Answer: The curve is a cardioid with cusp at the pole. It passes through \((a,0)\), \((2a,\tfrac{\pi}{2})\), \((a,\pi)\), and \((0,\tfrac{3\pi}{2})\).

The required area is \(a^2\left(\tfrac{\pi}{4}+1+\tfrac{\sqrt3}{8}\right)\).

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

• when \(\theta=0\), \(r=a\), so the curve passes through \((a,0)\)
• when \(\theta=\tfrac{\pi}{2}\), \(r=2a\), so it passes through \((2a,\tfrac{\pi}{2})\)
• when \(\theta=\pi\), \(r=a\), so it passes through \((a,\pi)\)
• when \(\theta=\tfrac{3\pi}{2}\), \(r=0\), so the curve has a cusp at the pole.

Hence \(C\) is a cardioid symmetric about the vertical axis, bulging upwards and touching the pole at the bottom.

To find the area between the curve and the half-lines \(\theta=\tfrac{\pi}{3}\) and \(\theta=\tfrac{2\pi}{3}\), use the polar area formula:

\(A=\tfrac12\int_{\pi/3}^{2\pi/3} r^2\,d\theta\).

Substitute \(r=a(1+\sin\theta)\):

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

Expand and simplify:

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

Using \(\sin^2\theta=\tfrac12(1-\cos2\theta)\),

\(A=\tfrac{a^2}{2}\int_{\pi/3}^{2\pi/3}\left(\tfrac32+2\sin\theta-\tfrac12\cos2\theta\right)d\theta\).

Integrate:

\(A=\tfrac{a^2}{2}\left[\tfrac32\theta-2\cos\theta-\tfrac14\sin2\theta\right]_{\pi/3}^{2\pi/3}\).

Now evaluate the limits:

At \(\theta=\tfrac{2\pi}{3}\), \(\cos\theta=-\tfrac12\) and \(\sin2\theta=\sin\tfrac{4\pi}{3}=-\tfrac{\sqrt3}{2}\).

At \(\theta=\tfrac{\pi}{3}\), \(\cos\theta=\tfrac12\) and \(\sin2\theta=\sin\tfrac{2\pi}{3}=\tfrac{\sqrt3}{2}\).

So

\(A=\tfrac{a^2}{2}\left(\pi+2+\tfrac{\sqrt3}{4}\right)=a^2\left(\tfrac{\pi}{4}+1+\tfrac{\sqrt3}{8}\right).\)