Edexcel M3 2009 June — Question 4 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2009
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of solid of revolution
DifficultyStandard +0.3 This is a standard M3 centre of mass question using volumes of revolution. Part (a) requires routine integration of π∫(1/x²)dx with given limits. Part (b) applies the standard formula for centre of mass of a solid of revolution, requiring integration of π∫x(1/x²)dx. Both parts follow textbook methods with straightforward algebra and no conceptual challenges beyond applying memorized formulas.
Spec4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids

The finite region bounded by the \(x\)-axis, the curve \(y = \frac{1}{x}\), the line \(x = \frac{1}{4}\) and the line \(x = 1\), is rotated through one complete revolution about the \(x\)-axis to form a uniform solid of revolution.
  1. Show that the volume of the solid is \(21\pi\). [4]
  2. Find the coordinates of the centre of mass of the solid. [5]

Part (a)
AnswerMarks
Volume \(= \int_1^4 \pi y^2 dx = \int_1^4 \pi \frac{1}{x} dx\)M1A1
\(= \left[\pi \times \frac{-1}{3x^3}\right]_1^4\)A1ft
\(= \pi\left(\frac{-1}{3} + \frac{64}{3}\right) = 21\pi\)A1
Part (b)
AnswerMarks
\(21\pi\bar{x} = \rho \int \pi y^2 x dx = \rho \int \pi \frac{1}{x^3} dx\)M1A1
\(21\pi\bar{x} = \pi \left[\frac{-1}{2x^2}\right]_1^4\)A1ft
\(\bar{x} = \frac{1}{21}\left(\frac{-1}{2} + \frac{16}{2}\right) = \frac{5}{14}\) or awrt 0.36A1
\(\bar{y} = 0\) by symmetryB1
## Part (a)

Volume $= \int_1^4 \pi y^2 dx = \int_1^4 \pi \frac{1}{x} dx$ | M1A1 |

$= \left[\pi \times \frac{-1}{3x^3}\right]_1^4$ | A1ft |

$= \pi\left(\frac{-1}{3} + \frac{64}{3}\right) = 21\pi$ | A1 |

## Part (b)

$21\pi\bar{x} = \rho \int \pi y^2 x dx = \rho \int \pi \frac{1}{x^3} dx$ | M1A1 |

$21\pi\bar{x} = \pi \left[\frac{-1}{2x^2}\right]_1^4$ | A1ft |

$\bar{x} = \frac{1}{21}\left(\frac{-1}{2} + \frac{16}{2}\right) = \frac{5}{14}$ or awrt 0.36 | A1 |

$\bar{y} = 0$ by symmetry | B1 |

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The finite region bounded by the $x$-axis, the curve $y = \frac{1}{x}$, the line $x = \frac{1}{4}$ and the line $x = 1$, is rotated through one complete revolution about the $x$-axis to form a uniform solid of revolution.

\begin{enumerate}[label=(\alph*)]
\item Show that the volume of the solid is $21\pi$. [4]
\item Find the coordinates of the centre of mass of the solid. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2009 Q4 [9]}}