| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | January |
| Marks | 7 |
| 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 solid of revolution centre of mass problem requiring setup of volume and moment integrals with substitution (due to the y² = 9(4-x) form), then evaluation and division. It's a standard M3/Further Mechanics topic but requires careful algebraic manipulation and integration technique beyond basic A-level, placing it moderately above average difficulty. |
| Spec | 1.08a 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 |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\text{Vol} = (\pi)\int y^2\,dx = (\pi)\int_0^4 9(4-x)\,dx = 9(\pi)\left[4x - \frac{1}{2}x^2\right]_0^4\) | M1 | Attempting volume integral, power to increase by 1 in at least one term. Limits not needed. |
| \(= 9(\pi)(16-8) = 72\pi\) | A1 | Correct result after substitution, with or without \(\pi\); need not be simplified. |
| \(\pi\int_0^4 9x(4-x)\,dx = 9\pi\left[2x^2 - \frac{1}{3}x^3\right]_0^4\) | M1A1 | Attempting \((\pi)\int y^2x\,dx\); correct integration with correct limits. |
| \(= 9\pi\left(32 - \frac{64}{3}\right) = 96\pi\) | dM1 | Substitute limits; depends on second M mark. |
| \(\bar{x} = \frac{96\pi}{72\pi} = \frac{4}{3}\) (Accept 1.3 or better) | dM1A1cao | Dividing results of integration; \(\pi\) in both or neither. [7] |
# Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\text{Vol} = (\pi)\int y^2\,dx = (\pi)\int_0^4 9(4-x)\,dx = 9(\pi)\left[4x - \frac{1}{2}x^2\right]_0^4$ | M1 | Attempting volume integral, power to increase by 1 in at least one term. Limits not needed. |
| $= 9(\pi)(16-8) = 72\pi$ | A1 | Correct result after substitution, with or without $\pi$; need not be simplified. |
| $\pi\int_0^4 9x(4-x)\,dx = 9\pi\left[2x^2 - \frac{1}{3}x^3\right]_0^4$ | M1A1 | Attempting $(\pi)\int y^2x\,dx$; correct integration with correct limits. |
| $= 9\pi\left(32 - \frac{64}{3}\right) = 96\pi$ | dM1 | Substitute limits; depends on second M mark. |
| $\bar{x} = \frac{96\pi}{72\pi} = \frac{4}{3}$ (Accept 1.3 or better) | dM1A1cao | Dividing results of integration; $\pi$ in both or neither. [7] |
---1.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[>=Stealth, scale=0.9]
% Coordinates
\coordinate (O) at (0,0);
\coordinate (X) at (5,0);
\coordinate (Y) at (0,4.5);
% The curve y^2 = 9(4-x), so y = 3*sqrt(4-x), x from 0 to 4
% At x=0, y=6 -> scale down: use x in [0,4], y=3*sqrt(4-x)
% We'll scale: 1 unit = 1 unit, but y at x=0 is 6, so let's scale y by 0.55 and x by 1
% Better: just plot with a reasonable scale
% Shaded region R
\fill[gray!30] (0,0) -- plot[domain=0:4, samples=80, smooth] (\x, {3*sqrt(4-\x)*0.6}) -- (4,0) -- cycle;
% Draw the curve
\draw[thick] plot[domain=0:4, samples=80, smooth] (\x, {3*sqrt(4-\x)*0.6});
% Axes
\draw[->] (O) -- (X) node[below] {$x$};
\draw[->] (O) -- (Y) node[left] {$y$};
% Origin label
\node[below left] at (O) {$O$};
% Region label
\node at (1.2, 1.2) {$R$};
% Curve label
\node[right] at (2.2, {3*sqrt(4-2.2)*0.6 + 0.3}) {$y^2 = 9(4-x)$};
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shaded region $R$ is bounded by the curve with equation $y ^ { 2 } = 9 ( 4 - x )$, the positive $x$-axis and the positive $y$-axis, as shown in Figure 1. A uniform solid $S$ is formed by rotating $R$ through $360 ^ { \circ }$ about the $x$-axis.\\
Use algebraic integration to find the $x$ coordinate of the centre of mass of $S$.\\
\hfill \mbox{\textit{Edexcel M3 2017 Q1 [7]}}