| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric surface area of revolution |
| Difficulty | Challenging +1.8 This is a Further Maths F3 question requiring surface area of revolution with parametric equations. While the formula application is standard, the integration involves non-trivial trigonometric manipulation (ln(sec θ + tan θ) and its derivative) and requires careful algebraic work to reach the exact form. The parametric context and exact answer requirement elevate it above typical A-level questions, but it follows a predictable template for F3 surface area problems. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=18.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta} = \frac{\sec\theta\tan\theta + \sec^2\theta}{\sec\theta + \tan\theta} - \cos\theta\) | B1 | Correct derivative. Do not condone missing brackets unless correct expression implied by subsequent work. Other forms possible e.g. \(\sec\theta - \cos\theta\), \(\tan\theta\sin\theta\) |
| Attempts \(\frac{dy}{d\theta}\) and then \(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2\) | M1 | \(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = \left(\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}-\cos\theta\right)^2 + (-\sin\theta)^2\) |
| Applies correct surface area formula \(S=(2\pi)\int\cos\theta\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta\) with their derivatives | M1 | With or without \(2\pi\). Note: \(\sqrt{\left(\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}-\cos\theta\right)^2+(-\sin\theta)^2}=\tan\theta\). Allow \(\pi\) in front of integral but must be an integral |
| \((2\pi)\int\sin\theta\,d\theta\) | A1 | Fully correct simplified integral with or without \(2\pi\) |
| \(=(2\pi)[-\cos\theta](+c)\) | A1 | Correct integration with or without \(2\pi\) |
| \((2\pi)[-\cos\theta]_0^{\frac{\pi}{4}} = (2\pi)\left(-\frac{1}{\sqrt{2}}+1\right)\) | dM1 | Applies limits 0 and \(\frac{\pi}{4}\). Must see evidence of both limits if necessary. Depends on both previous method marks |
| \(\text{TSA} = 2\pi\left(-\frac{1}{\sqrt{2}}+1\right)+\pi\times1^2+\pi\times\left(\frac{1}{\sqrt{2}}\right)^2\) | dM1 | Correct expressions for 2 "ends" added to curved surface area. Depends on previous method mark |
| \(=\frac{\pi}{2}(7-2\sqrt{2})\) | A1 | Correct answer in required form or correct values for \(p\) and \(q\) |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = \frac{\sec\theta\tan\theta + \sec^2\theta}{\sec\theta + \tan\theta} - \cos\theta$ | B1 | Correct derivative. Do not condone missing brackets unless correct expression implied by subsequent work. Other forms possible e.g. $\sec\theta - \cos\theta$, $\tan\theta\sin\theta$ |
| Attempts $\frac{dy}{d\theta}$ and then $\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2$ | M1 | $\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = \left(\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}-\cos\theta\right)^2 + (-\sin\theta)^2$ |
| Applies correct surface area formula $S=(2\pi)\int\cos\theta\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta$ with their derivatives | M1 | With or without $2\pi$. Note: $\sqrt{\left(\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}-\cos\theta\right)^2+(-\sin\theta)^2}=\tan\theta$. Allow $\pi$ in front of integral but must be an integral |
| $(2\pi)\int\sin\theta\,d\theta$ | A1 | Fully correct simplified integral with or without $2\pi$ |
| $=(2\pi)[-\cos\theta](+c)$ | A1 | Correct integration with or without $2\pi$ |
| $(2\pi)[-\cos\theta]_0^{\frac{\pi}{4}} = (2\pi)\left(-\frac{1}{\sqrt{2}}+1\right)$ | dM1 | Applies limits 0 and $\frac{\pi}{4}$. Must see evidence of both limits if necessary. Depends on both previous method marks |
| $\text{TSA} = 2\pi\left(-\frac{1}{\sqrt{2}}+1\right)+\pi\times1^2+\pi\times\left(\frac{1}{\sqrt{2}}\right)^2$ | dM1 | Correct expressions for 2 "ends" added to curved surface area. Depends on previous method mark |
| $=\frac{\pi}{2}(7-2\sqrt{2})$ | A1 | Correct answer in required form or correct values for $p$ and $q$ |
**Note:** Final answer should follow correct work. Final mark withheld following e.g. $\frac{dy}{d\theta}$ clearly seen as $+\sin\theta$ or $\int\sin\theta\,d\theta = +\cos\theta$. Without the "ends" the answer is $\frac{\pi}{2}(4-2\sqrt{2})$ (usually scores 6/8)
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cfc4afbd-3353-4f9f-b954-cb5178ebcf6c-06_624_872_210_543}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with parametric equations
$$x = \ln ( \sec \theta + \tan \theta ) - \sin \theta \quad y = \cos \theta \quad 0 \leqslant \theta \leqslant \frac { \pi } { 4 }$$
The curve $C$ is rotated through $2 \pi$ radians about the $x$-axis and is used to form a solid of revolution $S$.
Using calculus, show that the total surface area of $S$ is given by
$$\frac { \pi } { 2 } ( p + q \sqrt { 2 } )$$
where $p$ and $q$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F3 2022 Q2 [8]}}