| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Volumes of Revolution |
| Type | Volume requiring substitution or integration by parts |
| Difficulty | Challenging +1.2 This is a Further Maths volume of revolution question requiring integration of 1/(p+x³), substitution to simplify the result, and then optimization over a constrained parameter range. While it involves multiple steps and algebraic manipulation, the techniques are standard for FP2: set up πy² integral, integrate using standard methods, and apply boundary conditions. The conceptual demand is moderate—higher than typical A-level Core questions but routine for Further Maths students who have practiced volumes of revolution with parametric constraints. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi \int_{\sqrt{p}}^{\sqrt{3p}} \left(\frac{1}{\sqrt{p+x^2}}\right)^2 dx\) soi | B1 | Correct limits and function substituted in. Condone missing \(\pi\). |
| \(\int \left(\frac{1}{\sqrt{p+x^2}}\right) dx = \int \frac{1}{\sqrt{p} + x^2} dx\) | M1 | Writing in integrable form. If not seen condone incorrect constant outside or use of \(p\) for \(\sqrt{p}\) inside for M1 |
| \(= \frac{1}{\sqrt{p}} \tan^{-1}\left(\frac{x}{\sqrt{p}}\right)\) | A1 | |
| \(\left[\tan^{-1}\left(\frac{x}{\sqrt{p}}\right)\right]_{\sqrt{p}}^{\sqrt{3p}} = \tan^{-1}\left(\frac{\sqrt{3p}}{\sqrt{p}}\right) - \tan^{-1}\left(\frac{\sqrt{p}}{\sqrt{p}}\right)\) | M1 | Correct use of limits in an integrated expression of the form \(\tan^{-1}(bx)\) |
| \(V = \pi \times \frac{1}{\sqrt{p}}\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\pi^2}{12\sqrt{p}}\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 1, V \propto \frac{1}{\sqrt{p}} \Rightarrow V_{\max} = \frac{\pi^2}{12}\) | B1 | At least one of \(V_{\max}\) and \(V_{\min}\) must be clearly identified as such |
| \(\sqrt{3p} = \sqrt{48} \Rightarrow p = 16\) | M1 | Either seen |
| \(p = 16 \Rightarrow V_{\min} = \frac{\pi^2}{48}\) | A1 |
## (i)
$V = \pi \int_{\sqrt{p}}^{\sqrt{3p}} \left(\frac{1}{\sqrt{p+x^2}}\right)^2 dx$ soi | B1 | Correct limits and function substituted in. Condone missing $\pi$.
$\int \left(\frac{1}{\sqrt{p+x^2}}\right) dx = \int \frac{1}{\sqrt{p} + x^2} dx$ | M1 | Writing in integrable form. If not seen condone incorrect constant outside or use of $p$ for $\sqrt{p}$ inside for M1
$= \frac{1}{\sqrt{p}} \tan^{-1}\left(\frac{x}{\sqrt{p}}\right)$ | A1 |
$\left[\tan^{-1}\left(\frac{x}{\sqrt{p}}\right)\right]_{\sqrt{p}}^{\sqrt{3p}} = \tan^{-1}\left(\frac{\sqrt{3p}}{\sqrt{p}}\right) - \tan^{-1}\left(\frac{\sqrt{p}}{\sqrt{p}}\right)$ | M1 | Correct use of limits in an integrated expression of the form $\tan^{-1}(bx)$
$V = \pi \times \frac{1}{\sqrt{p}}\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\pi^2}{12\sqrt{p}}$ oe | A1 |
[5]
## (ii)
$p = 1, V \propto \frac{1}{\sqrt{p}} \Rightarrow V_{\max} = \frac{\pi^2}{12}$ | B1 | At least one of $V_{\max}$ and $V_{\min}$ must be clearly identified as such
$\sqrt{3p} = \sqrt{48} \Rightarrow p = 16$ | M1 | Either seen
$p = 16 \Rightarrow V_{\min} = \frac{\pi^2}{48}$ | A1 |
[3]
If neither identified then B0M1A0 or B1M1A0 are available.
---The region $R$ between the $x$-axis, the curve $y = \frac{1}{\sqrt{p+x^3}}$ and the lines $x = \sqrt{p}$ and $x = \sqrt{3p}$, where $p$ is a positive parameter, is rotated by $2\pi$ radians about the $x$-axis to form a solid of revolution $S$.
\begin{enumerate}[label=(\roman*)]
\item Find and simplify an algebraic expression, in terms of $p$, for the exact volume of $S$. [5]
\item Given that $R$ must lie entirely between the lines $x = 1$ and $x = \sqrt{48}$ find in exact form
\begin{itemize}
\item the greatest possible value of the volume of $S$
\item the least possible value of the volume of $S$. [3]
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q5 [8]}}