| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and tangent/normal |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question combining standard integration techniques. Part (i) requires basic area integration, part (ii) is a routine volume of revolution calculation using the standard formula, and part (iii) involves finding derivatives to get tangent gradients then using the angle between lines formula. All techniques are standard P1 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \int y \, dy = \int_1^2 \frac{y^2 - 1}{3} dy\) | M1 | Uses integration wrt \(y\) |
| \(= \left[\frac{y^3}{9} - \frac{y}{3}\right]_0^2 = \frac{4}{9}\) (allow 0.44 to 0.45) | A1, DM1, A1 | Integration correct. Use of limits 0 to 1. co |
| \(\text{[or } 2 - \int\sqrt{3x+1} dx = [2 - \frac{(3x+1)^{3/2}}{3 \times 3}]_0^1 = \frac{4}{9}]\) | B1, B1, M1A1 | B1 for everything but +3. B1 for +3. M1 for "2–" and use of limits 0 to 1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi \int y^3 dx = \pi \int(3x+1) dx\) | M1 | M1 for correct formula used with \(y^2\) and integration wrt \(x\). (does not need \(\pi\)) |
| \(= \pi\left(\frac{3x^2}{2} + x\right) \text{ from 0 to 1}\) | A1 | A1 integration correct, including \(\pi\). |
| \(\text{Vol of cylinder} = \pi \times 2^2 \times 1 = 4\pi\) | B1 | Or by integration of \(y^2 = 4\) |
| \(\rightarrow \text{Subtraction} \rightarrow 1.5\pi (4.71)\) | A1 | co |
$y = \sqrt{3x + 1}$
**(i)**
$A = \int y \, dy = \int_1^2 \frac{y^2 - 1}{3} dy$ | M1 | Uses integration wrt $y$
$= \left[\frac{y^3}{9} - \frac{y}{3}\right]_0^2 = \frac{4}{9}$ (allow 0.44 to 0.45) | A1, DM1, A1 | Integration correct. Use of limits 0 to 1. co
$\text{[or } 2 - \int\sqrt{3x+1} dx = [2 - \frac{(3x+1)^{3/2}}{3 \times 3}]_0^1 = \frac{4}{9}]$ | B1, B1, M1A1 | B1 for everything but +3. B1 for +3. M1 for "2–" and use of limits 0 to 1.
**(ii)**
$V = \pi \int y^3 dx = \pi \int(3x+1) dx$ | M1 | M1 for correct formula used with $y^2$ and integration wrt $x$. (does not need $\pi$)
$= \pi\left(\frac{3x^2}{2} + x\right) \text{ from 0 to 1}$ | A1 | A1 integration correct, including $\pi$.
$\text{Vol of cylinder} = \pi \times 2^2 \times 1 = 4\pi$ | B1 | Or by integration of $y^2 = 4$
$\rightarrow \text{Subtraction} \rightarrow 1.5\pi (4.71)$ | A1 | co
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{08729aab-586b-4210-94c9-77b1f6b1d873-4_719_670_264_735}
The diagram shows the curve $y = \sqrt { } ( 3 x + 1 )$ and the points $P ( 0,1 )$ and $Q ( 1,2 )$ on the curve. The shaded region is bounded by the curve, the $y$-axis and the line $y = 2$.\\
(i) Find the area of the shaded region.\\
(ii) Find the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
Tangents are drawn to the curve at the points $P$ and $Q$.\\
(iii) Find the acute angle, in degrees correct to 1 decimal place, between the two tangents.
\hfill \mbox{\textit{CAIE P1 2008 Q9 [12]}}