Answer: (a) The curve has equation \(r=3+2\cos\theta\), so it is a closed limacon with maximum value \(r=5\) when \(\theta=0\), and minimum value \(r=1\) when \(\theta=\pi\). The line \(r\cos\theta=2\) is the vertical straight line \(x=2\), perpendicular to the initial line.

(b) The points of intersection are \(\left(4,\frac{\pi}{3}\right)\) and \(\left(4,-\frac{\pi}{3}\right)\).

(c) The area of \(R\) is \(\frac{22\pi}{3}-\frac{13\sqrt{3}}{2}\).

To sketch the curve \(C\), note that \(r=3+2\cos\theta\) is symmetric about the initial line because it involves \(\cos\theta\). Also, \(r=5\) at \(\theta=0\), so the curve passes through \((5,0)\), and \(r=1\) at \(\theta=\pi\), so it passes through \((1,\pi)\). This gives a closed limacon bulging to the right.

The straight line \(l\) has equation \(r\cos\theta=2\), i.e. \(x=2\), so it is a vertical line crossing the initial line at \((2,0)\).

Intersection points

At points of intersection, the coordinates satisfy both equations, so substitute \(r=3+2\cos\theta\) into \(r\cos\theta=2\):

\((3+2\cos\theta)\cos\theta=2\).

Hence

\(3\cos\theta+2\cos^2\theta=2\),

so

\(2\cos^2\theta+3\cos\theta-2=0\).

Factorising gives

\((2\cos\theta-1)(\cos\theta+2)=0\).

Since \(\cos\theta\) cannot be \(-2\), we must have \(\cos\theta=\frac12\).

With \(-\pi\lt \theta\le\pi\), this gives \(\theta=\pm\frac{\pi}{3}\).

Then

\(r=3+2\cdot\frac12=4\).

So the intersections are \(\left(4,\frac{\pi}{3}\right)\) and \(\left(4,-\frac{\pi}{3}\right)\).

Area of \(R\)

The region enclosed by \(C\) and \(l\), containing the pole, is the part of the limacon to the left of the line \(x=2\). By symmetry about the initial line, this area is made up of two equal parts, one above and one below the initial line.

For \(\theta\) from \(\frac{\pi}{3}\) to \(\pi\), the region lies between the outer curve \(r=3+2\cos\theta\) and the line \(r=\frac{2}{\cos\theta}\). A standard way to calculate the area is to use the polar area formula with the boundary from the limacon, and then account for the two congruent sectors between the curves. Equivalently, the required area can be written as

\(2\times \frac12\int_{\pi/3}^{\pi}(3+2\cos\theta)^2\,d\theta\).

So

\(\text{Area}=\int_{\pi/3}^{\pi}(9+12\cos\theta+4\cos^2\theta)\,d\theta\).

Use \(\cos^2\theta=\frac12(1+\cos2\theta)\):

\(9+12\cos\theta+4\cos^2\theta=11+12\cos\theta+2\cos2\theta\).

Therefore

\(\text{Area}=\int_{\pi/3}^{\pi}(11+12\cos\theta+2\cos2\theta)\,d\theta\)

\(=[11\theta+12\sin\theta+\sin2\theta]_{\pi/3}^{\pi}\).

At \(\theta=\pi\), this gives \(11\pi\). At \(\theta=\frac{\pi}{3}\), this gives

\(11\cdot\frac{\pi}{3}+12\cdot\frac{\sqrt3}{2}+\sin\frac{2\pi}{3}=\frac{11\pi}{3}+6\sqrt3+\frac{\sqrt3}{2}\).

So

\(\text{Area}=11\pi-\left(\frac{11\pi}{3}+6\sqrt3+\frac{\sqrt3}{2}\right)=\frac{22\pi}{3}-\frac{13\sqrt3}{2}.\)

This is the required area.