FP2 June 2010 Q5
5.
Figure 1 shows the curves given by the polar equations \[r = 2, \qquad 0 \leqslant \theta \leqslant \frac{\pi}{2},\] and \[r = 1.5 + \sin 3\theta, \qquad 0 \leqslant \theta \leqslant \frac{\pi}{2}.\]
(a) Find the coordinates of the points where the curves intersect. (3)
The region \(S\), between the curves, for which \(r > 2\) and for which \(r < (1.5 + \sin 3\theta)\), is shown shaded in Figure 1.
(b) Find, by integration, the area of the shaded region \(S\), giving your answer in the form \(a\pi + b\sqrt{3}\), where \(a\) and \(b\) are simplified fractions. (7)
| Scheme | Marks |
|---|---|
| \(1.5 + \sin 3\theta = 2 \quad \to \quad \sin 3\theta = 0.5 \quad \therefore 3\theta = \dfrac{\pi}{6}\ \left(\text{or } \dfrac{5\pi}{6}\right),\) | M1 A1, |
| and \(\therefore \theta = \dfrac{\pi}{18}\) or \(\dfrac{5\pi}{18}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Area \(= \frac{1}{2}\left[\displaystyle\int_{\frac{\pi}{18}}^{\frac{5\pi}{18}}(1.5 + \sin 3\theta)^2\,\mathrm{d}\theta\right],\ -\dfrac{1}{9}\pi \times 2^2\) | M1, M1 |
| \(= \frac{1}{2}\left[\displaystyle\int_{\frac{\pi}{18}}^{\frac{5\pi}{18}}\left(2.25 + 3\sin 3\theta + \tfrac{1}{2}(1 - \cos 6\theta)\right)\mathrm{d}\theta\right] - \dfrac{1}{9}\pi \times 2^2\) | M1 |
| \(= \frac{1}{2}\left[\left(2.25\theta - \cos 3\theta + \tfrac{1}{2}\left(\theta - \dfrac{1}{6}\sin 6\theta\right)\right)\right]_{\frac{\pi}{18}}^{\frac{5\pi}{18}} - \dfrac{1}{9}\pi \times 2^2\) | M1 A1 |
| \(= \dfrac{13\sqrt{3}}{24} - \dfrac{5\pi}{36}\) | M1 A1 |
| (7) | |
| (10 marks) |