Answer:

• C₁ is the circle with radius \(a\) centred at the pole.
• C₂ is the curve \(r=2a\cos 2\theta\) for \(0\le \theta \le \frac{\pi}{4}\); it starts at \((2a,0)\), passes through the pole when \(\theta=\frac{\pi}{4}\), and lies inside the circle \(r=a\) for part of the interval.

\(\beta=\frac{\pi}{6}\).

The required area is \(a^{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)\).

For \(C_1\), the equation \(r=a\) is a circle centred at the pole with radius \(a\).

For \(C_2\), \(r=2a\cos 2\theta\) with \(0\le \theta \le \frac{\pi}{4}\). At \(\theta=0\), \(r=2a\), so the curve starts on the positive initial line at distance \(2a\) from the pole. At \(\theta=\frac{\pi}{4}\), \(\cos\frac{\pi}{2}=0\), so the curve meets the pole. Hence it forms one loop in this interval.

At the intersection point \((a,\beta)\), both curves have the same \(r\), so

\(a=2a\cos 2\beta\).

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

\(1=2\cos 2\beta\), so \(\cos 2\beta=\frac{1}{2}\).

With \(0\le \beta\le \frac{\pi}{4}\), we have \(2\beta=\frac{\pi}{3}\), hence

\(\beta=\frac{\pi}{6}\).

Now consider the enclosed region. From the initial line to \(\theta=\beta\), the boundary is the arc of \(C_1\), so this part contributes the sector area

\(\frac{1}{2}a^2\beta=\frac{1}{2}a^2\cdot\frac{\pi}{6}=\frac{\pi a^2}{12}.\)

From \(\theta=\beta\) to \(\theta=\frac{\pi}{4}\), the boundary is the arc of \(C_2\). The polar area formula gives

\(\frac{1}{2}\int_{\pi/6}^{\pi/4} r^2\,d\theta=\frac{1}{2}\int_{\pi/6}^{\pi/4} 4a^2\cos^2 2\theta\,d\theta.\)

So the total area is

\(\frac{\pi a^2}{12}+2a^2\int_{\pi/6}^{\pi/4}\cos^2 2\theta\,d\theta.\)

Use \(\cos^2 x=\frac{1}{2}(1+\cos 2x)\), so \(2\cos^2 2\theta=1+\cos 4\theta\). Therefore

\(\text{Area}=\frac{\pi a^2}{12}+a^2\int_{\pi/6}^{\pi/4}(1+\cos 4\theta)\,d\theta.\)

Integrating,

\(\text{Area}=\frac{\pi a^2}{12}+a^2\left[\theta+\frac{\sin 4\theta}{4}\right]_{\pi/6}^{\pi/4}.\)

Now evaluate the bracket:

at \(\theta=\frac{\pi}{4}\), \(\theta+\frac{\sin 4\theta}{4}=\frac{\pi}{4}+0\);

at \(\theta=\frac{\pi}{6}\), \(\theta+\frac{\sin 4\theta}{4}=\frac{\pi}{6}+\frac{\sin(2\pi/3)}{4}=\frac{\pi}{6}+\frac{\sqrt{3}}{8}.\)

Hence

\(\text{Area}=\frac{\pi a^2}{12}+a^2\left(\frac{\pi}{4}-\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)\)

\(=\frac{\pi a^2}{12}+a^2\left(\frac{\pi}{12}-\frac{\sqrt{3}}{8}\right)\)

\(=a^2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right).\)