| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Applied context: real-world solid |
| Difficulty | Standard +0.3 This is a multi-part volumes of revolution question with straightforward components: stating a parameter value, finding maxima using basic calculus (differentiation of sine and cubic), comparing models to given data, and integrating y² for volume. While it has many parts and real-world context, each step uses standard A-level techniques without requiring novel insight or complex manipulation. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 4\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Model A: (i) Widest point will be 4 (cm) from the base | B1 | |
| (ii) Width at widest point is 12 (cm) \((2\times('a'+2)\) ft\()\) | B1ft | |
| Model B: (i) \(y = 4 + \frac{x^3 - 64x}{100} \Rightarrow \frac{dy}{dx} = \frac{3x^2 - 64}{100}\) | M1 | |
| \(\frac{dy}{dx} = 0 \Rightarrow x = \pm\sqrt{\frac{64}{3}} = \pm\frac{8\sqrt{3}}{3} = \pm\text{awrt } 4.62\) | A1 | |
| So max width is at distance \(8 - \frac{8}{\sqrt{3}} = 8 - \frac{8\sqrt{3}}{3} \approx 3.38\) (cm) from base | A1 | |
| (ii) \(y\big | _{-4.61\ldots} = 4 + \frac{(-4.62\ldots)^3 - 64(-4.62\ldots)}{100} = \ldots\) | dM1 |
| \(= 5.97\ldots\) so diameter is approximately 11.9 (cm) \([2a + 3.94\ldots\text{ ft}]\) | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Model A and Model B both have diameters close to 12; Model B distance from base is closer to 3 than Model A so is more appropriate | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V_B = \pi\int_{-8}^{8} y^2\, dx = \pi\int_{-8}^{8}\left(4 + \frac{x^3 - 64x}{100}\right)^2 dx = \ldots\) | B1 | |
| Expands and simplifies integrand correctly | M1 | |
| \(= \frac{\{\pi\}}{10000}\left[160000x + \frac{x^7}{7} + 4096\frac{x^3}{3} + 800\frac{x^4}{4} - 51200\frac{x^2}{2} - 128\frac{x^5}{5}\right]_{(-8)}^{(8)}\) | dM1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(=\{\pi\}\left[16x+\frac{x^7}{70000}+\frac{256}{1875}x^3+\frac{1}{50}x^4-\frac{64}{25}x^2-\frac{8}{3125}x^5\right]_{(-8)}^{(8)}\) | ||
| \(=\frac{\{\pi\}}{10000}(620583.00...-(-2258983.01...))\approx\frac{2879566\pi}{10000}\) | M1 | 3.4 |
| \(=\text{awrt } 905\ (\text{cm}^3)\ \text{cso}\) | A1 | 1.1b |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Compares their volume to 900 or compares their volume \(+100\) to 1 litre or 1000 and comments appropriately. | B1ft | 3.5a |
| (1) |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 4$ | B1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Model A: (i) Widest point will be 4 (cm) from the base | B1 | |
| (ii) Width at widest point is 12 (cm) $(2\times('a'+2)$ ft$)$ | B1ft | |
| Model B: (i) $y = 4 + \frac{x^3 - 64x}{100} \Rightarrow \frac{dy}{dx} = \frac{3x^2 - 64}{100}$ | M1 | |
| $\frac{dy}{dx} = 0 \Rightarrow x = \pm\sqrt{\frac{64}{3}} = \pm\frac{8\sqrt{3}}{3} = \pm\text{awrt } 4.62$ | A1 | |
| So max width is at distance $8 - \frac{8}{\sqrt{3}} = 8 - \frac{8\sqrt{3}}{3} \approx 3.38$ (cm) from base | A1 | |
| (ii) $y\big|_{-4.61\ldots} = 4 + \frac{(-4.62\ldots)^3 - 64(-4.62\ldots)}{100} = \ldots$ | dM1 | |
| $= 5.97\ldots$ so diameter is approximately 11.9 (cm) $[2a + 3.94\ldots\text{ ft}]$ | A1ft | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Model A and Model B both have diameters close to 12; Model B distance from base is closer to 3 than Model A so is more appropriate | B1ft | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V_B = \pi\int_{-8}^{8} y^2\, dx = \pi\int_{-8}^{8}\left(4 + \frac{x^3 - 64x}{100}\right)^2 dx = \ldots$ | B1 | |
| Expands and simplifies integrand correctly | M1 | |
