| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring routine application of volume and centre of mass formulas for solids of revolution. Students must integrate sin²x (using double angle formula) and x·sin²x, both standard techniques. While it involves multiple steps and careful algebraic manipulation, it follows a well-practiced template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Vol} = \pi\int_0^{\pi/2} y^2\,dx = \pi\int_0^{\pi/2} \sin^2 x\,dx\) | M1 | Forming the required integral, limits not needed. \(\pi\) needed for this mark |
| \(= (\pi)\int_0^{\pi/2} \frac{1}{2}(1-\cos 2x)\,dx\) | M1 | Using \(\sin^2\theta = k(1\pm\cos 2\theta)\) with \(k=\pm\frac{1}{2}\) or \(\pm 2\), limits not needed, \(\pi\) not needed |
| \(= \frac{\pi}{2}\left[x - \frac{1}{2}\sin 2x\right]_0^{\pi/2} = \frac{\pi^2}{4}\) | DM1, A1cso (4) | Integration and substitution of correct limits. Must have used \(\sin^2\theta = k(1\pm\cos 2\theta)\) or \(\sin^2\theta = k(1\pm\sin 2\theta)\) with any \(k\). Correct given result with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pi\int_0^{\pi/2} y^2 x\,dx = \pi\int_0^{\pi/2} x\sin^2 x\,dx\) | M1 | Use \(\int y^2 x\,dx\), limits not needed. First 5 marks available w/wo \(\pi\) |
| \(= \pi\int_0^{\pi/2} \frac{1}{2}x(1-\cos 2x)\,dx = \frac{\pi}{2}\left[\frac{x^2}{2}\right]_0^{\pi/2} - \frac{\pi}{2}\int_0^{\pi/2} x\cos 2x\,dx\) | M1 | Attempt integration by parts (first stage) for any multiple of \(x\cos 2x\) or \(x\sin 2x\), or by using result from (a). \(\cos 2x \to \pm\frac{1}{2}\sin 2x\) or \(\sin 2x \to \pm\frac{1}{2}\cos 2x\) |
| \(= -\frac{\pi}{2}\left[x\times\frac{1}{2}\sin 2x\right]_0^{\pi/2} + \frac{\pi}{2}\int_0^{\pi/2}\frac{1}{2}\sin 2x\,dx + \frac{\pi^3}{16}\) | M1, B1 | For \(\frac{\pi^3}{16}\) (or \(\frac{\pi^2}{16}\) if working without \(\pi\)). Using result from (a) gets this mark for \(\frac{\pi^3}{8}\) or \(\frac{\pi^2}{8}\) |
| \(= 0 - \frac{\pi}{2}\left[\frac{1}{4}\cos 2x\right]_0^{\pi/2} + \frac{\pi^3}{16}\) | DM1 | Completing the integration. Depends on second M mark. Same condition on integration of trig function |
| \(= -\frac{\pi}{8}[-1-1] + \frac{\pi^3}{16} = \frac{\pi^3}{16} + \frac{\pi}{4}\) | A1 | Substituting correct limits to obtain \(\frac{\pi^3}{16}+\frac{\pi}{4}\) or \(\frac{\pi^2}{16}+\frac{1}{4}\) |
| \(\bar{x} = \frac{\pi^3+4\pi}{16} \div \frac{\pi^2}{4} = \frac{\pi^2+4}{4\pi}\) | M1, A1cso (7) | Using \(\bar{x} = (\pi)\int y^2 x\,dx \div (\pi)\int y^2\,dx\). \(\pi\) must be included for both or neither. Answer: \((\pi^2+4)/4\pi\), any equivalent in terms of \(\pi\) accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pi\int_0^{\pi/2} y^2 x\,dx = \pi\int_0^{\pi/2} x\sin^2 x\,dx\) | M1 | |
| \(= \pi\left[\frac{x}{2}\left(x-\frac{1}{2}\sin 2x\right)\right]_0^{\pi/2} - \pi\int\frac{1}{2}\left(x-\frac{1}{2}\sin 2x\right)dx\) | M1 | |
| \(= \frac{\pi^3}{8} - \frac{\pi}{2}\left[\frac{x^2}{2}+\frac{1}{4}\cos 2x\right]_0^{\pi/2}\) | B1, DM1 | |
| \(= \frac{\pi^3}{8} - \frac{\pi}{2}\left[\frac{\pi^2}{8}-\frac{1}{4}-\frac{1}{4}\right] = \frac{\pi^3}{16}+\frac{\pi}{4}\) | A1 | |
| \(\bar{x} = \frac{\pi^3+4\pi}{16} \div \frac{\pi^2}{4} = \frac{\pi^2+4}{4\pi}\) | M1, A1cso (7) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Vol} = \pi\int_0^{\pi/2} y^2\,dx = \pi\int_0^{\pi/2} \sin^2 x\,dx$ | M1 | Forming the required integral, limits not needed. $\pi$ needed for this mark |
