Question 9 - A-Level Maths

Edexcel CP AS 2019 June — Question 9 8 marks

Exam Board	Edexcel
Module	CP AS (Core Pure AS)
Year	2019
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Volumes of Revolution
Type	Rotation about x-axis: polynomial or root function
Difficulty	Standard +0.8 This question requires setting up and evaluating a volume of revolution integral with fractional powers, expanding (f(x))² correctly, integrating terms with fractional indices, applying limits that yield non-trivial values (1/8 and 8), then solving backwards for θ from a given volume. The algebraic manipulation and multi-step process elevates this above a standard volume question.
Spec	4.08d Volumes of revolution: about x and y axes

9. $$\mathrm { f } ( x ) = 2 x ^ { \frac { 1 } { 3 } } + x ^ { - \frac { 2 } { 3 } } \quad x > 0$$ The finite region bounded by the curve $y = \mathrm { f } ( x )$, the line $x = \frac { 1 } { 8 }$, 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 $\frac { 461 } { 2 }$ units cubed, use algebraic integration to find the angle $\theta$ through which the region is rotated.

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Question 9:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Overall strategy: attempt to integrate $y^2$ w.r.t. $x$, combine with volume of revolution formula ($\pi\int y^2\,dx$ or $\alpha\int y^2\,dx$), then attempt to find $\theta$	M1	Correct overall strategy
$y^2 = kx^{\frac{2}{3}} + \ldots + \frac{m}{x^{\frac{4}{3}}}$ or $y^2 = kx^{\frac{2}{3}} + \ldots + mx^{-\frac{4}{3}}$ (one or two more terms)	M1	Attempt to expand/find $y^2$
$y^2 = 4x^{\frac{2}{3}} + 4x^{-\frac{1}{3}} + x^{-\frac{4}{3}}$ or $y^2 = 4x^{\frac{2}{3}} + 2x^{-\frac{1}{3}} + x^{-\frac{4}{3}} + 2x^{-\frac{1}{3}}$ (oe)	A1	Correct $y^2$
$\int y^2\,dx = \int 4x^{\frac{2}{3}} + \frac{4}{x^{\frac{1}{3}}} + \frac{1}{x^{\frac{4}{3}}}\,dx = \alpha x^{\frac{5}{3}} + \beta x^{\frac{2}{3}} + \gamma x^{-\frac{1}{3}}$	M1	Correct integration attempt
$= \frac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \frac{3}{x^{\frac{1}{3}}}$ (oe)	A1ft, A1	Correct integrated expression
$\frac{\theta}{2}\left[\frac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \frac{3}{x^{\frac{1}{3}}}\right]_{\frac{1}{8}}^{8} = \frac{461}{2}$ or $\pi\left[\ldots\right]_{\frac{1}{8}}^{8} = \ldots$ followed by $\frac{\theta}{2\pi}\times\ldots = \frac{461}{2}$	M1	Applies limits and forms equation in $\theta$
$\theta = \frac{40}{9}$ (radians)	A1	Correct final answer

# Question 9:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Overall strategy: attempt to integrate $y^2$ w.r.t. $x$, combine with volume of revolution formula ($\pi\int y^2\,dx$ or $\alpha\int y^2\,dx$), then attempt to find $\theta$ | M1 | Correct overall strategy |
| $y^2 = kx^{\frac{2}{3}} + \ldots + \frac{m}{x^{\frac{4}{3}}}$ or $y^2 = kx^{\frac{2}{3}} + \ldots + mx^{-\frac{4}{3}}$ (one or two more terms) | M1 | Attempt to expand/find $y^2$ |
| $y^2 = 4x^{\frac{2}{3}} + 4x^{-\frac{1}{3}} + x^{-\frac{4}{3}}$ or $y^2 = 4x^{\frac{2}{3}} + 2x^{-\frac{1}{3}} + x^{-\frac{4}{3}} + 2x^{-\frac{1}{3}}$ (oe) | A1 | Correct $y^2$ |
| $\int y^2\,dx = \int 4x^{\frac{2}{3}} + \frac{4}{x^{\frac{1}{3}}} + \frac{1}{x^{\frac{4}{3}}}\,dx = \alpha x^{\frac{5}{3}} + \beta x^{\frac{2}{3}} + \gamma x^{-\frac{1}{3}}$ | M1 | Correct integration attempt |
| $= \frac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \frac{3}{x^{\frac{1}{3}}}$ (oe) | A1ft, A1 | Correct integrated expression |
| $\frac{\theta}{2}\left[\frac{12x^{\frac{5}{3}}}{5} + 6x^{\frac{2}{3}} - \frac{3}{x^{\frac{1}{3}}}\right]_{\frac{1}{8}}^{8} = \frac{461}{2}$ or $\pi\left[\ldots\right]_{\frac{1}{8}}^{8} = \ldots$ followed by $\frac{\theta}{2\pi}\times\ldots = \frac{461}{2}$ | M1 | Applies limits and forms equation in $\theta$ |
| $\theta = \frac{40}{9}$ (radians) | A1 | Correct final answer |

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9.

$$\mathrm { f } ( x ) = 2 x ^ { \frac { 1 } { 3 } } + x ^ { - \frac { 2 } { 3 } } \quad x > 0$$

The finite region bounded by the curve $y = \mathrm { f } ( x )$, the line $x = \frac { 1 } { 8 }$, 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 $\frac { 461 } { 2 }$ units cubed, use algebraic integration to find the angle $\theta$ through which the region is rotated.

\hfill \mbox{\textit{Edexcel CP AS 2019 Q9 [8]}}

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