| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate volume |
| Difficulty | Challenging +1.2 This is a standard P4/Further Pure volume of revolution question with parametric equations. Part (a) requires applying the formula V = π∫y²dx/dθ dθ, using the identity sin²2θ = (1-cos4θ)/2, and finding limits (tan k = √3 gives k = π/3). Part (b) is routine integration. While it requires multiple techniques (parametric volume formula, trig identities, finding parameter limits), these are all standard P4 procedures with no novel insight needed. Slightly above average due to the algebraic manipulation required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\theta = \sqrt{3} \Rightarrow k = \frac{\pi}{3}\) (or 60°) | B1 | Allow \(x = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3}\). May be implied by their integral |
| \(V = (\pi)\int y^2 dx = (\pi)\int (2\sin 2\theta)^2 \sec^2\theta \, d\theta\) | M1A1 | Condone bracketing errors |
| \(4(\pi)\int \sin^2 2\theta \sec^2\theta \, d\theta = 4(\pi)\int 4\sin^2\theta\cos^2\theta \times \frac{1}{\cos^2\theta} \, d\theta\) | dM1 | Uses \(\sin 2\theta = 2\sin\theta\cos\theta\); depends on first M |
| \(= 16(\pi)\int \sin^2\theta \, d\theta\) e.g. \(16(\pi)\int(1-\cos^2\theta) \, d\theta\) | A1 | No requirement for limits yet |
| \(\sin^2\theta = \frac{1-\cos 2\theta}{2} \Rightarrow 16(\pi)\int\sin^2\theta \, d\theta = 16(\pi)\int\frac{1-\cos 2\theta}{2} \, d\theta\) | dM1 | Attempts to use identity; depends on first M |
| Volume \(= \displaystyle\int_0^{\frac{\pi}{3}} 8\pi(1-\cos 2\theta) \, d\theta\) | A1 Cso | Fully correct integral with both limits and \(d\theta\); 8 and/or \(\pi\) can be either side of integral sign |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(1-\cos 2\theta) \, d\theta \rightarrow \theta - \frac{\sin 2\theta}{2}\) | B1 | |
| Volume \(= \displaystyle\int_0^{\frac{\pi}{3}} 8\pi(1-\cos 2\theta) \, d\theta = \left[8\pi\theta - 4\pi\sin 2\theta\right]_0^{\frac{\pi}{3}} = \frac{8}{3}\pi^2 - 2\sqrt{3}\pi\) | M1 A1 | Uses limit \(\frac{\pi}{3}\) (not 60°); limit of 0 may not be seen |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = \sqrt{3} \Rightarrow k = \frac{\pi}{3}$ (or 60°) | B1 | Allow $x = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3}$. May be implied by their integral |
| $V = (\pi)\int y^2 dx = (\pi)\int (2\sin 2\theta)^2 \sec^2\theta \, d\theta$ | M1A1 | Condone bracketing errors |
| $4(\pi)\int \sin^2 2\theta \sec^2\theta \, d\theta = 4(\pi)\int 4\sin^2\theta\cos^2\theta \times \frac{1}{\cos^2\theta} \, d\theta$ | dM1 | Uses $\sin 2\theta = 2\sin\theta\cos\theta$; depends on first M |
| $= 16(\pi)\int \sin^2\theta \, d\theta$ e.g. $16(\pi)\int(1-\cos^2\theta) \, d\theta$ | A1 | No requirement for limits yet |
| $\sin^2\theta = \frac{1-\cos 2\theta}{2} \Rightarrow 16(\pi)\int\sin^2\theta \, d\theta = 16(\pi)\int\frac{1-\cos 2\theta}{2} \, d\theta$ | dM1 | Attempts to use identity; depends on first M |
| Volume $= \displaystyle\int_0^{\frac{\pi}{3}} 8\pi(1-\cos 2\theta) \, d\theta$ | A1 **Cso** | Fully correct integral with both limits and $d\theta$; 8 and/or $\pi$ can be either side of integral sign |
**(7 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(1-\cos 2\theta) \, d\theta \rightarrow \theta - \frac{\sin 2\theta}{2}$ | B1 | |
| Volume $= \displaystyle\int_0^{\frac{\pi}{3}} 8\pi(1-\cos 2\theta) \, d\theta = \left[8\pi\theta - 4\pi\sin 2\theta\right]_0^{\frac{\pi}{3}} = \frac{8}{3}\pi^2 - 2\sqrt{3}\pi$ | M1 A1 | Uses limit $\frac{\pi}{3}$ (not 60°); limit of 0 may not be seen |
**(3 marks)**
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{216f5735-a7ad-4d70-9da9-ae1f098a97d9-20_714_714_269_616}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with parametric equations
$$x = \tan \theta \quad y = 2 \sin 2 \theta \quad \theta \geqslant 0$$
The finite region, shown shaded in Figure 3, is bounded by the curve, the $x$-axis and the line with equation $x = \sqrt { 3 }$
The region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
\begin{enumerate}[label=(\alph*)]
\item Show that the exact volume of this solid of revolution is given by
$$\int _ { 0 } ^ { k } p ( 1 - \cos 2 \theta ) d \theta$$
where $p$ and $k$ are constants to be found.
\item Hence find, by algebraic integration, the exact volume of this solid of revolution.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q9 [10]}}