| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: Cartesian curve |
| Difficulty | Challenging +1.3 This is a standard FP2 surface of revolution question with a prescribed hyperbolic substitution. Part (a) requires applying the surface area formula (5 marks suggests straightforward differentiation and simplification). Part (b) involves routine hyperbolic integration techniques that are well-practiced in FP2. While the algebra requires care and the topic is advanced, the question follows a predictable template with no novel problem-solving required—students who have practiced similar examples will find this mechanical. |
| Spec | 4.07f Inverse hyperbolic: logarithmic forms4.08d Volumes of revolution: about x and y axes4.08h Integration: inverse trig/hyperbolic substitutions |
\begin{enumerate}[label=(\alph*)]
\item The arc of the curve $y^2 = x^2 + 8$ between the points where $x = 0$ and $x = 6$ is rotated through $2\pi$ radians about the $x$-axis. Show that the area $S$ of the curved surface formed is given by
$$S = 2\sqrt{2}\pi \int_0^6 \sqrt{x^2 + 4} \, dx$$ [5 marks]
\item By means of the substitution $x = 2 \sinh \theta$, show that
$$S = \pi(24\sqrt{5} + 4\sqrt{2} \sinh^{-1} 3)$$ [8 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2011 Q5 [13]}}