| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| 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 C4 volumes of revolution question with parametric curves. Part (a) requires integration of a parametric volume formula (likely involving sin²θ or cos²θ terms), which is routine but requires careful algebraic manipulation. Part (b) is straightforward elimination of the parameter. The multi-step nature and parametric setup make it slightly above average difficulty, but it follows standard C4 techniques without requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = -1 \Rightarrow \theta = -\frac{\pi}{4}, x = 1 \Rightarrow \theta = \frac{\pi}{4}\) | B1 | |
| \(\frac{dx}{d\theta} = \sec^2 \theta\) | M1 | |
| Volume \(= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(\cos^2\theta)^2 \times \sec^2\theta \, d\theta = \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos^2\theta \, d\theta\) | A1 | |
| \(= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(\frac{1}{4} + \frac{1}{2}\cos 2\theta) \, d\theta\) | M1 | |
| \(= \pi[\frac{1}{4}\theta + \frac{1}{4}\sin 2\theta]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\) | M1 A1 | |
| \(= \pi[(\frac{\pi}{8} + \frac{1}{4}) - (-\frac{\pi}{8} - \frac{1}{4})]\) | M1 | |
| \(= \pi(\frac{\pi}{4} + \frac{1}{2}) = \frac{1}{4}\pi(\pi + 2)\) | A1 | (11) |
| (b) \(y = \cos^2\theta = \frac{1}{\sec^2\theta} = \frac{1}{1 + \tan^2\theta} \therefore y = \frac{1}{1 + x^2}\) | M2 A1 | (11) |
**(a)** $x = -1 \Rightarrow \theta = -\frac{\pi}{4}, x = 1 \Rightarrow \theta = \frac{\pi}{4}$ | B1 |
$\frac{dx}{d\theta} = \sec^2 \theta$ | M1 |
Volume $= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(\cos^2\theta)^2 \times \sec^2\theta \, d\theta = \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos^2\theta \, d\theta$ | A1 |
$= \pi\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(\frac{1}{4} + \frac{1}{2}\cos 2\theta) \, d\theta$ | M1 |
$= \pi[\frac{1}{4}\theta + \frac{1}{4}\sin 2\theta]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$ | M1 A1 |
$= \pi[(\frac{\pi}{8} + \frac{1}{4}) - (-\frac{\pi}{8} - \frac{1}{4})]$ | M1 |
$= \pi(\frac{\pi}{4} + \frac{1}{2}) = \frac{1}{4}\pi(\pi + 2)$ | A1 | (11)
**(b)** $y = \cos^2\theta = \frac{1}{\sec^2\theta} = \frac{1}{1 + \tan^2\theta} \therefore y = \frac{1}{1 + x^2}$ | M2 A1 | (11)
---\begin{enumerate}[label=(\alph*)]
\item Show that the volume of the solid formed is $\frac { 1 } { 4 } \pi ( \pi + 2 )$.
\item Find a cartesian equation for the curve.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [11]}}