| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume requiring substitution or integration by parts |
| Difficulty | Standard +0.3 This is a standard C4 volumes of revolution question where the setup is straightforward (formula application) and the key technique—recognizing that squaring y eliminates the square root and creates a rational function requiring logarithmic integration—is a well-practiced skill at this level. The substitution or partial fractions approach is routine for C4 students, making this slightly easier than average. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume \(= \pi\int_0^2\left(\sqrt{\frac{2x}{3x^2+4}}\right)^2 dx\) | B1 | Use of \(V=\pi\int y^2\,dx\); \(\pi\) required; limits and \(dx\) ignored for this mark |
| \(= (\pi)\left[\frac{1}{3}\ln(3x^2+4)\right]_0^2\) | M1 | For \(\pm k\ln(3x^2+4)\) where \(k\) is a constant |
| A1 | For \(\frac{1}{3}\ln(3x^2+4)\) | |
| \(= (\pi)\left[\left(\frac{1}{3}\ln 16\right)-\left(\frac{1}{3}\ln 4\right)\right]\) | dM1 | Substitutes limits of 2 and 0 and subtracts correct way round |
| Volume \(= \frac{1}{3}\pi\ln 4\) | A1 oe isw | \(\frac{1}{3}\pi\ln 4\) or \(\frac{2}{3}\pi\ln 2\); \(\pi\) required for this mark |
# Question 4: Volume of Revolution
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume $= \pi\int_0^2\left(\sqrt{\frac{2x}{3x^2+4}}\right)^2 dx$ | B1 | Use of $V=\pi\int y^2\,dx$; $\pi$ required; limits and $dx$ ignored for this mark |
| $= (\pi)\left[\frac{1}{3}\ln(3x^2+4)\right]_0^2$ | M1 | For $\pm k\ln(3x^2+4)$ where $k$ is a constant |
| | A1 | For $\frac{1}{3}\ln(3x^2+4)$ |
| $= (\pi)\left[\left(\frac{1}{3}\ln 16\right)-\left(\frac{1}{3}\ln 4\right)\right]$ | dM1 | Substitutes limits of 2 and 0 and subtracts correct way round |
| Volume $= \frac{1}{3}\pi\ln 4$ | A1 oe isw | $\frac{1}{3}\pi\ln 4$ or $\frac{2}{3}\pi\ln 2$; $\pi$ required for this mark |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8c963567-d751-4898-b7a7-7095d90514f0-06_606_1185_237_383}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve with equation
$$y = \sqrt { } \left( \frac { 2 x } { 3 x ^ { 2 } + 4 } \right) , x \geqslant 0$$
The finite region $S$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the line $x = 2$
The region $S$ is rotated $360 ^ { \circ }$ about the $x$-axis.\\
Use integration to find the exact value of the volume of the solid generated, giving your answer in the form $k \ln a$, where $k$ and $a$ are constants.\\
\hfill \mbox{\textit{Edexcel C4 2012 Q4 [5]}}