| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate volume |
| Difficulty | Standard +0.8 This is a multi-part parametric volume of revolution question requiring: (a) integration using a double angle identity, (b) deriving the volume integral formula with parametric equations (requiring dx/dθ and substitution of limits), and (c) evaluating to reach an exact form. The combination of parametric calculus, trigonometric manipulation, and exact value work makes this moderately challenging, though each individual step follows standard C4 techniques. |
| Spec | 1.08d Evaluate definite integrals: between limits4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\int \sin^2\theta\,d\theta = \frac{1}{2}\int(1-\cos 2\theta)\,d\theta = \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\) \((+C)\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(x = \tan\theta \Rightarrow \frac{dx}{d\theta} = \sec^2\theta\) | ||
| \(\pi\int y^2\,dx = \pi\int y^2 \frac{dx}{d\theta}\,d\theta = \pi\int (2\sin 2\theta)^2 \sec^2\theta\,d\theta\) | M1 A1 | |
| \(= \pi\int \frac{(2\times 2\sin\theta\cos\theta)^2}{\cos^2\theta}\,d\theta\) | M1 | |
| \(= 16\pi\int \sin^2\theta\,d\theta\) | A1 | \(k = 16\pi\) |
| \(x=0 \Rightarrow \tan\theta=0 \Rightarrow \theta=0,\quad x=\frac{1}{\sqrt{3}} \Rightarrow \tan\theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6}\) | B1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(V = 16\pi\left[\frac{1}{2}\theta - \frac{\sin 2\theta}{4}\right]_0^{\frac{\pi}{6}}\) | M1 | |
| \(= 16\pi\left[\left(\frac{\pi}{12} - \frac{1}{4}\sin\frac{\pi}{3}\right) - (0-0)\right]\) | M1 | Use of correct limits |
| \(= 16\pi\left(\frac{\pi}{12} - \frac{\sqrt{3}}{8}\right) = \frac{4}{3}\pi^2 - 2\pi\sqrt{3}\) | A1 | \(p = \frac{4}{3},\ q = -2\) (3) |
## Question 8:
**Part (a):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\int \sin^2\theta\,d\theta = \frac{1}{2}\int(1-\cos 2\theta)\,d\theta = \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta$ $(+C)$ | M1 A1 | (2) |
**Part (b):**
| Working | Marks | Notes |
|---------|-------|-------|
| $x = \tan\theta \Rightarrow \frac{dx}{d\theta} = \sec^2\theta$ | | |
| $\pi\int y^2\,dx = \pi\int y^2 \frac{dx}{d\theta}\,d\theta = \pi\int (2\sin 2\theta)^2 \sec^2\theta\,d\theta$ | M1 A1 | |
| $= \pi\int \frac{(2\times 2\sin\theta\cos\theta)^2}{\cos^2\theta}\,d\theta$ | M1 | |
| $= 16\pi\int \sin^2\theta\,d\theta$ | A1 | $k = 16\pi$ |
| $x=0 \Rightarrow \tan\theta=0 \Rightarrow \theta=0,\quad x=\frac{1}{\sqrt{3}} \Rightarrow \tan\theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6}$ | B1 | (5) |
**Part (c):**
| Working | Marks | Notes |
|---------|-------|-------|
| $V = 16\pi\left[\frac{1}{2}\theta - \frac{\sin 2\theta}{4}\right]_0^{\frac{\pi}{6}}$ | M1 | |
| $= 16\pi\left[\left(\frac{\pi}{12} - \frac{1}{4}\sin\frac{\pi}{3}\right) - (0-0)\right]$ | M1 | Use of correct limits |
| $= 16\pi\left(\frac{\pi}{12} - \frac{\sqrt{3}}{8}\right) = \frac{4}{3}\pi^2 - 2\pi\sqrt{3}$ | A1 | $p = \frac{4}{3},\ q = -2$ (3) |
**Total: [10]**8. (a) Using the identity $\cos 2 \theta = 1 - 2 \sin ^ { 2 } \theta$, find $\int \sin ^ { 2 } \theta \mathrm {~d} \theta$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-15_516_580_383_680}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows part of the curve $C$ with parametric equations
$$x = \tan \theta , \quad y = 2 \sin 2 \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
The finite shaded region $S$ shown in Figure 4 is bounded by $C$, the line $x = \frac { 1 } { \sqrt { 3 } }$ and the $x$-axis. This shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.\\
(b) Show that the volume of the solid of revolution formed is given by the integral
$$k \int _ { 0 } ^ { \frac { \pi } { 6 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$
where $k$ is a constant.\\
(c) Hence find the exact value for this volume, giving your answer in the form $p \pi ^ { 2 } + q \pi \sqrt { } 3$, where $p$ and $q$ are constants.
\hfill \mbox{\textit{Edexcel C4 2009 Q8 [10]}}