A2 June 2022 Paper 1 Q5
5
(a) Sketch the polar curve \(r = a(1 - \cos\theta)\), \(0 \leqslant \theta \lt 2\pi\), where \(a\) is a positive constant. [2]
(b) Determine the exact area of the region enclosed by the curve. [5]
| Scheme | Marks | AO |
|---|---|---|
 | M1 A1 | 1.1 1.1 |
| [2] |
Notes
M1: symmetrical loop about the initial line
A1: correct shape with cusp at O
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle A = \int_0^{2\pi} \frac{1}{2}a^2(1 - \cos\theta)^2\,\mathrm{d}\theta\) | M1 | 1.1a |
| \(\displaystyle = \frac{1}{2}a^2\int_0^{2\pi}\left(1 - 2\cos\theta + \frac{1}{2}[1 + \cos 2\theta]\right)\mathrm{d}\theta\) | M1 M1 | 1.1 3.1a |
| \(= \frac{1}{4}a^2\left[3\theta - 4\sin\theta + \frac{1}{2}\sin 2\theta\right]_0^{2\pi}\) | B1 | 1.1 |
| \(= \dfrac{3}{2}\pi a^2\) | A1cao | 1.1 |
| [5] |
Notes
M1: correct integral and limits, condone missing \(\mathrm{d}\theta\)
limits can be soi by later work
may see \(\int_0^{\pi} a^2(1 - \cos\theta)^2\,\mathrm{d}\theta\)
M1: Expanding correctly
M1: substituting for \(\cos^2\theta\)
B1: \(k\left[3\theta - 4\sin\theta + \dfrac{1}{2}\sin 2\theta\right]\)