AQA C3 2009 January — Question 2 4 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2009
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about x-axis: polynomial or root function
DifficultyModerate -0.3 This is a straightforward volumes of revolution question requiring the standard formula V = π∫y² dx with a simple power function. The integration of (x-2)^5 is routine, and finding exact values at the limits involves basic arithmetic. Slightly easier than average due to the direct application of a standard technique with no complications.
Spec4.08d Volumes of revolution: about x and y axes

2 The diagram shows the curve with equation \(y = \sqrt { ( x - 2 ) ^ { 5 } }\) for \(x \geqslant 2\). \includegraphics[max width=\textwidth, alt={}, center]{59b896ae-60ce-49ea-9c70-0f76fc5fffae-2_885_1125_854_461} The shaded region \(R\) is bounded by the curve \(y = \sqrt { ( x - 2 ) ^ { 5 } }\), the \(x\)-axis and the lines \(x = 3\) and \(x = 4\). Find the exact value of the volume of the solid formed when the region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(V = (\pi)\int y^2\, dx = (\pi)\int(x-2)^5\, dx\)M1
\(= (\pi)\left[\frac{(x-2)^6}{6}\right]_3^4\)A1 Limits not required
\(= (\pi)\left(\frac{2^6}{6} - \frac{1}{6}\right)\)m1 Correct substitution into \((\pi)k(x-2)^6\)
\(= 10.5\pi\)A1 Allow equivalent fraction \(\left(\frac{63}{6}\pi\right)\); AWRT 10.5 or \(10.5\pi\), m1, A0
Total: 4 marks
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = (\pi)\int y^2\, dx = (\pi)\int(x-2)^5\, dx$ | M1 | |
| $= (\pi)\left[\frac{(x-2)^6}{6}\right]_3^4$ | A1 | Limits not required |
| $= (\pi)\left(\frac{2^6}{6} - \frac{1}{6}\right)$ | m1 | Correct substitution into $(\pi)k(x-2)^6$ |
| $= 10.5\pi$ | A1 | Allow equivalent fraction $\left(\frac{63}{6}\pi\right)$; AWRT 10.5 or $10.5\pi$, m1, A0 |

**Total: 4 marks**

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2 The diagram shows the curve with equation $y = \sqrt { ( x - 2 ) ^ { 5 } }$ for $x \geqslant 2$.\\
\includegraphics[max width=\textwidth, alt={}, center]{59b896ae-60ce-49ea-9c70-0f76fc5fffae-2_885_1125_854_461}

The shaded region $R$ is bounded by the curve $y = \sqrt { ( x - 2 ) ^ { 5 } }$, the $x$-axis and the lines $x = 3$ and $x = 4$.

Find the exact value of the volume of the solid formed when the region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis.

\hfill \mbox{\textit{AQA C3 2009 Q2 [4]}}