| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate volume |
| Difficulty | Challenging +1.2 This is a standard parametric volume of revolution question requiring systematic application of the formula V = π∫y²dx/dt dt, followed by routine integration using standard trigonometric identities. Part (a) involves algebraic manipulation to reach the given form, and part (b) requires integrating tan²t and sin²t using well-known identities. While it requires multiple steps and careful algebra, the techniques are all standard C3/C4 material with no novel insight needed. |
| 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 |
| \(\int(\tan^2 t \cdot \sin^2 t)\,dt = \int(\sec^2 t - 1)\sin^2 t\,dt\) | M1 | Uses \(1 + \tan^2 t = \sec^2 t\) |
| M1 | Uses \(\cos 2t = 1 - 2\sin^2 t\) | |
| \(= \sin^2 t \tan t - \frac{3}{2}t + \frac{3}{4}\sin 2t\) | M1 A1 | Integration by parts giving \(\sin^2 t \tan t - \int 2\sin t \cos t \tan t\,dt - \int\frac{1-\cos 2t}{2}\,dt\) |
| Applies limit of \(\frac{\pi}{3}\): \(-\left(\frac{3}{4}\tan\left(\frac{\pi}{3}\right) - \left(\frac{\pi}{2}\right) + \frac{3}{4}\sin\left(\frac{2\pi}{3}\right)\right) - (0)\) | ddM1 | Applies limit of \(\frac{\pi}{3}\) |
| \(V = 4\pi\left(\frac{9\sqrt{3}}{8} - \frac{\pi}{2}\right)\) or \(\pi\left(\frac{9\sqrt{3}}{2} - 2\pi\right)\) oe | A1 | Two term exact answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(1 + \tan^2 t = \sec^2 t\) and \(\cos 2t = 1 - 2\sin^2 t\) | M1 M1 | |
| \(= \sin^2 t(\tan t - t) - t + \frac{1}{2}\sin 2t - \frac{t}{2}\cos 2t + \frac{1}{4}\sin 2t\) | M1 A1 | Needs parts twice |
| \(V = 4\pi\left(\frac{9\sqrt{3}}{8} - \frac{\pi}{2}\right)\) or \(\pi\left(\frac{9\sqrt{3}}{2} - 2\pi\right)\) oe | ddM1 A1 |
# Question 12(b):
**Way 1 – Integration by Parts**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(\tan^2 t \cdot \sin^2 t)\,dt = \int(\sec^2 t - 1)\sin^2 t\,dt$ | M1 | Uses $1 + \tan^2 t = \sec^2 t$ |
| | M1 | Uses $\cos 2t = 1 - 2\sin^2 t$ |
| $= \sin^2 t \tan t - \frac{3}{2}t + \frac{3}{4}\sin 2t$ | M1 A1 | Integration by parts giving $\sin^2 t \tan t - \int 2\sin t \cos t \tan t\,dt - \int\frac{1-\cos 2t}{2}\,dt$ |
| Applies limit of $\frac{\pi}{3}$: $-\left(\frac{3}{4}\tan\left(\frac{\pi}{3}\right) - \left(\frac{\pi}{2}\right) + \frac{3}{4}\sin\left(\frac{2\pi}{3}\right)\right) - (0)$ | ddM1 | Applies limit of $\frac{\pi}{3}$ |
| $V = 4\pi\left(\frac{9\sqrt{3}}{8} - \frac{\pi}{2}\right)$ or $\pi\left(\frac{9\sqrt{3}}{2} - 2\pi\right)$ oe | A1 | Two term exact answer |
**Way 2 – Integration by Parts (alternative)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $1 + \tan^2 t = \sec^2 t$ and $\cos 2t = 1 - 2\sin^2 t$ | M1 M1 | |
| $= \sin^2 t(\tan t - t) - t + \frac{1}{2}\sin 2t - \frac{t}{2}\cos 2t + \frac{1}{4}\sin 2t$ | M1 A1 | Needs parts twice |
| $V = 4\pi\left(\frac{9\sqrt{3}}{8} - \frac{\pi}{2}\right)$ or $\pi\left(\frac{9\sqrt{3}}{2} - 2\pi\right)$ oe | ddM1 A1 | |
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{423eb549-0873-4185-8faf-12dedafcd108-19_568_956_221_502}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with parametric equations
$$x = \tan t , \quad y = 2 \sin ^ { 2 } t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$
The finite region $S$, shown shaded in Figure 3, is bounded by the curve $C$, the line $x = \sqrt { 3 }$ and the $x$-axis. This shaded 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 volume of the solid of revolution formed is given by
$$4 \pi \int _ { 0 } ^ { \frac { \pi } { 3 } } \left( \tan ^ { 2 } t - \sin ^ { 2 } t \right) \mathrm { d } t$$
\item Hence use integration to find the exact value for this volume.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q12 [12]}}