Answer: The curve passes through the origin when \(1+2\cos\theta=0\), so \(\cos\theta=-\tfrac12\) and hence \(\theta=\pm\tfrac{2\pi}{3}\). It also passes through \(r=3\) when \(\theta=0\), so the sketch is a right-facing limacon with an inner loop, symmetric about the polar axis.

The required region is the part of the curve between the lines \(\theta=-\tfrac{\pi}{3}\) and \(\theta=\tfrac{\pi}{3}\). By symmetry about the polar axis,

\(\text{Area}=2\times \frac12\int_{0}^{\pi/3}(1+2\cos\theta)^2\,d\theta=\int_{0}^{\pi/3}(1+4\cos\theta+4\cos^2\theta)\,d\theta.\)

Using \(\cos^2\theta=\tfrac12(1+\cos 2\theta)\),

\(\text{Area}=\int_{0}^{\pi/3}(3+4\cos\theta+2\cos 2\theta)\,d\theta.\)

So

\(\text{Area}=[3\theta+4\sin\theta+\sin 2\theta]_{0}^{\pi/3}=\pi+2\sqrt3+\frac{\sqrt3}{2}=\pi+\frac{5\sqrt3}{2}.\)

Hence the area bounded by the curve and the two half-lines is \(\pi+\tfrac{5\sqrt3}{2}\).

For the sketch, note that \(r=1+2\cos\theta\).

• When \(\theta=0\), \(r=3\), so the curve passes through \((3,0)\).
• When \(r=0\), \(1+2\cos\theta=0\), so \(\cos\theta=-\tfrac12\), giving \(\theta=\pm\tfrac{2\pi}{3}\). Thus the curve passes through the origin.
• The equation depends on \(\cos\theta\), so the curve is symmetric about the polar axis and opens to the right, forming a limacon with an inner loop.

Only the part from \(\theta=-\tfrac{\pi}{3}\) to \(\theta=\tfrac{\pi}{3}\) is needed for the area.

Using the polar area formula,

\(A=\frac12\int_{-\pi/3}^{\pi/3}(1+2\cos\theta)^2\,d\theta.\)

Since the integrand is even,

\(A=2\times\frac12\int_0^{\pi/3}(1+2\cos\theta)^2\,d\theta=\int_0^{\pi/3}(1+4\cos\theta+4\cos^2\theta)\,d\theta.\)

Now use \(\cos^2\theta=\tfrac12(1+\cos 2\theta)\):

\(1+4\cos\theta+4\cos^2\theta=1+4\cos\theta+2(1+\cos 2\theta)=3+4\cos\theta+2\cos 2\theta.\)

Hence

\(A=\int_0^{\pi/3}(3+4\cos\theta+2\cos 2\theta)\,d\theta.\)

Integrating term by term gives

\(A=[3\theta+4\sin\theta+\sin 2\theta]_0^{\pi/3}.\)

So

\(A=3\cdot\frac{\pi}{3}+4\sin\frac{\pi}{3}+\sin\frac{2\pi}{3}=\pi+4\cdot\frac{\sqrt3}{2}+\frac{\sqrt3}{2}=\pi+\frac{5\sqrt3}{2}.\)

Therefore the required area is \(\pi+\tfrac{5\sqrt3}{2}\).