| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Standard +0.8 This is a Further Maths M3 question requiring volume of revolution (standard but with exponential function) and centre of mass calculation using integration. Part (a) is routine verification, but part (b) requires setting up and evaluating the moment integral ∫x·πy²dx with exponential functions, which involves integration by parts and careful algebraic manipulation. The 6-mark allocation and multi-step nature place it above average difficulty. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks |
|---|---|
| \(\text{Vol} = \pi \int_1^2 y^2 dx = \pi \int_1^2 e^{2x} dx\) | M1 |
| \(= \frac{1}{2}\pi[e^{2x}]_1^2\) | M1A1 |
| \(= \frac{1}{2}\pi[e^4 - e^2]\) | A1 |
| Answer | Marks |
|---|---|
| \(C \text{ of } M = \frac{\int_1^2 \pi x y^2 dx}{\text{vol}}\) | |
| \(\int_1^2 e^{2x} x dx = [\frac{1}{2}xe^{2x}]_1^2 - \int_1^2 \frac{1}{2}e^{2x} dx\) | M1A1 |
| \(= [\frac{1}{2}xe^{2x}]_1^2 - [\frac{1}{4}e^{2x}]_1^2\) | M1 |
| \(= \frac{1}{2} \times 2e^4 - \frac{1}{2} \times 1e^2 - (\frac{1}{4}e^4 - \frac{1}{4}e^2)\) | |
| \(= (\frac{3}{4}e^4 - \frac{1}{4}e^2)\) | A1 |
| \(C \text{ of } M = \frac{\pi(\frac{3}{4}e^4 - \frac{1}{4}e^2)}{\frac{1}{2}\pi(e^4 - e^2)} = 1.656...\) | M1A1 |
| \(= 1.66\) (3 sf) | A1 |
## Part (a):
$\text{Vol} = \pi \int_1^2 y^2 dx = \pi \int_1^2 e^{2x} dx$ | M1 |
$= \frac{1}{2}\pi[e^{2x}]_1^2$ | M1A1 |
$= \frac{1}{2}\pi[e^4 - e^2]$ | A1 |
## Part (b):
$C \text{ of } M = \frac{\int_1^2 \pi x y^2 dx}{\text{vol}}$ | |
$\int_1^2 e^{2x} x dx = [\frac{1}{2}xe^{2x}]_1^2 - \int_1^2 \frac{1}{2}e^{2x} dx$ | M1A1 |
$= [\frac{1}{2}xe^{2x}]_1^2 - [\frac{1}{4}e^{2x}]_1^2$ | M1 |
$= \frac{1}{2} \times 2e^4 - \frac{1}{2} \times 1e^2 - (\frac{1}{4}e^4 - \frac{1}{4}e^2)$ | |
$= (\frac{3}{4}e^4 - \frac{1}{4}e^2)$ | A1 |
$C \text{ of } M = \frac{\pi(\frac{3}{4}e^4 - \frac{1}{4}e^2)}{\frac{1}{2}\pi(e^4 - e^2)} = 1.656...$ | M1A1 |
$= 1.66$ (3 sf) | A1 |\includegraphics{figure_2}
The region $R$ is bounded by the curve with equation $y = e^x$, the line $x = 1$, the line $x = 2$ and the $x$-axis as shown in Figure 2. A uniform solid $S$ is formed by rotating $R$ through $2\pi$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume of $S$ is $\frac{1}{2}\pi (e^4 - e^2)$.
[4]
\item Find, to 3 significant figures, the $x$-coordinate of the centre of mass of $S$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q3 [10]}}