| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with implicit or parametric curves |
| Difficulty | Standard +0.3 This is a straightforward volume of revolution question requiring students to identify the ellipse equation from a diagram, then apply the standard formula V = π∫y²dx. While it involves an ellipse (slightly beyond basic C3), the integration is routine after substituting y² from the ellipse equation, and the 'show that' format provides a target answer. The multi-step nature and ellipse context place it slightly above average, but it requires no novel insight or complex manipulation. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Writes any equation of a non-circular ellipse | M1 | |
| \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) | A1 | Accept \(3^2\) for 9 and \(2^2\) for 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{y^2}{4} = 1 - \frac{x^2}{9}\), so \(y^2 = 4 - \frac{4x^2}{9}\) | B1F | Correct expression for \(y^2\) in terms of \(x\); FT their ellipse equation |
| \(\text{Volume} = \pi\int_{-3}^{3}\left(4 - \frac{4x^2}{9}\right)dx\) | M1 | Uses volume of revolution formula; condone missing \(\pi\), \(dx\) and missing/incorrect limits; condone \(\int(my^2+c)\) |
| \(\pi\int_{-3}^{3}\left(4-\frac{4x^2}{9}\right)dx\) fully correct | R1 | Condone missing brackets |
| \(= \pi\left[4x - \frac{4x^3}{27}\right]_{-3}^{3} = \pi(12-4) - \pi(-12+4) = 16\pi\) | A1 | Correct volume in exact form |
## Question 10:
### Part 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Writes any equation of a non-circular ellipse | M1 | |
| $\frac{x^2}{9} + \frac{y^2}{4} = 1$ | A1 | Accept $3^2$ for 9 and $2^2$ for 4 |
### Part 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{y^2}{4} = 1 - \frac{x^2}{9}$, so $y^2 = 4 - \frac{4x^2}{9}$ | B1F | Correct expression for $y^2$ in terms of $x$; FT their ellipse equation |
| $\text{Volume} = \pi\int_{-3}^{3}\left(4 - \frac{4x^2}{9}\right)dx$ | M1 | Uses volume of revolution formula; condone missing $\pi$, $dx$ and missing/incorrect limits; condone $\int(my^2+c)$ |
| $\pi\int_{-3}^{3}\left(4-\frac{4x^2}{9}\right)dx$ fully correct | R1 | Condone missing brackets |
| $= \pi\left[4x - \frac{4x^3}{27}\right]_{-3}^{3} = \pi(12-4) - \pi(-12+4) = 16\pi$ | A1 | Correct volume in exact form |
---10 The diagram below shows an ellipse $E$
The coordinate axes are the lines of symmetry of $E$\\
\includegraphics[max width=\textwidth, alt={}, center]{fd9715c4-9ce1-4608-aed6-f3d4f71208b5-14_645_780_450_630}
10
\begin{enumerate}[label=(\alph*)]
\item Write down an equation of $E$
10
\item The region bounded by the $x$-axis and the ellipse $E$ for $y \geq 0$ is shaded in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{fd9715c4-9ce1-4608-aed6-f3d4f71208b5-15_643_775_408_635}
A solid $S$ is formed by rotating the shaded region through $360 ^ { \circ }$ about the $x$-axis. Show that the volume of $S$ is $a \pi$ where $a$ is an integer to be found.\\
\includegraphics[max width=\textwidth, alt={}, center]{fd9715c4-9ce1-4608-aed6-f3d4f71208b5-16_2488_1732_219_139}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q10 [6]}}