| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with exact answer required |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring students to apply the standard formula V = π∫y²dx, simplify the algebraic expression (3x+1)²/x, and integrate term-by-term. While it involves some algebraic manipulation and produces a logarithmic answer, it follows a completely standard method with no conceptual challenges, making it slightly easier than average. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \pi \int_1^3 \frac{(3x+1)^2}{x} dx\) | M1 | |
| \(= \pi \int_1^3 \frac{9x^2 + 6x + 1}{x} dx = \int_1^3 (9x + 6 + \frac{1}{x}) dx\) | A1 | |
| \(= \pi[\frac{9}{2}x^2 + 6x + \ln | x | ]_1^3\) |
| \(= \pi[(\frac{81}{2} + 18 + \ln 3) - (\frac{9}{2} + 6 + 0)]\) | M1 | |
| \(= \pi(48 + \ln 3)\) | A1 | (6) |
$= \pi \int_1^3 \frac{(3x+1)^2}{x} dx$ | M1 |
$= \pi \int_1^3 \frac{9x^2 + 6x + 1}{x} dx = \int_1^3 (9x + 6 + \frac{1}{x}) dx$ | A1 |
$= \pi[\frac{9}{2}x^2 + 6x + \ln|x|]_1^3$ | M1 A1 |
$= \pi[(\frac{81}{2} + 18 + \ln 3) - (\frac{9}{2} + 6 + 0)]$ | M1 |
$= \pi(48 + \ln 3)$ | A1 | (6)
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\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{80bef9d4-b84c-4d3a-a093-67a466c6d1b9-02_615_791_146_532}
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\caption{Figure 1}
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Figure 1 shows the curve with equation $y = \frac { 3 x + 1 } { \sqrt { x } } , x > 0$.\\
The shaded region is bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 3$.\\
Find the volume of the solid formed when the shaded region is rotated through $2 \pi$ radians about the $x$-axis, giving your answer in the form $\pi ( a + \ln b )$, where $a$ and $b$ are integers.\\
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\hfill \mbox{\textit{Edexcel C4 Q1 [6]}}