| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with implicit or parametric curves |
| Difficulty | Challenging +1.2 This is a standard parametric volumes of revolution question requiring conversion of the volume formula to parameter form, finding limits from x = √2, and integrating sin³θcos²θ using standard trigonometric identities. While it involves multiple steps and careful algebraic manipulation, the techniques are all routine A-level methods with no novel insight required. The integration in part (b) is straightforward using sin²θ = 1-cos²θ substitution. Slightly above average difficulty due to the parametric setup and multi-step nature, but well within typical Further Maths Pure content. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05l Double angle formulae: and compound angle formulae4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Show volume of solid of revolution is \(k \int_{\pi/4}^{\pi/2} \sin^3\theta\cos^2\theta \, d\theta\) | (5) | M1 Attempts \(V = \pi \int y \, dx = \pi \int y\frac{dx}{d\theta} d\theta\) where \(\frac{dx}{d\theta} = \pm k\sin\theta\); A1 \(\int (\sin 2\theta)^2 (-2\sin\theta) d\theta\); M1 Attempts to use \(\sin 2\theta = 2\sin\theta\cos\theta\) within integral which may be implied by B1 Finds correct limits stating \(x = 0 \Rightarrow \theta = \frac{\pi}{2}\), \(x = \sqrt{2} \Rightarrow \theta = \frac{\pi}{4}\) or correct value for \(k\); A1 Achieves printed answer including \(d\theta\) with correct limits and \(8\pi\) in place with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Find exact value for volume in form \(p\pi\sqrt{2}\) | (5) | B1 States \(u = \cos\theta \Rightarrow \frac{du}{d\theta} = -\sin\theta\); M1 Substitutes fully including for \(d\theta\) using \(u = \cos\theta\) and \(\sin^2\theta = \pm 1 \pm \cos^2\theta\) to produce integral just in terms of \(u\); M1 Multiplies out to form polynomial in \(u\) and integrates with \(u^n \to u^{n+1}\) for at least one of their powers of \(u\); M1 All methods must have been scored. It is for using limits \(0\) and \(\frac{\sqrt{2}}{2}\) and subtracting or for using limits \(\frac{\pi}{2}\) and \(\frac{\pi}{4}\) if return to \(\cos\theta\); A1 \(V = \frac{7\sqrt{2}}{15}\pi\) |
**Part (a):**
Show volume of solid of revolution is $k \int_{\pi/4}^{\pi/2} \sin^3\theta\cos^2\theta \, d\theta$ | (5) | M1 Attempts $V = \pi \int y \, dx = \pi \int y\frac{dx}{d\theta} d\theta$ where $\frac{dx}{d\theta} = \pm k\sin\theta$; A1 $\int (\sin 2\theta)^2 (-2\sin\theta) d\theta$; M1 Attempts to use $\sin 2\theta = 2\sin\theta\cos\theta$ within integral which may be implied by B1 Finds correct limits stating $x = 0 \Rightarrow \theta = \frac{\pi}{2}$, $x = \sqrt{2} \Rightarrow \theta = \frac{\pi}{4}$ or correct value for $k$; A1 Achieves printed answer including $d\theta$ with correct limits and $8\pi$ in place with no errors
**Part (b):**
Find exact value for volume in form $p\pi\sqrt{2}$ | (5) | B1 States $u = \cos\theta \Rightarrow \frac{du}{d\theta} = -\sin\theta$; M1 Substitutes fully including for $d\theta$ using $u = \cos\theta$ and $\sin^2\theta = \pm 1 \pm \cos^2\theta$ to produce integral just in terms of $u$; M1 Multiplies out to form polynomial in $u$ and integrates with $u^n \to u^{n+1}$ for at least one of their powers of $u$; M1 All methods must have been scored. It is for using limits $0$ and $\frac{\sqrt{2}}{2}$ and subtracting or for using limits $\frac{\pi}{2}$ and $\frac{\pi}{4}$ if return to $\cos\theta$; A1 $V = \frac{7\sqrt{2}}{15}\pi$
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-21_645_935_301_589}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows part of the curve $C$ with parametric equations
$$x = 2 \cos \theta \quad y = \sin 2 \theta \quad 0 \leq \theta \leq \frac { \pi } { 2 }$$
The region $R$, shown shaded in figure 5, is bounded by the curve $C$, the line $x = \sqrt { 2 }$ and the $x$-axis. This shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid revolution.\\
a. Show that the volume of the solid of revolution formed is given by the integral.
$$k \int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 2 } } \sin ^ { 3 } \theta \cos ^ { 2 } \theta \mathrm {~d} \theta$$
where $k$ is a constant.\\
\includegraphics[max width=\textwidth, alt={}, center]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-22_164_1148_54_118}\\
b. Hence, find the exact value for this volume, giving your answer in the form $p \pi \sqrt { 2 }$ where $p$ is a constant.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q12 [10]}}