AS June 2019 Paper 1 Q9
9.
\[\mathrm{f}(x) = 2x^{\frac{1}{3}} + x^{-\frac{2}{3}} \qquad x > 0\]The finite region bounded by the curve \(y = \mathrm{f}(x)\), the line \(x = \dfrac{1}{8}\), the \(x\)-axis and the line \(x = 8\) is rotated through \(\theta\) radians about the \(x\)-axis to form a solid of revolution.
Given that the volume of the solid formed is \(\dfrac{461}{2}\) units cubed, use algebraic integration to find the angle \(\theta\) through which the region is rotated. (8)
| Scheme | Marks | AO |
|---|---|---|
| A correct overall strategy, an attempt at integrating \(y^2\) with respect to \(x\) combine in some way with the volume of revolution formula (use of \(\pi\displaystyle\int y^2\,\mathrm{d}x\) or \(\alpha\displaystyle\int y^2\,\mathrm{d}x\) for any variable \(\alpha\) is fine) followed by attempt to find an angle/form an equation in \(\theta\) | M1 | 3.1a |
| \(y^2 = kx^{\frac{2}{3}} + \ldots + \dfrac{m}{x^{\frac{4}{3}}}\) or \(y^2 = kx^{\frac{2}{3}} + \ldots + mx^{-\frac{4}{3}}\) where … is one or two more terms. | M1 | 1.1b |
| \(y^2 = 4x^{\frac{2}{3}} + 4x^{-\frac{1}{3}} + x^{-\frac{4}{3}}\) or \(y^2 = 4x^{\frac{2}{3}} + 2x^{-\frac{1}{3}} + x^{-\frac{4}{3}} + 2x^{-\frac{1}{3}}\) (oe) | A1 | 1.1b |
| \(\displaystyle\int y^2\,\mathrm{d}x = \int 4x^{\frac{2}{3}} + \frac{4}{x^{\frac{1}{3}}} + \frac{1}{x^{\frac{4}{3}}}\,\mathrm{d}x = \alpha x^{\frac{5}{3}} + \beta x^{\frac{2}{3}} + \gamma x^{-\frac{1}{3}}\) | M1 | 1.1b |
| \(= \dfrac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \dfrac{3}{x^{\frac{1}{3}}}\) (oe) | A1ft A1 | 1.1b 1.1b |
| \[\frac{\theta}{2}\left[\frac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \frac{3}{x^{\frac{1}{3}}}\right]_{\frac{1}{8}}^{8} = \frac{461}{2}\]\[\begin{aligned}&\Rightarrow \frac{\theta}{2}\Biggl[\left(\frac{12 \times 8^{\frac{5}{3}}}{5} + 6 \times 8^{\frac{2}{3}} - \frac{3}{8^{\frac{1}{3}}}\right)\\ &\qquad - \left(\frac{12 \times \left(\frac{1}{8}\right)^{\frac{5}{3}}}{5} + 6 \times \left(\frac{1}{8}\right)^{\frac{2}{3}} - \frac{3}{\left(\frac{1}{8}\right)^{\frac{1}{3}}}\right)\Biggr] = \frac{461}{2} \Rightarrow \theta = \ldots\end{aligned}\]OR\[\begin{aligned}&\pi\left[\frac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \frac{3}{x^{\frac{1}{3}}}\right]_{\frac{1}{8}}^{8} = \pi\Biggl[\left(\frac{12 \times 8^{\frac{5}{3}}}{5} + 6 \times 8^{\frac{2}{3}} - \frac{3}{8^{\frac{1}{3}}}\right)\\ &\qquad - \left(\frac{12 \times \left(\frac{1}{8}\right)^{\frac{5}{3}}}{5} + 6 \times \left(\frac{1}{8}\right)^{\frac{2}{3}} - \frac{3}{\left(\frac{1}{8}\right)^{\frac{1}{3}}}\right)\Biggr] = \ldots\end{aligned}\]followed by \(\dfrac{\theta}{2\pi} \times \ldots = \dfrac{461}{2} \Rightarrow \theta = \ldots\) | M1 | 3.1a |
| \(\theta = \dfrac{40}{9}\) (radians) | A1 | 1.1b |
| (8) | ||
| (8 marks) |
Notes
M1: A correct overall strategy, either finding full volume rotated by \(2\pi\) first, then performing some kind of scaling, or using \(\alpha\displaystyle\int y^2\,\mathrm{d}x\) for a variable \(\alpha\) (ideally \(\dfrac{\theta}{2}\), but for the strategy accept with any variable multiple), to form an equation in just the angle.
M1: Attempting to square \(y\) to a three or four term expression. Look for correct powers on first and last term with some term(s) in the middle.
A1: Correct expansion in three or four terms – award when first seen.
M1: Integrates \(y^2\) w.r.t. \(x\). Must have at least two terms in their \(y^2\) with fractional indices. Power to be increased by 1 in at least two terms.
A1ft: Two terms of integral correct. Follow through on their expansion. Need not be simplified.
A1: Fully correct integral. Need not be simplified. May still be four terms
M1: Either: Substitutes limits and subtracts correct way round (must be seen or implied by the answer), and equates to \(\dfrac{461}{2}\) if using \(\frac{1}{2}\theta\displaystyle\int y^2\,\mathrm{d}x\) and proceeds to find \(\theta\).
Or: Substitutes limits and subtracts correct way round (seen or implied) and multiplies by \(\pi\) to get the full volume AND then multiplies the result by \(\dfrac{\theta}{2\pi}\) before equating to \(\dfrac{461}{2}\).
The method must be correct for this mark – so they must be using \(\dfrac{\theta}{2}\displaystyle\int y^2\,\mathrm{d}x\) directly or \(\pi\displaystyle\int y^2\,\mathrm{d}x\) and scale by \(\dfrac{\theta}{2\pi}\) when setting equal to \(\dfrac{461}{2}\)
A1: Correct angle found. Accept \(\dfrac{40}{9}\), awrt 4.44 or awrt 255° (as long as the degrees units are made clear – do not accept just 255) isw once a correct value of \(\theta\) is found.
Special case The question specified that algebraic integration must be used, so use of a calculator to find the integral cannot score the marks for integration but may be allowed the strategy and answer marks. A maximum of M1M0A0M0A0A0M1A1 is available in such cases.
Expanding \(y^2\) first but showing no integration can score the second M and first A (if earned) as well.
Note that \(\displaystyle\int_{1/8}^{8}\left(2x^{\frac{1}{3}} + x^{-\frac{2}{3}}\right)^2\mathrm{d}x = \frac{4149}{40} = 103.725\) but just this alone is worth no marks. There must be an attempt to incorporate this within a strategy to gain access to marks.