| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| 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 rotation about the y-axis with a simple quadratic curve. Students need to rearrange to x² = 2-y, find limits (y from 0 to 2), apply the standard formula V = π∫x²dy, and multiply by density. The integration is elementary and the context parts (b) and (c) require only brief written responses. This is easier than average as it's a direct application of a standard technique with no conceptual challenges. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{V=\}\pi\int_0^2\left[(2-y)^{\frac{1}{2}}\right]^2\,dy\) or \(\{V=\}\pi\int_0^2(2-y)\,dy\) | B1 | Sets up correct expression including limits |
| Integrates to form \(\alpha y \pm \beta y^2\) | M1 | |
| Correct integration \(2y-\frac{1}{2}y^2\) | A1 | |
| Uses \(y\) limits correctly: \(\pi\left[2y-\frac{1}{2}y^2\right]_0^2 = \pi\left(2(2)-\frac{1}{2}(2^2)\right)-0=\ldots\{2\pi \text{ or } 6.28\ldots\}\) | M1 | Must be changed expression |
| mass \(=\) their volume \(\times 900\) | M1 | |
| Mass \(= 5700\) kg (2 s.f.) | A1 | Note: incorrect upper limit of \(\sqrt{2}\) leads to 5200 kg, scores B0 M1 A1 M1 M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. surface will not be smooth / pile won't follow shape of curve / pile will not be solid / equation of curves may not be suitable model / concrete likely to be uneven / pile unlikely to be symmetrical | B1 | Must refer to model, not value of density |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Makes comparison between their mass and 5500, draws conclusion e.g. 200 difference is a lot therefore not good model; or mass of 5700 is very close to 5500 kg therefore a good model; or finds percentage error and draws conclusion | B1ft | Follow through on (a). If using calculation it must be correct. Not sufficient to say \(5700>5500\). Ignore contradictory comments |
# Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{V=\}\pi\int_0^2\left[(2-y)^{\frac{1}{2}}\right]^2\,dy$ or $\{V=\}\pi\int_0^2(2-y)\,dy$ | B1 | Sets up correct expression including limits |
| Integrates to form $\alpha y \pm \beta y^2$ | M1 | |
| Correct integration $2y-\frac{1}{2}y^2$ | A1 | |
| Uses $y$ limits correctly: $\pi\left[2y-\frac{1}{2}y^2\right]_0^2 = \pi\left(2(2)-\frac{1}{2}(2^2)\right)-0=\ldots\{2\pi \text{ or } 6.28\ldots\}$ | M1 | Must be changed expression |
| mass $=$ their volume $\times 900$ | M1 | |
| Mass $= 5700$ kg (2 s.f.) | A1 | Note: incorrect upper limit of $\sqrt{2}$ leads to 5200 kg, scores B0 M1 A1 M1 M1 A0 |
# Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. surface will not be smooth / pile won't follow shape of curve / pile will not be solid / equation of curves may not be suitable model / concrete likely to be uneven / pile unlikely to be symmetrical | B1 | Must refer to model, not value of density |
# Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Makes comparison between their mass and 5500, draws conclusion e.g. 200 difference is a lot therefore not good model; or mass of 5700 is very close to 5500 kg therefore a good model; or finds percentage error and draws conclusion | B1ft | Follow through on (a). If using calculation it must be correct. Not sufficient to say $5700>5500$. Ignore contradictory comments |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ab572f1e-2828-4ab3-b148-605f35ccd1db-14_385_526_447_420}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ab572f1e-2828-4ab3-b148-605f35ccd1db-14_485_433_388_1187}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A large pile of concrete waste is created on a building site.\\
Figure 1 shows a central vertical cross-section of the concrete waste.\\
The curve $C$, shown in Figure 2, has equation
$$y + x ^ { 2 } = 2 \quad 0 \leqslant x \leqslant \sqrt { 2 }$$
The region $R$, shown shaded in Figure 2, is bounded by the $y$-axis, the $x$-axis and the curve $C$.
The volume of concrete waste is modelled by the volume of revolution formed when $R$ is rotated through $360 ^ { \circ }$ about the $y$-axis. The units are metres.
The density of the concrete waste is $900 \mathrm { kgm } ^ { - 3 }$\\
(a) Use the model to estimate the mass of the concrete waste. Give your answer to 2 significant figures.\\
(b) Give a limitation of the model.
The mass of the concrete waste is approximately 5500 kg .\\
(c) Use this information and your answer to part (a) to evaluate the model, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel CP AS 2023 Q5 [8]}}