M3 June 2011 Q2

Figure 1: shaded region R in the first quadrant under the curve y = 9 - x squared, meeting the x-axis at 3 — Figure 1

The shaded region \(R\) is bounded by the curve with equation \(y = 9 - x^2\), 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.

Find the \(x\)-coordinate of the centre of mass of \(S\). (9)

Mark scheme

Scheme	Marks
\(V = \pi\displaystyle\int_0^3\left(9 - x^2\right)^2\mathrm{d}x = \pi\int_0^3\left(81 - 18x^2 + x^4\right)\mathrm{d}x\)	M1
\(= \pi\left[81x - 6x^3 + \dfrac{x^5}{5}\right]_0^3 = \dfrac{648}{5}\pi\)	M1 A1
\(\displaystyle\int_0^3 \pi\left(9 - x^2\right)^2 x\,\mathrm{d}x\)
\(= \dfrac{\pi}{6}\left[-\left(9 - x^2\right)^3\right]_0^3\)	M1 A1
\(= \dfrac{\pi}{6}\left[0 + (9)^3\right]\)	M1
\(= \dfrac{243}{2}\pi\)	A1
\(\bar{x} = \dfrac{\frac{243}{2}}{\frac{648}{5}} = \dfrac{15}{16} \quad (\text{accept } 0.94)\)	M1 A1
	(9)
	(9 marks)

OR

\(\pi\displaystyle\int_0^3\left(81x - 18x^3 + x^5\right)\mathrm{d}x\)
\(= \pi\left[\dfrac{81}{2}x^2 - \dfrac{9}{2}x^4 + \dfrac{1}{6}x^6\right]_0^3\)	M1 A1
\(= \pi\left[\dfrac{81}{2} \times 3^2 - \dfrac{9}{2} \times 3^4 + \dfrac{1}{6} \times 3^6\right]\)	M1
\(= \dfrac{243}{2}\pi\)	A1