OCR FP2 2007 June — Question 1 4 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2007
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea enclosed by polar curve
DifficultyStandard +0.3 This is a straightforward application of the polar area formula A = ½∫r²dθ with r = 2sin(3θ). It requires knowing the standard formula, squaring to get 4sin²(3θ), using the double angle identity to integrate, and evaluating between given limits. While it's a Further Maths topic, the execution is mechanical with no problem-solving insight needed, making it slightly easier than average overall.
Spec4.09c Area enclosed: by polar curve

1 The equation of a curve, in polar coordinates, is $$r = 2 \sin 3 \theta , \quad \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi .$$ Find the exact area of the region enclosed by the curve between \(\theta = 0\) and \(\theta = \frac { 1 } { 3 } \pi\).

AnswerMarks
Correct formula with correct \(r\)M1
Rewrite as \(a + b\cos 6\theta\)M1
Integrate their expression correctlyA1√
Get \(\frac{1}{3}\pi\)A1
Correct formula with correct $r$ | M1 | 
Rewrite as $a + b\cos 6\theta$ | M1 | 
Integrate their expression correctly | A1√ | 
Get $\frac{1}{3}\pi$ | A1 |
1 The equation of a curve, in polar coordinates, is

$$r = 2 \sin 3 \theta , \quad \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi .$$

Find the exact area of the region enclosed by the curve between $\theta = 0$ and $\theta = \frac { 1 } { 3 } \pi$.

\hfill \mbox{\textit{OCR FP2 2007 Q1 [4]}}