A2 October 2020 Paper 1 Q10
10 In this question you must show detailed reasoning.
The region in the first quadrant bounded by curve \(y = \cosh\frac{1}{2}x^2\), the \(y\)-axis, and the line \(y = 2\) is rotated through \(360^\circ\) about the \(y\)-axis.
Find the exact volume of revolution generated, expressing your answer in a form involving a logarithm. [7]
| Scheme | Marks | AO |
|---|---|---|
| DR \(V = \displaystyle\int_1^2 \pi x^2\,\mathrm{d}y = \int_1^2 2\pi\operatorname{arcosh} y\,\mathrm{d}y\) | M1 | 1.1 |
| let \(u = 2\pi\operatorname{arcosh} y,\ u' = 2\pi/\sqrt{(y^2 - 1)}\) \(v' = 1,\ v = y\) | M1 | 3.1a |
| \(V = \left[2\pi y\operatorname{arcosh} y\right]_1^2 - \displaystyle\int_1^2 2\pi\frac{y}{\sqrt{y^2 - 1}}\,\mathrm{d}y\) | A1 | 2.1 |
| \(= 2\pi\left[y\operatorname{arcosh} y - \sqrt{y^2 - 1}\right]_1^2\) | M1 A1 | 1.1 1.1 |
| \(= 2\pi\left(2\ln(2 + \sqrt{3}) - \sqrt{3}\right)\) | M1 A1cao | 1.1 3.2a |
| [7] |
Notes
M1: \(\displaystyle\int_1^2 2\pi\operatorname{arcosh} y\,\mathrm{d}y\)
M1: integration by parts
A1: condone missing \(2\pi\) and incorrect limits
M1: subst \(u = y^2 - 1\) or inspection
A1: A1 for \(\sqrt{y^2 - 1}\)
M1: use of \(\operatorname{arcosh} x = \ln\left[x + \sqrt{(x^2 - 1)}\right]\)