| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| 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 | Challenging +1.2 This is a volume of revolution question with an implicit curve, requiring rearrangement to x² = f(y) form and application of the standard formula V = π∫x²dy. While the implicit equation looks intimidating, the algebraic manipulation is straightforward (isolating x²), and the integration itself is routine (polynomial and standard trig integral). The y-intercepts are given, eliminating any root-finding difficulty. Part (b) is a simple interpretation. This is moderately above average difficulty due to the implicit curve context and multi-step nature, but remains a standard Core Pure 2 exercise without requiring novel insight. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\pi}{16}\int_{-1.545}^{1.257}\left(6-3y^2+y\cos\left(\frac{5}{2}y\right)\right)\{dy\}\) leading to \(\pi\int_{-1.545}^{1.257}\left(\frac{3}{8}-\frac{3}{16}y^2+\frac{1}{16}y\cos\left(\frac{5}{2}y\right)\right)\{dy\}\) | B1 | Selects correct volume of revolution formula with correct limits in evidence |
| \(\int x^2\,dy = \frac{1}{16}\int 6-3y^2+y\cos\left(\frac{5}{2}y\right)dy \to Ky - Ly^3 + ...\) | M1 | Attempts to integrate with correct form for constant and \(y^2\) term |
| \(\int y\cos\left(\frac{5}{2}y\right)dy = Ay\sin\left(\frac{5}{2}y\right)+B\cos\left(\frac{5}{2}y\right)(+c)\), specifically \(= \frac{2}{5}y\sin\left(\frac{5}{2}y\right)+\frac{4}{25}\cos\left(\frac{5}{2}y\right)(+c)\) | M1 | Applies integration by parts fully on \(y\cos\left(\frac{5}{2}y\right)\) term. Form must be correct, allow slips in coefficients |
| \(\int x^2\,dy = \frac{3}{8}y - \frac{1}{16}y^3 + \frac{1}{40}y\sin\left(\frac{5}{2}y\right)+\frac{1}{100}\cos\left(\frac{5}{2}y\right)(+c)\) | A1 | Correct integration of \(x^2\) equation |
| \(\int_{-1.545}^{1.257} x^2\,dy = \frac{1}{16}\left[6y-y^3+\frac{2}{5}y\sin\left(\frac{5}{2}y\right)+\frac{4}{25}\cos\left(\frac{5}{2}y\right)\right]_{-1.545}^{1.257}\) \(= \frac{1}{16}(5.3954... -(-6.1101...))\) \(= (0.3372...) - (-0.3818...)\) | M1 | Applies correct limits. Condone working in degrees \((5.74...) - (-5.38...)\) if integration is correct |
| Volume \(= \pi \times \frac{11.505...}{16} = 2.26 \text{ cm}^3\) | A1 | Correct volume including units. Accept awrt \(2.26\text{ cm}^3\). Allow \(0.719\pi\text{ cm}^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max volume for 100 berries is \(100 \times 2.26 \approx 226\) | B1ft | Attempts to estimate volume of juice from 100 berries — their (a) multiplied by 100 |
| Reason e.g. not all berries will become juice (skin, flesh, seeds); not all will be as big as the largest; \(150 < 200\) or \(300 > 200\). If value is less than 220 — not likely to produce \(200\text{ cm}^3\); between 220 and 250 — either way; greater than 250 — likely to produce \(200\text{ cm}^3\) | B1ft | Draws suitable conclusion with reason given |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\pi}{16}\int_{-1.545}^{1.257}\left(6-3y^2+y\cos\left(\frac{5}{2}y\right)\right)\{dy\}$ leading to $\pi\int_{-1.545}^{1.257}\left(\frac{3}{8}-\frac{3}{16}y^2+\frac{1}{16}y\cos\left(\frac{5}{2}y\right)\right)\{dy\}$ | B1 | Selects correct volume of revolution formula with correct limits in evidence |
| $\int x^2\,dy = \frac{1}{16}\int 6-3y^2+y\cos\left(\frac{5}{2}y\right)dy \to Ky - Ly^3 + ...$ | M1 | Attempts to integrate with correct form for constant and $y^2$ term |
| $\int y\cos\left(\frac{5}{2}y\right)dy = Ay\sin\left(\frac{5}{2}y\right)+B\cos\left(\frac{5}{2}y\right)(+c)$, specifically $= \frac{2}{5}y\sin\left(\frac{5}{2}y\right)+\frac{4}{25}\cos\left(\frac{5}{2}y\right)(+c)$ | M1 | Applies integration by parts fully on $y\cos\left(\frac{5}{2}y\right)$ term. Form must be correct, allow slips in coefficients |
| $\int x^2\,dy = \frac{3}{8}y - \frac{1}{16}y^3 + \frac{1}{40}y\sin\left(\frac{5}{2}y\right)+\frac{1}{100}\cos\left(\frac{5}{2}y\right)(+c)$ | A1 | Correct integration of $x^2$ equation |
| $\int_{-1.545}^{1.257} x^2\,dy = \frac{1}{16}\left[6y-y^3+\frac{2}{5}y\sin\left(\frac{5}{2}y\right)+\frac{4}{25}\cos\left(\frac{5}{2}y\right)\right]_{-1.545}^{1.257}$ $= \frac{1}{16}(5.3954... -(-6.1101...))$ $= (0.3372...) - (-0.3818...)$ | M1 | Applies correct limits. Condone working in degrees $(5.74...) - (-5.38...)$ if integration is correct |
| Volume $= \pi \times \frac{11.505...}{16} = 2.26 \text{ cm}^3$ | A1 | Correct volume including units. Accept awrt $2.26\text{ cm}^3$. Allow $0.719\pi\text{ cm}^3$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max volume for 100 berries is $100 \times 2.26 \approx 226$ | B1ft | Attempts to estimate volume of juice from 100 berries — their (a) multiplied by 100 |
| Reason e.g. not all berries will become juice (skin, flesh, seeds); not all will be as big as the largest; $150 < 200$ or $300 > 200$. If value is less than 220 — not likely to produce $200\text{ cm}^3$; between 220 and 250 — either way; greater than 250 — likely to produce $200\text{ cm}^3$ | B1ft | Draws suitable conclusion with reason given |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{59a57888-8aa8-4ed8-b704-ebf3980c0344-20_557_558_408_756}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
John picked 100 berries from a plant.\\
The largest berry picked was approximately 2.8 cm long.\\
The shape of this berry is modelled by rotating the curve with equation
$$16 x ^ { 2 } + 3 y ^ { 2 } - y \cos \left( \frac { 5 } { 2 } y \right) = 6 \quad x \geqslant 0$$
shown in Figure 2, about the $y$-axis through $2 \pi$ radians, where the units are cm .\\
Given that the $y$ intercepts of the curve are - 1.545 and 1.257 to four significant figures,\\
(a) use algebraic integration to determine, according to the model, the volume of this berry.
Given that the 100 berries John picked were then squeezed for juice,\\
(b) use your answer to part (a) to decide whether, in reality, there is likely to be enough juice to fill a $200 \mathrm {~cm} ^ { 3 }$ cup, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel CP2 2023 Q7 [8]}}