Answer: The point of intersection is \(\left(\frac1{\sqrt2},\frac\pi3\right)\). The exact area is \(\frac\pi{12}+\frac{\sqrt3}{2}-1\).

At the intersection,

\(\frac1{\sqrt2}=\sqrt{\sin\frac\theta2}\).

Squaring gives \(\frac12=\sin\frac\theta2\). Since \(0\leqslant\theta\leqslant\pi\), this gives \(\frac\theta2=\frac\pi6\), so \(\theta=\frac\pi3\).

Thus the intersection is \(\left(\frac1{\sqrt2},\frac\pi3\right)\).

For the sketch, \(C_1\) is the circle centred at the pole with radius \(\frac1{\sqrt2}\). The curve \(C_2\) starts at the pole when \(\theta=0\), passes through the intersection at \(\theta=\frac\pi3\), and reaches \(r=1\) when \(\theta=\pi\).

For \(0\leqslant\theta\leqslant\frac\pi3\), the enclosed region lies between \(C_1\) and \(C_2\). Hence

\(A=\frac12\int_0^{\pi/3}\left[\left(\frac1{\sqrt2}\right)^2-\left(\sqrt{\sin\frac\theta2}\right)^2\right]d\theta\).

So

\(A=\frac12\int_0^{\pi/3}\left(\frac12-\sin\frac\theta2\right)d\theta\).

Therefore

\(A=\frac12\left[\frac\theta2+2\cos\frac\theta2\right]_0^{\pi/3}\).

This gives

\(A=\frac12\left(\frac\pi6+\sqrt3-2\right)=\frac\pi{12}+\frac{\sqrt3}{2}-1\).