Challenging +1.2 This is a Further Maths polar coordinates question requiring integration by parts after substitution. While it involves multiple techniques (substitution, integration by parts, polar area formula), the substitution is given explicitly and the integration follows a standard pattern. The algebraic manipulation is moderately involved but straightforward for FM students.
7 The curve \(C\) has equation
$$r = 10 \ln ( 1 + \theta )$$
where \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). Draw a sketch of \(C\).
Use the substitution \(w = \ln ( 1 + \theta )\) to show that the area of the sector bounded by the line \(\theta = \frac { 1 } { 2 } \pi\) and the arc of \(C\) joining the origin to the point where \(\theta = \frac { 1 } { 2 } \pi\) is
$$50 \left( b ^ { 2 } - 2 b + 2 \right) \mathrm { e } ^ { b } - 100$$
where \(b = \ln \left( 1 + \frac { 1 } { 2 } \pi \right)\).
7 The curve $C$ has equation
$$r = 10 \ln ( 1 + \theta )$$
where $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$. Draw a sketch of $C$.
Use the substitution $w = \ln ( 1 + \theta )$ to show that the area of the sector bounded by the line $\theta = \frac { 1 } { 2 } \pi$ and the arc of $C$ joining the origin to the point where $\theta = \frac { 1 } { 2 } \pi$ is
$$50 \left( b ^ { 2 } - 2 b + 2 \right) \mathrm { e } ^ { b } - 100$$
where $b = \ln \left( 1 + \frac { 1 } { 2 } \pi \right)$.
\hfill \mbox{\textit{CAIE FP1 2006 Q7 [8]}}