| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Standard +0.3 This is a standard volumes of revolution question with three routine parts: finding area by integration, volume about x-axis using πy² formula, and volume about y-axis using πx² formula (requiring rearrangement). All techniques are textbook exercises for P1 level with straightforward polynomial integration and no conceptual surprises. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Area} = \int \frac{1}{2}(x^4-1)\,dx = \frac{1}{2}\left[\frac{x^5}{5} - x\right]\) | *B1 | |
| \(\frac{1}{2}\left[\frac{1}{5}-1\right] - 0 = \left(-\right)\frac{2}{5}\) | DM1A1 | Apply limits \(0 \to 1\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Vol} = \pi\int y^2\,dx = \frac{1}{4}(\pi)\int(x^8 - 2x^4 + 1)\,dx\) | M1 | If middle term missed out can only gain the M marks |
| \(\frac{1}{4}(\pi)\left[\frac{x^9}{9} - \frac{2x^5}{5} + x\right]\) | *A1 | |
| \(\frac{1}{4}(\pi)\left[\left(\frac{1}{9} - \frac{2}{5} + 1\right) - 0\right]\) | DM1 | |
| \(\frac{8\pi}{45}\) or \(0.559\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Vol} = \pi\int x^2\,dy = (\pi)\int(2y+1)^{1/2}\,dy\) | M1 | Condone use of \(x\) if integral is correct |
| \((\pi)\left[\frac{(2y+1)^{3/2}}{3/2}\right][\div 2]\) | *A1A1 | Expect \((\pi)\left[\frac{(2y+1)^{3/2}}{3}\right]\) |
| \((\pi)\left[\frac{1}{3} - 0\right]\) | DM1 | Apply \(-\frac{1}{2} \to 0\) |
| \(\frac{\pi}{3}\) or \(1.05\) | A1 | |
| Total | 5 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Area} = \int \frac{1}{2}(x^4-1)\,dx = \frac{1}{2}\left[\frac{x^5}{5} - x\right]$ | *B1 | |
| $\frac{1}{2}\left[\frac{1}{5}-1\right] - 0 = \left(-\right)\frac{2}{5}$ | DM1A1 | Apply limits $0 \to 1$ |
| **Total** | **3** | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Vol} = \pi\int y^2\,dx = \frac{1}{4}(\pi)\int(x^8 - 2x^4 + 1)\,dx$ | M1 | If middle term missed out can only gain the M marks |
| $\frac{1}{4}(\pi)\left[\frac{x^9}{9} - \frac{2x^5}{5} + x\right]$ | *A1 | |
| $\frac{1}{4}(\pi)\left[\left(\frac{1}{9} - \frac{2}{5} + 1\right) - 0\right]$ | DM1 | |
| $\frac{8\pi}{45}$ or $0.559$ | A1 | |
| **Total** | **4** | |
---
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Vol} = \pi\int x^2\,dy = (\pi)\int(2y+1)^{1/2}\,dy$ | M1 | Condone use of $x$ if integral is correct |
| $(\pi)\left[\frac{(2y+1)^{3/2}}{3/2}\right][\div 2]$ | *A1A1 | Expect $(\pi)\left[\frac{(2y+1)^{3/2}}{3}\right]$ |
| $(\pi)\left[\frac{1}{3} - 0\right]$ | DM1 | Apply $-\frac{1}{2} \to 0$ |
| $\frac{\pi}{3}$ or $1.05$ | A1 | |
| **Total** | **5** | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{5201a3d5-7733-4d10-9de5-0c2255e3ad60-18_401_584_264_776}
The diagram shows part of the curve $y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)$, defined for $x \geqslant 0$.\\
(i) Find, showing all necessary working, the area of the shaded region.\\
(ii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
(iii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $y$-axis.\\
\hfill \mbox{\textit{CAIE P1 2017 Q10 [12]}}