| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Applied context: real-world solid |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring standard application of the formula V = π∫y² dx with a simple rational function, followed by a similarity scaling calculation. The integration is routine (substitution or recognition of standard form), and part (b) is a direct application of the volume scale factor k³. Slightly easier than average due to the simple function and standard technique. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Volume} = \pi\int_{-\frac{1}{4}}^{\frac{1}{2}}\left(\frac{1}{3(1+2x)}\right)^2 dx = \frac{\pi}{9}\int_{-\frac{1}{4}}^{\frac{1}{2}}\frac{1}{(1+2x)^2}\,dx\) | B1 | Use of \(V=\pi\int y^2\,dx\). Can be implied. Ignore limits. |
| \(= \left(\frac{\pi}{9}\right)\int_{-\frac{1}{4}}^{\frac{1}{2}}(1+2x)^{-2}\,dx\) | M1 | Moving power to top. (Do not allow power of \(-1\).) Can be implied. Ignore limits and \(\frac{\pi}{9}\) |
| \(= \left(\frac{\pi}{9}\right)\left[\frac{(1+2x)^{-1}}{(-1)(2)}\right]_{-\frac{1}{4}}^{\frac{1}{2}}\) | M1 | Integrating to give \(\pm p(1+2x)^{-1}\) |
| \(= \left(\frac{\pi}{9}\right)\left[-\frac{1}{2}(1+2x)^{-1}\right]_{-\frac{1}{4}}^{\frac{1}{2}}\) | A1 | \(-\frac{1}{2}(1+2x)^{-1}\) |
| \(= \left(\frac{\pi}{9}\right)\left[\left(\frac{-1}{2(2)}\right)-\left(\frac{-1}{2(\frac{1}{2})}\right)\right] = \left(\frac{\pi}{9}\right)\left[-\frac{1}{4}-(-1)\right]\) | ||
| \(= \dfrac{\pi}{12}\) | A1 aef | Use of limits to give exact values of \(\frac{\pi}{12}\) or \(\frac{3\pi}{36}\) or \(\frac{2\pi}{24}\) or aef |
| Total: 5 marks | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB = \frac{1}{2}-\left(-\frac{1}{4}\right) = \frac{3}{4}\) units; as \(\frac{3}{4}\) units \(\equiv 3\) cm, scale factor \(k = \frac{3}{\frac{3}{4}} = 4\) | ||
| Volume of paperweight \(= (4)^3\left(\frac{\pi}{12}\right)\) | M1 | \((4)^3 \times\) (their answer to part (a)) |
| \(V = \dfrac{16\pi}{3}\ \text{cm}^3 = 16.755\ldots\ \text{cm}^3\) | A1 | \(\frac{16\pi}{3}\) or awrt 16.8 or \(\frac{64\pi}{12}\) or aef |
| Total: 2 marks | [2] |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Volume} = \pi\int_{-\frac{1}{4}}^{\frac{1}{2}}\left(\frac{1}{3(1+2x)}\right)^2 dx = \frac{\pi}{9}\int_{-\frac{1}{4}}^{\frac{1}{2}}\frac{1}{(1+2x)^2}\,dx$ | B1 | Use of $V=\pi\int y^2\,dx$. Can be implied. Ignore limits. |
| $= \left(\frac{\pi}{9}\right)\int_{-\frac{1}{4}}^{\frac{1}{2}}(1+2x)^{-2}\,dx$ | M1 | Moving power to top. (Do not allow power of $-1$.) Can be implied. Ignore limits and $\frac{\pi}{9}$ |
| $= \left(\frac{\pi}{9}\right)\left[\frac{(1+2x)^{-1}}{(-1)(2)}\right]_{-\frac{1}{4}}^{\frac{1}{2}}$ | M1 | Integrating to give $\pm p(1+2x)^{-1}$ |
| $= \left(\frac{\pi}{9}\right)\left[-\frac{1}{2}(1+2x)^{-1}\right]_{-\frac{1}{4}}^{\frac{1}{2}}$ | A1 | $-\frac{1}{2}(1+2x)^{-1}$ |
| $= \left(\frac{\pi}{9}\right)\left[\left(\frac{-1}{2(2)}\right)-\left(\frac{-1}{2(\frac{1}{2})}\right)\right] = \left(\frac{\pi}{9}\right)\left[-\frac{1}{4}-(-1)\right]$ | | |
| $= \dfrac{\pi}{12}$ | A1 aef | Use of limits to give exact values of $\frac{\pi}{12}$ or $\frac{3\pi}{36}$ or $\frac{2\pi}{24}$ or aef |
| **Total: 5 marks** | **[5]** | |
---
# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB = \frac{1}{2}-\left(-\frac{1}{4}\right) = \frac{3}{4}$ units; as $\frac{3}{4}$ units $\equiv 3$ cm, scale factor $k = \frac{3}{\frac{3}{4}} = 4$ | | |
| Volume of paperweight $= (4)^3\left(\frac{\pi}{12}\right)$ | M1 | $(4)^3 \times$ (their answer to part (a)) |
| $V = \dfrac{16\pi}{3}\ \text{cm}^3 = 16.755\ldots\ \text{cm}^3$ | A1 | $\frac{16\pi}{3}$ or awrt 16.8 or $\frac{64\pi}{12}$ or aef |
| **Total: 2 marks** | **[2]** | |
---2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{d366e541-15f6-4fb5-9afb-faf6120f1a1c-03_502_917_296_548}
\end{center}
\end{figure}
The curve with equation $y = \frac { 1 } { 3 ( 1 + 2 x ) } , x > - \frac { 1 } { 2 }$, is shown in Figure 1.\\
The region bounded by the lines $x = - \frac { 1 } { 4 } , x = \frac { 1 } { 2 }$, the $x$-axis and the curve is shown shaded in Figure 1.
This region is rotated through 360 degrees about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the exact value of the volume of the solid generated.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{d366e541-15f6-4fb5-9afb-faf6120f1a1c-03_383_447_1411_753}
\end{center}
\end{figure}
Figure 2 shows a paperweight with axis of symmetry $A B$ where $A B = 3 \mathrm {~cm}$. $A$ is a point on the top surface of the paperweight, and $B$ is a point on the base of the paperweight. The paperweight is geometrically similar to the solid in part (a).
\item Find the volume of this paperweight.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q2 [7]}}