| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 5 |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This is a Further Maths polar coordinates question requiring students to identify a region bounded by a rose curve and calculate its area using the polar area integral formula. While the integration itself is straightforward (∫sin²(3θ)dθ using double angle formula), students must correctly set up the polar area formula and work with the three-petaled rose curve r=4sin(3θ), which is less routine than Cartesian area problems. The exact answer requirement and Further Maths context place it moderately above average difficulty. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Lower half of loop shaded | B1 [1] | Below axis of symmetry |
| (ii) \(A = \frac{1}{2}\int_0^{\frac{\pi}{2}} r^2 d\theta = 8\int_0^{\frac{\pi}{2}} \sin^2 3\theta d\theta\) | M1 [4] | Use of correct formula |
| \(= 4\int_0^{\frac{\pi}{2}} (1-\cos 6\theta)d\theta\) | M1 [4] | Use trig identity and integrate |
| \(= 4[\theta - \frac{\sin 6\theta}{6}]_0^{\frac{\pi}{2}}\) | A1 [4] | |
| \(= \frac{2}{3}\pi\) | A1 [4] |
**(i)** Lower half of loop shaded | B1 [1] | Below axis of symmetry
**(ii)** $A = \frac{1}{2}\int_0^{\frac{\pi}{2}} r^2 d\theta = 8\int_0^{\frac{\pi}{2}} \sin^2 3\theta d\theta$ | M1 [4] | Use of correct formula
$= 4\int_0^{\frac{\pi}{2}} (1-\cos 6\theta)d\theta$ | M1 [4] | Use trig identity and integrate
$= 4[\theta - \frac{\sin 6\theta}{6}]_0^{\frac{\pi}{2}}$ | A1 [4] |
$= \frac{2}{3}\pi$ | A1 [4]
---9 The diagram below shows the curve $r = 4 \sin 3 \theta$ for $0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c03cae53-eb00-496b-948f-ccff676bc03c-3_311_775_1713_644}\\
(i) On the diagram in your Printed Answer Booklet, shade the region $R$ for which
$$r \leqslant 4 \sin 3 \theta \text { and } 0 \leqslant \theta \leqslant \frac { 1 } { 6 } \pi .$$
In this question you must show detailed reasoning.\\
(ii) Find the exact area of the region $R$.
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q9 [5]}}