C4 January 2010 Q8
8.
(a) Using the substitution \(x = 2\cos u\), or otherwise, find the exact value of \[\int_1^{\sqrt{2}} \frac{1}{x^2\sqrt{(4 - x^2)}}\,\mathrm{d}x\] (7)
Figure 3 shows a sketch of part of the curve with equation \(y = \dfrac{4}{x(4 - x^2)^{\frac{1}{4}}},\quad 0 < x < 2\).
The shaded region \(S\), shown in Figure 3, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 1\) and \(x = \sqrt{2}\). The shaded region \(S\) is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = -2\sin u\) | B1 |
| \(\displaystyle\int \frac{1}{x^2\sqrt{4 - x^2}}\,\mathrm{d}x = \int \frac{1}{(2\cos u)^2\sqrt{4 - (2\cos u)^2}} \times -2\sin u\,\mathrm{d}u\) | M1 |
| \(\displaystyle = \int \frac{-2\sin u}{4\cos^2 u\sqrt{4\sin^2 u}}\,\mathrm{d}u\) Use of \(1 - \cos^2 u = \sin^2 u\) | M1 |
| \(\displaystyle = -\frac{1}{4}\int \frac{1}{\cos^2 u}\,\mathrm{d}u\) \(\pm k\displaystyle\int \frac{1}{\cos^2 u}\,\mathrm{d}u\) | M1 |
| \(= -\dfrac{1}{4}\tan u\quad (+C)\) \(\pm k\tan u\) | M1 |
| \(x = \sqrt{2} \Rightarrow \sqrt{2} = 2\cos u \Rightarrow u = \dfrac{\pi}{4}\) | |
| \(x = 1 \Rightarrow 1 = 2\cos u \Rightarrow u = \dfrac{\pi}{3}\) | M1 |
| \(\left[-\dfrac{1}{4}\tan u\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}} = -\dfrac{1}{4}\left(\tan\dfrac{\pi}{4} - \tan\dfrac{\pi}{3}\right)\) | |
| \(= -\dfrac{1}{4}\left(1 - \sqrt{3}\right)\quad \left(= \dfrac{\sqrt{3} - 1}{4}\right)\) | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(V = \pi\displaystyle\int_1^{\sqrt{2}} \left(\frac{4}{x(4 - x^2)^{\frac{1}{4}}}\right)^2\mathrm{d}x\) | M1 |
| \(= 16\pi\displaystyle\int_1^{\sqrt{2}} \frac{1}{x^2\sqrt{4 - x^2}}\,\mathrm{d}x\) \(16\pi \times\) integral in (a) | M1 |
| \(= 16\pi\left(\dfrac{\sqrt{3} - 1}{4}\right)\) \(16\pi \times\) their answer to part (a) | A1ft |
| (3) | |
| (10 marks) |