Answer: The curve is a cardioid with cusp at the pole. It passes through \((a,0)\), \((2a,\tfrac{\pi}{2})\), \((a,\pi)\), and \((0,\tfrac{3\pi}{2})\). Hence it is symmetric about the vertical axis and opens upwards.

The required area is \(a^2\left(\tfrac{\pi}{4}+1+\tfrac{\sqrt{3}}{8}\right)\).

For \(r=a(1+\sin\theta)\), some key points are:

• when \(\theta=0\), \(r=a\), so the point is \((a,0)\);
• when \(\theta=\tfrac{\pi}{2}\), \(r=2a\), so the point is \((2a,\tfrac{\pi}{2})\);
• when \(\theta=\pi\), \(r=a\), so the point is \((a,\pi)\);
• when \(\theta=\tfrac{3\pi}{2}\), \(r=0\), so the curve passes through the pole.

So the sketch is a cardioid symmetric about the vertical axis, with a cusp at the pole and its furthest point at \((2a,\tfrac{\pi}{2})\).

The area enclosed between the radii \(\theta=\tfrac{\pi}{3}\) and \(\theta=\tfrac{2\pi}{3}\) is

\(\displaystyle A=\frac{1}{2}\int_{\pi/3}^{2\pi/3} r^2\,d\theta=\frac{a^2}{2}\int_{\pi/3}^{2\pi/3}(1+\sin\theta)^2\,d\theta.\)

Expand the square:

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

Use \(\sin^2\theta=\frac{1-\cos 2\theta}{2}\):

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

Integrating term by term gives

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

Now evaluate at the limits:

At \(\theta=\tfrac{2\pi}{3}\),

\(\displaystyle \frac{3}{2}\theta-2\cos\theta-\frac{1}{4}\sin 2\theta=\pi+1+\frac{\sqrt{3}}{8}.\)

At \(\theta=\tfrac{\pi}{3}\),

\(\displaystyle \frac{3}{2}\theta-2\cos\theta-\frac{1}{4}\sin 2\theta=\frac{\pi}{2}-1-\frac{\sqrt{3}}{8}.\)

Therefore

\(\displaystyle A=\frac{a^2}{2}\left(\frac{\pi}{2}+2+\frac{\sqrt{3}}{4}\right)=a^2\left(\frac{\pi}{4}+1+\frac{\sqrt{3}}{8}\right).\)