| $= \frac{\{\pi\}}{10000}\left[160000x + \frac{x^7}{7} + 4096\frac{x^3}{3} + 800\frac{x^4}{4} - 51200\frac{x^2}{2} - 128\frac{x^5}{5}\right]_{(-8)}^{(8)}$ | dM1 | |
## Question (continuing from previous parts):
---
### Part (d):
| Working | Mark | Guidance |
|---|---|---|
| $=\{\pi\}\left[16x+\frac{x^7}{70000}+\frac{256}{1875}x^3+\frac{1}{50}x^4-\frac{64}{25}x^2-\frac{8}{3125}x^5\right]_{(-8)}^{(8)}$ | | |
| $=\frac{\{\pi\}}{10000}(620583.00...-(-2258983.01...))\approx\frac{2879566\pi}{10000}$ | M1 | 3.4 |
| $=\text{awrt } 905\ (\text{cm}^3)\ \text{cso}$ | A1 | 1.1b |
| | **(5)** | |
**Notes for (d):**
- **B1:** Applies $\pi\int_{-8}^{8}y^2\,dx$ to the model. Must have $\pi$ and correct limits, with $y$ substituted in.
- Alternatively attempts to square $y$ first and then substitute in.
- **M1:** Attempts to expand $y^2$; must include at least a constant and $x^6$ terms as long as a clear attempt at $y^2$. (Limits not required.)
- **dM1:** Attempts the integration, must first be rearranged to an integrable form then power increasing by at least 1 in at least two terms. (Limits not required.)
- **M1:** Applies correct limits to their integral following an attempt at $y^2$ with at least a constant and $x^6$ terms. If no working shown, allow this mark if the correct answer appears from a calculator. (So M0dM0M1 is possible.)
- **A1:** awrt 905 cso — must come from a fully correct solution.
- **Note:** For answers appearing from calculator, B1M0dM0M1A0 is possible; algebraic integration must be seen to score other marks.
---
### Part (e):
| Working | Mark | Guidance |
|---|---|---|
| Compares their volume to 900 or compares their volume $+100$ to 1 litre or 1000 and comments appropriately. | B1ft | 3.5a |
| | **(1)** | |
**Notes for (e):**
- **B1ft:** Compares their volume to 900 or compares their volume $+100$ to 1 litre or 1000 and comments appropriately. Correct answer in (d) needs to conclude that it is suitable.
---
**Total: (15 marks)**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{545661a6-8d78-488c-b73b-ab2ced60debf-28_663_531_210_258}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{545661a6-8d78-488c-b73b-ab2ced60debf-28_394_903_431_900}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 1 shows a sketch of a 16 cm tall vase which has a flat circular base with diameter 8 cm and a circular opening of diameter 8 cm at the top.
A student measures the circular cross-section halfway up the vase to be 8 cm in diameter.\\
The student models the shape of the vase by rotating a curve, shown in Figure 2, through $360 ^ { \circ }$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item State the value of $a$ that should be used when setting up the model.
Two possible equations are suggested for the curve in the model.
$$\begin{array} { l l }
\text { Model A } & y = a - 2 \sin \left( \frac { 45 } { 2 } x \right) ^ { \circ } \\
\text { Model B } & y = a + \frac { x ( x - 8 ) ( x + 8 ) } { 100 }
\end{array}$$
For each model,
\item \begin{enumerate}[label=(\roman*)]
\item find the distance from the base at which the widest part of the vase occurs,
\item find the diameter of the vase at this widest point.
The widest part of the vase has diameter 12 cm and is just over 3 cm from the base.
\end{enumerate}\item Using this information and making your reasoning clear, suggest which model is more appropriate.
\item Using algebraic integration, find the volume for the vase predicted by Model B. You must make your method clear.
The student pours water from a full one litre jug into the vase and finds that there is 100 ml left in the jug when the vase is full.
\item Comment on the suitability of Model B in light of this information.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP AS 2022 Q8 [15]}}