C4 June 2009 Q8
8.
(a) Using the identity \(\cos 2\theta = 1 - 2\sin^2\theta\), find \(\displaystyle\int \sin^2\theta\,\mathrm{d}\theta\). (2)
Figure 4 shows part of the curve \(C\) with parametric equations \[x = \tan\theta, \qquad y = 2\sin 2\theta, \qquad 0 \leqslant \theta < \frac{\pi}{2}\]
The finite shaded region \(S\) shown in Figure 4 is bounded by \(C\), the line \(x = \dfrac{1}{\sqrt{3}}\) and the \(x\)-axis. This shaded region is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Show that the volume of the solid of revolution formed is given by the integral \[k\int_0^{\frac{\pi}{6}} \sin^2\theta\,\mathrm{d}\theta\] where \(k\) is a constant. (5)
(c) Hence find the exact value for this volume, giving your answer in the form \(p\pi^2 + q\pi\sqrt{3}\), where \(p\) and \(q\) are constants. (3)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \sin^2\theta\,\mathrm{d}\theta = \frac{1}{2}\int (1 - \cos 2\theta)\,\mathrm{d}\theta = \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta\quad (+C)\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(x = \tan\theta \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \sec^2\theta\) | |
| \(\displaystyle\pi\int y^2\,\mathrm{d}x = \pi\int y^2\frac{\mathrm{d}x}{\mathrm{d}\theta}\,\mathrm{d}\theta = \pi\int (2\sin 2\theta)^2\sec^2\theta\,\mathrm{d}\theta\) | M1 A1 |
| \(\displaystyle = \pi\int \frac{(2 \times 2\sin\theta\cos\theta)^2}{\cos^2\theta}\,\mathrm{d}\theta\) | M1 |
| \(\displaystyle = 16\pi\int \sin^2\theta\,\mathrm{d}\theta\) \(k = 16\pi\) | A1 |
| \(x = 0 \Rightarrow \tan\theta = 0 \Rightarrow \theta = 0,\quad x = \dfrac{1}{\sqrt{3}} \Rightarrow \tan\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = \dfrac{\pi}{6}\) | B1 |
| \(\left(V = 16\pi\displaystyle\int_0^{\frac{\pi}{6}} \sin^2\theta\,\mathrm{d}\theta\right)\) | |
| (5) |
| Scheme | Marks |
|---|---|
| \(V = 16\pi\left[\dfrac{1}{2}\theta - \dfrac{\sin 2\theta}{4}\right]_0^{\frac{\pi}{6}}\) | M1 |
| \(= 16\pi\left[\left(\dfrac{\pi}{12} - \dfrac{1}{4}\sin\dfrac{\pi}{3}\right) - (0 - 0)\right]\) Use of correct limits | M1 |
| \(= 16\pi\left(\dfrac{\pi}{12} - \dfrac{\sqrt{3}}{8}\right) = \dfrac{4}{3}\pi^2 - 2\pi\sqrt{3}\) \(p = \dfrac{4}{3},\ q = -2\) | A1 |
| (3) | |
| (10 marks) |