| $= (\pi)\int_0^{\pi/2} \frac{1}{2}(1-\cos 2x)\,dx$ | M1 | Using $\sin^2\theta = k(1\pm\cos 2\theta)$ with $k=\pm\frac{1}{2}$ or $\pm 2$, limits not needed, $\pi$ not needed |
| $= \frac{\pi}{2}\left[x - \frac{1}{2}\sin 2x\right]_0^{\pi/2} = \frac{\pi^2}{4}$ | DM1, A1cso (4) | Integration and substitution of correct limits. Must have used $\sin^2\theta = k(1\pm\cos 2\theta)$ or $\sin^2\theta = k(1\pm\sin 2\theta)$ with any $k$. Correct given result with no errors |
---
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pi\int_0^{\pi/2} y^2 x\,dx = \pi\int_0^{\pi/2} x\sin^2 x\,dx$ | M1 | Use $\int y^2 x\,dx$, limits not needed. First 5 marks available w/wo $\pi$ |
| $= \pi\int_0^{\pi/2} \frac{1}{2}x(1-\cos 2x)\,dx = \frac{\pi}{2}\left[\frac{x^2}{2}\right]_0^{\pi/2} - \frac{\pi}{2}\int_0^{\pi/2} x\cos 2x\,dx$ | M1 | Attempt integration by parts (first stage) for any multiple of $x\cos 2x$ or $x\sin 2x$, or by using result from (a). $\cos 2x \to \pm\frac{1}{2}\sin 2x$ or $\sin 2x \to \pm\frac{1}{2}\cos 2x$ |
| $= -\frac{\pi}{2}\left[x\times\frac{1}{2}\sin 2x\right]_0^{\pi/2} + \frac{\pi}{2}\int_0^{\pi/2}\frac{1}{2}\sin 2x\,dx + \frac{\pi^3}{16}$ | M1, B1 | For $\frac{\pi^3}{16}$ (or $\frac{\pi^2}{16}$ if working without $\pi$). Using result from (a) gets this mark for $\frac{\pi^3}{8}$ or $\frac{\pi^2}{8}$ |
| $= 0 - \frac{\pi}{2}\left[\frac{1}{4}\cos 2x\right]_0^{\pi/2} + \frac{\pi^3}{16}$ | DM1 | Completing the integration. Depends on second M mark. Same condition on integration of trig function |
| $= -\frac{\pi}{8}[-1-1] + \frac{\pi^3}{16} = \frac{\pi^3}{16} + \frac{\pi}{4}$ | A1 | Substituting correct limits to obtain $\frac{\pi^3}{16}+\frac{\pi}{4}$ or $\frac{\pi^2}{16}+\frac{1}{4}$ |
| $\bar{x} = \frac{\pi^3+4\pi}{16} \div \frac{\pi^2}{4} = \frac{\pi^2+4}{4\pi}$ | M1, A1cso (7) | Using $\bar{x} = (\pi)\int y^2 x\,dx \div (\pi)\int y^2\,dx$. $\pi$ must be included for both or neither. Answer: $(\pi^2+4)/4\pi$, any equivalent in terms of $\pi$ accepted |
**ALT(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pi\int_0^{\pi/2} y^2 x\,dx = \pi\int_0^{\pi/2} x\sin^2 x\,dx$ | M1 | |
| $= \pi\left[\frac{x}{2}\left(x-\frac{1}{2}\sin 2x\right)\right]_0^{\pi/2} - \pi\int\frac{1}{2}\left(x-\frac{1}{2}\sin 2x\right)dx$ | M1 | |
| $= \frac{\pi^3}{8} - \frac{\pi}{2}\left[\frac{x^2}{2}+\frac{1}{4}\cos 2x\right]_0^{\pi/2}$ | B1, DM1 | |
| $= \frac{\pi^3}{8} - \frac{\pi}{2}\left[\frac{\pi^2}{8}-\frac{1}{4}-\frac{1}{4}\right] = \frac{\pi^3}{16}+\frac{\pi}{4}$ | A1 | |
| $\bar{x} = \frac{\pi^3+4\pi}{16} \div \frac{\pi^2}{4} = \frac{\pi^2+4}{4\pi}$ | M1, A1cso (7) | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d93ae982-9395-4311-9972-be727b3ce954-14_510_723_269_607}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the finite region $R$ which is bounded by part of the curve with equation $y = \sin x$, the $x$-axis and the line with equation $x = \frac { \pi } { 2 }$. A uniform solid $S$ is formed by rotating $R$ through $2 \pi$ radians about the $x$-axis.
Using algebraic integration,
\begin{enumerate}[label=(\alph*)]
\item show that the volume of $S$ is $\frac { \pi ^ { 2 } } { 4 }$
\item find, in terms of $\pi$, the $x$ coordinate of the centre of mass of $S$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q5 [11]}}