Answer: The circle is the curve with constant radius \(r=a\), so it is centred at the pole with radius \(a\). The cardioid \(r=a(1-\cos\theta)\) has a cusp at the pole when \(\theta=0\), is symmetric about the initial line, and extends furthest to the left when \(\theta=\pi\).

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

The area inside both curves is

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

Let the required region be the set of points lying inside the circle \(r=a\) and inside the cardioid \(r=a(1-\cos\theta)\).

First find where the curves meet:

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

so \(1-\cos\theta=1\), hence \(\cos\theta=0\).

Therefore \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\), so the intersection points are

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

For \(-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}\), the cardioid lies inside the circle, and for \(\frac{\pi}{2}\le \theta \le \frac{3\pi}{2}\), the circle lies inside the cardioid. So the common region can be found as the area of the left-hand semicircle together with the two equal cardioid parts on the right.

Thus

\(\displaystyle A=\frac12\pi a^2+2\times\frac12\int_0^{\pi/2} a^2(1-\cos\theta)^2\,d\theta\)

so

\(\displaystyle A=\frac12\pi a^2+a^2\int_0^{\pi/2}(1-2\cos\theta+\cos^2\theta)\,d\theta\).

Using \(\cos^2\theta=\frac12(1+\cos2\theta)\),

\(\displaystyle A=\frac12\pi a^2+a^2\int_0^{\pi/2}\left(\frac32-2\cos\theta+\frac12\cos2\theta\right)d\theta\).

Integrating,

\(\displaystyle A=\frac12\pi a^2+a^2\left[\frac{3\theta}{2}-2\sin\theta+\frac14\sin2\theta\right]_0^{\pi/2}\).

Evaluating the limits gives

\(\displaystyle A=\frac12\pi a^2+a^2\left(\frac34\pi-2\right)=\left(\frac54\pi-2\right)a^2.\)