Challenging +1.2 This is a standard polar area calculation requiring the formula A = ½∫r²dθ, followed by a substitution (u = 1 + tan θ) that is strongly suggested by the form of r. The algebraic manipulation is straightforward once the substitution is made, and the integration is routine. While it requires competence with polar coordinates and substitution, it follows a predictable template for FP3 polar area questions.
3 The diagram shows a sketch of a curve \(C\), the pole \(O\) and the initial line.
\includegraphics[max width=\textwidth, alt={}, center]{c4bce668-61f1-4be0-97ee-c635df7e1fc6-2_380_735_1827_648}
The polar equation of \(C\) is
$$r = 2 \sqrt { 1 + \tan \theta } , \quad - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }$$
Show that the area of the shaded region, bounded by the curve \(C\) and the initial line, is \(\frac { \pi } { 2 } - \ln 2\).
(4 marks)
3 The diagram shows a sketch of a curve $C$, the pole $O$ and the initial line.\\
\includegraphics[max width=\textwidth, alt={}, center]{c4bce668-61f1-4be0-97ee-c635df7e1fc6-2_380_735_1827_648}
The polar equation of $C$ is
$$r = 2 \sqrt { 1 + \tan \theta } , \quad - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }$$
Show that the area of the shaded region, bounded by the curve $C$ and the initial line, is $\frac { \pi } { 2 } - \ln 2$.\\
(4 marks)
\hfill \mbox{\textit{AQA FP3 2012 Q3 [4]}}