| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about x-axis, region between two curves |
| Difficulty | Standard +0.8 This question requires setting up and evaluating the difference of two volumes of revolution (line minus curve), involving integration of both a linear function squared and a reciprocal function squared. While the setup is conceptually straightforward for students who understand volumes of revolution, the algebraic manipulation and careful handling of the subtraction adds moderate complexity beyond routine single-function rotation problems. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Vol} = \pi\int(5-x)^2\,dx - \pi\int\frac{16}{x^2}\,dx\) | M1* | Use of volume formula at least once, condone omission of \(\pi\) and limits and \(dx\) |
| DM1 | Subtracting volumes somewhere must be after squaring | |
| \(\int(5-x)^2\,dx = \frac{(5-x)^3}{3} \div -1\) | B1 B1 | B1 without \(\div(-1)\). B1 for \(\div(-1)\) |
| (or \(25x - 10x^2/2 + \frac{1}{3}x^3\)) | (B2,1,0) | \(-1\) for each incorrect term |
| \(\int\frac{16}{x^2}\,dx = -\frac{16}{x}\) | B1 | |
| Use of limits 1 and 4 in an integrated expression and subtracted | DM1 | Must have used "\(y^2\)" at least once. Need to see values substituted |
| \(\rightarrow 9\pi\) or \(28.3\) | A1 |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Vol} = \pi\int(5-x)^2\,dx - \pi\int\frac{16}{x^2}\,dx$ | M1* | Use of volume formula at least once, condone omission of $\pi$ and limits and $dx$ |
| | DM1 | Subtracting volumes somewhere must be after squaring |
| $\int(5-x)^2\,dx = \frac{(5-x)^3}{3} \div -1$ | B1 B1 | **B1** without $\div(-1)$. **B1** for $\div(-1)$ |
| (or $25x - 10x^2/2 + \frac{1}{3}x^3$) | (B2,1,0) | $-1$ for each incorrect term |
| $\int\frac{16}{x^2}\,dx = -\frac{16}{x}$ | B1 | |
| Use of limits 1 and 4 in an integrated expression and subtracted | DM1 | Must have used "$y^2$" at least once. Need to see values substituted |
| $\rightarrow 9\pi$ or $28.3$ | A1 | |
---6\\
\begin{tikzpicture}[>=stealth, thick, scale=0.95]
% --- Shaded region between line x+y=5 and curve y=4/x, from x=1 to x=4 ---
\fill[gray!60]
plot[domain=1:4, samples=80, smooth] ({\x}, {5-\x})
--
plot[domain=4:1, samples=80, smooth] ({\x}, {4/\x})
-- cycle;
% --- Curve y = 4/x ---
\draw[thick] plot[domain=0.55:5.2, samples=100, smooth] ({\x}, {4/\x});
% --- Line x + y = 5 ---
\draw[thick] (0.3, 4.7) -- (4.7, 0.3);
% --- Axes ---
\draw[->] (-0.3, 0) -- (5.5, 0) node[below] {$x$};
\draw[->] (0, -0.3) -- (0, 5.5) node[left] {$y$};
% --- Origin ---
\node[below left] at (0, 0) {$O$};
% --- Points A and B ---
\fill (1, 4) circle (1.8pt);
\node[above right] at (1, 4) {$A\,(1,\,4)$};
\fill (4, 1) circle (1.8pt);
\node[above right] at (4, 1) {$B\,(4,\,1)$};
% --- Labels for curve and line ---
\node[above right] at (2.8, 2.6) {$x + y = 5$};
\node[below left] at (2.2, 1.5) {$y = \dfrac{4}{x}$};
\end{tikzpicture}
The diagram shows the straight line $x + y = 5$ intersecting the curve $y = \frac { 4 } { x }$ at the points $A ( 1,4 )$ and $B ( 4,1 )$. Find, showing all necessary working, the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
\hfill \mbox{\textit{CAIE P1 2017 Q6 [7]}}