| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.3 This is a standard polar area calculation using the formula A = ½∫r²dθ with straightforward limits. While it requires knowing the polar area formula and integrating (2+sin3θ)², the expansion and integration of sin²3θ using double angle formulas is routine for FP2 students. The question is slightly easier than average because it's a direct application of a standard technique with no conceptual complications. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Area \(= \frac{1}{2}\int_0^{\frac{\pi}{2}}(4 + 4\sin 3\theta + \sin^2 3\theta)\, d\theta\) | M1 | |
| \(= \frac{1}{2}\left[4\theta - \frac{4\cos 3\theta}{3} + \frac{\theta}{2} - \frac{\sin 6\theta}{12}\right]_0^{\frac{\pi}{2}}\) | M1 A1 M1 A1 | |
| \(= \frac{1}{2}\left(2\pi + \frac{\pi}{4}\right) - \frac{1}{2}\left(-\frac{4}{3}\right)\) | M1 | |
| \(= \frac{9\pi}{8} + \frac{2}{3}\) | A1 (7) |
# Question 5:
| Working | Marks | Notes |
|---------|-------|-------|
| Area $= \frac{1}{2}\int_0^{\frac{\pi}{2}}(4 + 4\sin 3\theta + \sin^2 3\theta)\, d\theta$ | M1 | |
| $= \frac{1}{2}\left[4\theta - \frac{4\cos 3\theta}{3} + \frac{\theta}{2} - \frac{\sin 6\theta}{12}\right]_0^{\frac{\pi}{2}}$ | M1 A1 M1 A1 | |
| $= \frac{1}{2}\left(2\pi + \frac{\pi}{4}\right) - \frac{1}{2}\left(-\frac{4}{3}\right)$ | M1 | |
| $= \frac{9\pi}{8} + \frac{2}{3}$ | A1 (7) | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{affb668f-4d43-4fa8-a5b7-d536a58126b9-3_529_668_223_660}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$, shown in Figure 1, has polar equation, $r = 2 + \sin 3 \theta , 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$\\
Use integration to calculate the exact value of the area enclosed by $C$, the line $\theta = 0$ and the line $\theta = \frac { \pi } { 2 }$.\\
\hfill \mbox{\textit{Edexcel FP2 Q5 [7]}}