| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| 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/Further Mechanics question requiring application of the centre of mass formula for a solid of revolution. The function y = (1/2)√x is straightforward to integrate, requiring only basic power rule integration (x^(1/2) becomes x^(3/2), etc.). While it involves multiple steps and careful bookkeeping of the formula, it's a routine textbook exercise with no conceptual surprises—slightly easier than average due to the simple integrand. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\text{Vol} = \int_2^4 (\pi) \times \frac{1}{4}x \, dx\) | ||
| \(= (\pi)\left[\frac{1}{8}x^2\right]_2^4\) | M1 | Use \(\text{Vol} = (\pi)\int_2^4 y^2 \, dx\) and attempt integration. Limits not needed |
| \(= (\pi)\left[2 - \frac{1}{2}\right] = \frac{3(\pi)}{2}\) | A1 | Correct volume following substitution of correct limits. Can be decimal or implied by correct final answer |
| \(\int_2^4 (\pi) \times \frac{1}{4}x^2 \, dx\) | ||
| \(= (\pi)\left[\frac{1}{12}x^3\right]_2^4\) | M1 | Use \((\pi)\int_2^4 xy^2 \, dx\) and attempt integration. Limits not needed |
| \(= (\pi)\frac{1}{12}[64-8] = \frac{56}{12}(\pi)\) | A1 | Correct result following substitution of correct limits |
| \(\bar{x} = \frac{56}{12}\pi \times \frac{2}{3\pi} = \frac{28}{9}\) | M1A1 (6) | M1: Use \(\frac{\int \pi xy^2 \, dx}{\int \pi y^2 \, dx}\); A1: Correct \(x\) coordinate, must be exact. Give A0 if decimal equivalent of 56/12 given |
# Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\text{Vol} = \int_2^4 (\pi) \times \frac{1}{4}x \, dx$ | | |
| $= (\pi)\left[\frac{1}{8}x^2\right]_2^4$ | M1 | Use $\text{Vol} = (\pi)\int_2^4 y^2 \, dx$ and attempt integration. Limits not needed |
| $= (\pi)\left[2 - \frac{1}{2}\right] = \frac{3(\pi)}{2}$ | A1 | Correct volume following substitution of correct limits. Can be decimal or implied by correct final answer |
| $\int_2^4 (\pi) \times \frac{1}{4}x^2 \, dx$ | | |
| $= (\pi)\left[\frac{1}{12}x^3\right]_2^4$ | M1 | Use $(\pi)\int_2^4 xy^2 \, dx$ and attempt integration. Limits not needed |
| $= (\pi)\frac{1}{12}[64-8] = \frac{56}{12}(\pi)$ | A1 | Correct result following substitution of correct limits |
| $\bar{x} = \frac{56}{12}\pi \times \frac{2}{3\pi} = \frac{28}{9}$ | M1A1 (6) | M1: Use $\frac{\int \pi xy^2 \, dx}{\int \pi y^2 \, dx}$; A1: Correct $x$ coordinate, must be exact. Give A0 if decimal equivalent of 56/12 given |
---\begin{enumerate}
\item The region enclosed by the curve with equation $y = \frac { 1 } { 2 } \sqrt { x }$, the $x$-axis and the lines $x = 2$ and $x = 4$, is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid $S$. Use algebraic integration to find the exact value of the $x$ coordinate of the centre of mass of $S$.\\
(6)
\end{enumerate}
"都 D\\
$\_\_\_\_$ 1\\
\hfill \mbox{\textit{Edexcel M3 2017 Q1 [6]}}