) has a cusp at the pole, is symmetric about the initial line, and opens to the left, with maximum radius at \(\theta=\pi\).

The points of intersection satisfy \(a=a(1-\cos\theta)\), so \(\cos\theta=0\). Hence \(\theta=\frac{\pi}{2}\) or \(\frac{3\pi}{2}\).

Therefore the intersection points are \(\left(a,\frac{\pi}{2}\right)\) and \(\left(a,\frac{3\pi}{2}\right)\).

The common area is

\(\displaystyle A=\frac{1}{2}\pi a^2+2\times \frac{1}{2}\int_0^{\pi/2} a^2(1-\cos\theta)^2\,d\theta.\)

So

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\int_0^{\pi/2}(1-2\cos\theta+\cos^2\theta)\,d\theta.\)

Using \(\cos^2\theta=\frac{1}{2}(1+\cos 2\theta)\),

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\int_0^{\pi/2}\left(\frac{3}{2}-2\cos\theta+\frac{1}{2}\cos 2\theta\right)d\theta.\)

Hence

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\left[\frac{3\theta}{2}-2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\pi/2}\)

\(\displaystyle =\frac{1}{2}\pi a^2+a^2\left(\frac{3\pi}{4}-2\right)=\left(\frac{5}{4}\pi-2\right)a^2.\)

Sketch the circle \(r=a\), which is a circle of radius \(a\) centred at the pole.

Sketch the cardioid \(r=a(1-\cos\theta)\). It has a cusp at the pole, is symmetric about the initial line, and opens to the left since its greatest value occurs when \(\theta=\pi\).

To find the intersections, equate the two values of \(r\):

\(\displaystyle a=a(1-\cos\theta).\)

Since \(a\gt 0\), divide through by \(a\):

\(\displaystyle 1=1-\cos\theta,\)

so

\(\displaystyle \cos\theta=0.\)

Therefore

\(\displaystyle \theta=\frac{\pi}{2},\frac{3\pi}{2}.\)

At each point, \(r=a\), so the intersection points are

\(\displaystyle \left(a,\frac{\pi}{2}\right)\text{ and }\left(a,\frac{3\pi}{2}\right).\)

For the common region, the whole left semicircle of the circle lies inside the cardioid, contributing area \(\frac{1}{2}\pi a^2\).

In the first and fourth quadrants, the cardioid lies inside the circle from \(\theta=0\) to \(\theta=\frac{\pi}{2}\), and by symmetry these two parts are equal. So

\(\displaystyle A=\frac{1}{2}\pi a^2+2\times \frac{1}{2}\int_0^{\pi/2} a^2(1-\cos\theta)^2\,d\theta.\)

Thus

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\int_0^{\pi/2}(1-2\cos\theta+\cos^2\theta)\,d\theta.\)

Now use

\(\displaystyle \cos^2\theta=\frac{1}{2}(1+\cos 2\theta),\)

giving

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\int_0^{\pi/2}\left(\frac{3}{2}-2\cos\theta+\frac{1}{2}\cos 2\theta\right)d\theta.\)

Integrating,

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\left[\frac{3\theta}{2}-2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\pi/2}.\)

Substituting the limits,

\(\displaystyle A=\frac{1}{2}\pi a^2+a^2\left(\frac{3\pi}{4}-2\right).\)

Hence

\(\displaystyle A=\left(\frac{5}{4}\pi-2\right)a^2.\)