Answer: (a) The curve is the spiral segment \(r=\theta\) for \(0\le \theta \le \frac{\pi}{2}\). It starts at the pole when \(\theta=0\) and moves outwards as \(\theta\) increases, with the tangent at the pole along the initial line. Draw also the ray \(\theta=\alpha\) from the origin, with \(0\lt \alpha\lt \frac{\pi}{2}\).

(b) The required angle is \(\alpha=\pi\,2^{-4/3}\). Equivalently, \(\alpha=\frac{\pi}{\sqrt[3]{16}}\).

For a polar curve, the area enclosed between \(\theta=a\) and \(\theta=b\) is

\(\displaystyle \frac12\int_a^b r^2\,d\theta\).

Here \(r=\theta\), so the total area of \(R\) is

\(\displaystyle \frac12\int_0^{\pi/2} \theta^2\,d\theta = \frac12\left[\frac{\theta^3}{3}\right]_0^{\pi/2} = \frac{1}{6}\left(\frac{\pi}{2}\right)^3 = \frac{\pi^3}{48}.\)

If the line \(\theta=\alpha\) divides \(R\) into two equal areas, then the area from \(0\) to \(\alpha\) is half of \(R\):

\(\displaystyle \frac12\int_0^{\alpha} \theta^2\,d\theta = \frac12\cdot \frac{\pi^3}{96} = \frac{\pi^3}{96}.\)

Now

\(\displaystyle \frac12\int_0^{\alpha} \theta^2\,d\theta = \frac12\left[\frac{\theta^3}{3}\right]_0^{\alpha} = \frac{\alpha^3}{6}.\)

So

\(\displaystyle \frac{\alpha^3}{6} = \frac{\pi^3}{96}.\)

Hence

\(\displaystyle \alpha^3 = \frac{\pi^3}{16},\)

and therefore

\(\displaystyle \alpha = \pi\left(\frac{1}{16}\right)^{1/3} = \pi\,2^{-4/3}.\)