Standard +0.3 This is a straightforward volumes of revolution question requiring the standard formula V = π∫(y²)dx with clear bounds. The curve equation y² = x - 2 is already in convenient form, making the integral π∫₂⁵(x-2)dx minus the cylinder π(1)²(3). While it requires careful setup of the region (subtracting the cylinder from the full rotation), the calculus itself is routine polynomial integration with no algebraic complications.
9
\includegraphics[max width=\textwidth, alt={}, center]{5b8ddd32-c884-48a0-ad51-5582ef0d5128-10_540_1113_260_516}
The diagram shows part of the curve with equation \(y ^ { 2 } = x - 2\) and the lines \(x = 5\) and \(y = 1\). The shaded region enclosed by the curve and the lines is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
Find the volume obtained.
Throughout Q9: \(1 <\) *their* \(3 < 5\). Sight of \(x=3\)
Volume \(= [\pi]\int(x-2)\,[dx]\)
M1
M1 for showing intention to integrate \((x-2)\). Condone missing \(\pi\) or using \(2\pi\)
\([\pi]\left[\frac{1}{2}x^2 - 2x\right]\) or \([\pi]\left[\frac{1}{2}(x-2)^2\right]\)
A1
Correct integral. Condone missing \(\pi\) or using \(2\pi\)
\(=[\pi]\left[\left(\frac{5^2}{2}-2\times5\right)-\left(\frac{\textit{their}\,3^2}{2}-2\times\textit{their}\,3\right)\right]\) \(=[\pi]\left[\frac{5}{2}+\frac{3}{2}\right]\) as minimum requirement for *their* values
M1
Correct use of '*their* 3' and 5 in integrated expression. Condone missing \(\pi\) or using \(2\pi\). Condone \(+c\)
9\\
\includegraphics[max width=\textwidth, alt={}, center]{5b8ddd32-c884-48a0-ad51-5582ef0d5128-10_540_1113_260_516}
The diagram shows part of the curve with equation $y ^ { 2 } = x - 2$ and the lines $x = 5$ and $y = 1$. The shaded region enclosed by the curve and the lines is rotated through $360 ^ { \circ }$ about the $x$-axis.
Find the volume obtained.\\
\hfill \mbox{\textit{CAIE P1 2021 Q9 [6]}}