FP3 June 2009 Q8
8. A curve, which is part of an ellipse, has parametric equations \[x = 3\cos\theta, \quad y = 5\sin\theta, \qquad 0 \leqslant \theta \leqslant \frac{\pi}{2}.\]
The curve is rotated through \(2\pi\) radians about the \(x\)-axis.
(a) Show that the area of the surface generated is given by the integral \[k\pi\int_0^{\alpha} \sqrt{(16c^2 + 9)}\,\mathrm{d}c, \quad \text{where } c = \cos\theta,\] and where \(k\) and \(\alpha\) are constants to be found. (6)
(b) Using the substitution \(c = \dfrac{3}{4}\sinh u\), or otherwise, evaluate the integral, showing all of your working and giving the final answer to 3 significant figures. (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = -3\sin\theta,\ \dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 5\cos\theta\) | B1 |
| so \(S = 2\pi\displaystyle\int 5\sin\theta\sqrt{(-3\sin\theta)^2 + (5\cos\theta)^2}\,\mathrm{d}\theta\) | M1 |
| \(\therefore S = 2\pi\displaystyle\int 5\sin\theta\sqrt{9 - 9\cos^2\theta + 25\cos^2\theta}\,\mathrm{d}\theta\) | M1 |
| Let \(c = \cos\theta,\ \dfrac{\mathrm{d}c}{\mathrm{d}\theta} = -\sin\theta\), limits 0 and \(\tfrac{\pi}{2}\) become 1 and 0 | M1 |
| So \(S = k\pi\displaystyle\int_0^{\alpha} \sqrt{16c^2 + 9}\,\mathrm{d}c\), where \(k = 10\), and \(\alpha\) is 1 | A1, A1 |
| (6) |
| Scheme | Marks |
|---|---|
| Let \(c = \tfrac{3}{4}\sinh u\). Then \(\dfrac{\mathrm{d}c}{\mathrm{d}u} = \dfrac{3}{4}\cosh u\) | M1 |
| So \(S = k\pi\displaystyle\int_{?}^{?} \sqrt{9\sinh^2 u + 9}\ \dfrac{3}{4}\cosh u\,\mathrm{d}u\) | A1 |
| \(= k\pi\displaystyle\int_{?}^{?} \dfrac{9}{4}\cosh^2 u\,\mathrm{d}u = k\pi\displaystyle\int_{?}^{?} \dfrac{9}{8}(\cosh 2u + 1)\,\mathrm{d}u\) | M1 |
| \(= k\pi\left[\dfrac{9}{16}\sinh 2u + \dfrac{9}{8}u\right]_0^{d}\) | A1 |
| \(= \dfrac{45\pi}{4}\left[\dfrac{20}{9} + \ln 3\right]\) i.e. \(\underline{117}\) | B1 |
| (5) | |
| (11 marks) |