| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about x-axis: rational or reciprocal function |
| Difficulty | Standard +0.2 This is a straightforward application of the volume of revolution formula V = π∫y² dx with a simple rational function. The integration requires only basic substitution (linear denominator) and results in a logarithm. While it involves algebraic manipulation to reach the final simplified form, it's a standard textbook exercise with no conceptual challenges beyond direct formula application. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume \(= \pi\int_a^b \left(\frac{1}{2x+1}\right)^2 dx = \pi\int_a^b \frac{1}{(2x+1)^2}\,dx\) | B1 | Use of \(V = \pi\int y^2\,dx\). Can be implied. Ignore limits. |
| \(= \pi\int_a^b (2x+1)^{-2}\,dx\) | ||
| \(= (\pi)\left[\frac{(2x+1)^{-1}}{(-1)(2)}\right]_a^b = (\pi)\left[-\frac{1}{2}(2x+1)^{-1}\right]_a^b\) | M1 | Integrating to give \(\pm p(2x+1)^{-1}\) |
| A1 | \(-\frac{1}{2}(2x+1)^{-1}\) | |
| \(= (\pi)\left[\left(\frac{-1}{2(2b+1)}\right) - \left(\frac{-1}{2(2a+1)}\right)\right]\) | dM1 | Substitutes limits of \(b\) and \(a\) and subtracts the correct way round |
| \(= \frac{\pi}{2}\left[\frac{-2a-1+2b+1}{(2a+1)(2b+1)}\right] = \frac{\pi}{2}\left[\frac{2(b-a)}{(2a+1)(2b+1)}\right]\) | ||
| \(= \dfrac{\pi(b-a)}{(2a+1)(2b+1)}\) | A1 aef | \(\dfrac{\pi(b-a)}{(2a+1)(2b+1)}\) [5 marks total] |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume $= \pi\int_a^b \left(\frac{1}{2x+1}\right)^2 dx = \pi\int_a^b \frac{1}{(2x+1)^2}\,dx$ | B1 | Use of $V = \pi\int y^2\,dx$. Can be implied. Ignore limits. |
| $= \pi\int_a^b (2x+1)^{-2}\,dx$ | | |
| $= (\pi)\left[\frac{(2x+1)^{-1}}{(-1)(2)}\right]_a^b = (\pi)\left[-\frac{1}{2}(2x+1)^{-1}\right]_a^b$ | M1 | Integrating to give $\pm p(2x+1)^{-1}$ |
| | A1 | $-\frac{1}{2}(2x+1)^{-1}$ |
| $= (\pi)\left[\left(\frac{-1}{2(2b+1)}\right) - \left(\frac{-1}{2(2a+1)}\right)\right]$ | dM1 | Substitutes limits of $b$ and $a$ and subtracts the correct way round |
| $= \frac{\pi}{2}\left[\frac{-2a-1+2b+1}{(2a+1)(2b+1)}\right] = \frac{\pi}{2}\left[\frac{2(b-a)}{(2a+1)(2b+1)}\right]$ | | |
| $= \dfrac{\pi(b-a)}{(2a+1)(2b+1)}$ | A1 aef | $\dfrac{\pi(b-a)}{(2a+1)(2b+1)}$ [5 marks total] |
> **Note:** $\pi$ is not required for the middle three marks. Equivalent forms allowed: $\dfrac{\pi b - \pi a}{(2a+1)(2b+1)}$, $\dfrac{-\pi(a-b)}{(2a+1)(2b+1)}$, $\dfrac{\pi(b-a)}{4ab+2a+2b+1}$, $\dfrac{\pi b-\pi a}{4ab+2a+2b+1}$3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ac7d862f-d10d-45ed-9077-ae4c7413cbf6-04_493_490_278_712}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The curve shown in Figure 2 has equation $y = \frac { 1 } { ( 2 x + 1 ) }$. The finite region bounded by the curve, the $x$-axis and the lines $x = a$ and $x = b$ is shown shaded in Figure 2. This region is rotated through $360 ^ { \circ }$ about the $x$-axis to generate a solid of revolution.
Find the volume of the solid generated. Express your answer as a single simplified fraction, in terms of $a$ and $b$.\\
\hfill \mbox{\textit{Edexcel C4 2008 Q3 [5]}}