| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Session | Specimen |
| 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 question requires converting parametric equations to Cartesian form using trigonometric identities (part a) and then applying the volume of revolution formula (part b). While it involves multiple steps and parametric curves (a Further Maths topic), the algebraic manipulation is relatively straightforward once you recognize to use sin θ = -(y+1), and the volume integration is standard. The question is more challenging than average A-level due to the parametric setup, but the techniques are well-practiced in Core Pure modules. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \cos\theta + \sin\theta\cos\theta = -y\cos\theta\) | M1 | Obtains \(x\) in terms of \(y\) and \(\cos\theta\) |
| \(\sin\theta = -y - 1\) | M1 | Obtains an equation connecting \(y\) and \(\sin\theta\) |
| \(\left(\frac{x}{-y}\right)^2 = 1-(-y-1)^2\) | M1 | Uses Pythagoras to obtain an equation in \(x\) and \(y\) only |
| \(x^2 = -(y^4 + 2y^3)\) | A1* | Obtains printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \pi\int x^2\,dy = \pi\int -(y^4+2y^3)\,dy\) | M1 | Uses the correct volume of revolution formula with the given expression |
| \(= \pi\left[-\left(\frac{y^5}{5}+\frac{y^4}{2}\right)\right]\) | A1 | Correct integration |
| \(= -\pi\left[\left(\frac{(0)^5}{5}+\frac{(0)^4}{2}\right) - \left(\frac{(-2)^5}{5}+\frac{(-2)^4}{2}\right)\right]\) | M1 | Correct use of correct limits |
| \(= 1.6\pi\text{ cm}^3\) or \(\text{awrt } 5.03\text{ cm}^3\) | A1 | Correct volume |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \cos\theta + \sin\theta\cos\theta = -y\cos\theta$ | M1 | Obtains $x$ in terms of $y$ and $\cos\theta$ |
| $\sin\theta = -y - 1$ | M1 | Obtains an equation connecting $y$ and $\sin\theta$ |
| $\left(\frac{x}{-y}\right)^2 = 1-(-y-1)^2$ | M1 | Uses Pythagoras to obtain an equation in $x$ and $y$ only |
| $x^2 = -(y^4 + 2y^3)$ | A1* | Obtains printed answer |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \pi\int x^2\,dy = \pi\int -(y^4+2y^3)\,dy$ | M1 | Uses the correct volume of revolution formula with the given expression |
| $= \pi\left[-\left(\frac{y^5}{5}+\frac{y^4}{2}\right)\right]$ | A1 | Correct integration |
| $= -\pi\left[\left(\frac{(0)^5}{5}+\frac{(0)^4}{2}\right) - \left(\frac{(-2)^5}{5}+\frac{(-2)^4}{2}\right)\right]$ | M1 | Correct use of correct limits |
| $= 1.6\pi\text{ cm}^3$ or $\text{awrt } 5.03\text{ cm}^3$ | A1 | Correct volume |
**(4 marks)**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b36bdc3-a68d-4982-bf23-f780773df5cc-14_259_327_214_868}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the image of a gold pendant which has height 2 cm . The pendant is modelled by a solid of revolution of a curve $C$ about the $y$-axis. The curve $C$ has parametric equations
$$x = \cos \theta + \frac { 1 } { 2 } \sin 2 \theta , \quad y = - ( 1 + \sin \theta ) \quad 0 \leqslant \theta \leqslant 2 \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that a Cartesian equation of the curve $C$ is
$$x ^ { 2 } = - \left( y ^ { 4 } + 2 y ^ { 3 } \right)$$
\item Hence, using the model, find, in $\mathrm { cm } ^ { 3 }$, the volume of the pendant.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 Q7 [8]}}