Edexcel C4 — Question 1 6 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about x-axis: polynomial or root function
DifficultyStandard +0.3 This is a straightforward volume of revolution question requiring students to find the roots of a quadratic, set up the integral π∫y² dx with correct limits, expand (x²-2x)² algebraically, and integrate term-by-term. While it involves multiple steps, each is routine for C4 and follows a standard template with no conceptual surprises, making it slightly easier than average.
Spec4.08d Volumes of revolution: about x and y axes
The region bounded by the curve \(y = x^2 - 2x\) and the \(x\)-axis is rotated through \(2\pi\) radians about the \(x\)-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi\). [6]
AnswerMarks Guidance
\(x(x-2)=0, x=0,2 \therefore\) crosses \(x\)-axis at \((0,0)\) and \((2,0)\)M1
\(\text{volume} = \pi \int_0^2 (x^2-2x)^2 \, dx\)M1
\(= \pi \int_0^2 (x^4-4x^3+4x^2) \, dx\)A1
\(= \pi\left[\frac{1}{5}x^5 - x^4 + \frac{4}{3}x^3\right]_0^2\)M1 A1
\(= \pi\left(\left(\frac{32}{5}-16+\frac{32}{3}\right)-(0)\right) = \frac{16}{15}\pi\)M1 A1 (6 marks) 
$x(x-2)=0, x=0,2 \therefore$ crosses $x$-axis at $(0,0)$ and $(2,0)$ | M1 |

$\text{volume} = \pi \int_0^2 (x^2-2x)^2 \, dx$ | M1 |

$= \pi \int_0^2 (x^4-4x^3+4x^2) \, dx$ | A1 |

$= \pi\left[\frac{1}{5}x^5 - x^4 + \frac{4}{3}x^3\right]_0^2$ | M1 A1 |

$= \pi\left(\left(\frac{32}{5}-16+\frac{32}{3}\right)-(0)\right) = \frac{16}{15}\pi$ | M1 A1 | **(6 marks)**

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The region bounded by the curve $y = x^2 - 2x$ and the $x$-axis is rotated through $2\pi$ radians about the $x$-axis.

Find the volume of the solid formed, giving your answer in terms of $\pi$. [6]

\hfill \mbox{\textit{Edexcel C4  Q1 [6]}}
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