M3 June 2017 Q1
1.
A uniform lamina is in the shape of the region \(R\). Region \(R\) is bounded by the curve with equation \(y = 4 - x^2\), the positive \(x\)-axis and the positive \(y\)-axis, as shown shaded in Figure 1.
Use algebraic integration to find the \(x\) coordinate of the centre of mass of the lamina. (7)
| Scheme | Marks |
|---|---|
| Area \(= \displaystyle\int_0^2 \left(4 - x^2\right)\mathrm{d}x = \left[4x - \frac{1}{3}x^3\right]_0^2,\ \ = \frac{16}{3}\) | M1,A1 |
| \(\displaystyle\int xy\,\mathrm{d}x = \int_0^2 \left(4x - x^3\right)\mathrm{d}x\) | M1 |
| \(= \left[2x^2 - \dfrac{1}{4}x^4\right]_0^2,\ \ = 4\) | dM1, A1 |
| \(\bar{x} = \dfrac{\int xy\,\mathrm{d}x}{\int y\,\mathrm{d}x} = 4 \div \dfrac{16}{3},\ = \dfrac{3}{4}\) oe | M1,A1cso |
| (7) | |
| (7 marks) |
Notes
Use of volumes scores 0/7
Ignore any work for \(\bar{y}\) whether before or after \(\bar{x}\)
M1 Using \(\displaystyle\int y\,\mathrm{d}x\) with lower limit 0, an attempt at the upper limit and an attempt at algebraic integration.
A1 Correct result for the area. May be implied by a correct final answer. No need to show substitution of either limit providing algebraic integration and attached limits are correct.
M1 Using \(\displaystyle\int xy\,\mathrm{d}x\) with the same limits as the first integral
dM1 Attempting the algebraic integration, including substitution of their limits. Depends on the second M mark.
A1 Correct result for this integral. May be implied by a correct final answer.
M1 Using \(\bar{x} = \dfrac{\int xy\,\mathrm{d}x}{\int y\,\mathrm{d}x}\). If a constant for mass/unit area is seen it must be with both integrals or neither.
A1cso Correct final answer.