| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| 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 standard C4 volumes of revolution question with parametric equations. Part (a) is routine substitution, part (b) requires applying the standard formula with dx/dt (shown, not derived), and part (c) uses a given substitution to evaluate a trigonometric integral. While it requires multiple techniques, all steps follow predictable patterns for this topic with no novel insight needed, making it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \((a)\) \(x = 0 \Rightarrow t = 0\) at \(O\) | B1 | |
| \(y = 0 \Rightarrow t = 0\) (at \(O\)) or \(\frac{2}{3}; \therefore t = \frac{2}{3}\) at \(A\) | B1 | |
| \((b)\) \(=\) volume when region above \(x\)-axis is rotated through \(2\pi\) | ||
| \(\frac{dx}{dt} = 3\cos t\) | M1 | |
| \(\therefore\) volume \(= \pi \int_0^{2/3} (2\sin 2t)^2 \times 3\cos t \, dt = \int_0^{2/3} 12\pi \sin^2 2t \cos t \, dt\) | M1 A1 | |
| \((c)\) \(t = 0 \Rightarrow u = 0, t = \frac{2}{3} \Rightarrow u = 1, \frac{du}{dt} = \cos t\) | B1 | |
| \(\sin^2 2t = 4\sin^2 t \cos^2 t = 4\sin^2 t(1 - \sin^2 t)\) | M1 | |
| \(\therefore = \int_0^1 12\pi \times 4u^2(1 - u^2) \, du\) | M1 | |
| \(= 48\pi \int_0^1 (u^2 - u^4) \, du\) | A1 | |
| \(= 48\pi[\frac{1}{3}u^3 - \frac{1}{5}u^5]_0^1\) | M1 A1 | |
| \(= 48\pi(\frac{1}{3} - \frac{1}{5}) - [0)] = \frac{32}{5}\pi\) | M1 A1 | (13 marks) |
$(a)$ $x = 0 \Rightarrow t = 0$ at $O$ | B1 |
$y = 0 \Rightarrow t = 0$ (at $O$) or $\frac{2}{3}; \therefore t = \frac{2}{3}$ at $A$ | B1 |
$(b)$ $=$ volume when region above $x$-axis is rotated through $2\pi$ | |
$\frac{dx}{dt} = 3\cos t$ | M1 |
$\therefore$ volume $= \pi \int_0^{2/3} (2\sin 2t)^2 \times 3\cos t \, dt = \int_0^{2/3} 12\pi \sin^2 2t \cos t \, dt$ | M1 A1 |
$(c)$ $t = 0 \Rightarrow u = 0, t = \frac{2}{3} \Rightarrow u = 1, \frac{du}{dt} = \cos t$ | B1 |
$\sin^2 2t = 4\sin^2 t \cos^2 t = 4\sin^2 t(1 - \sin^2 t)$ | M1 |
$\therefore = \int_0^1 12\pi \times 4u^2(1 - u^2) \, du$ | M1 |
$= 48\pi \int_0^1 (u^2 - u^4) \, du$ | A1 |
$= 48\pi[\frac{1}{3}u^3 - \frac{1}{5}u^5]_0^1$ | M1 A1 |
$= 48\pi(\frac{1}{3} - \frac{1}{5}) - [0)] = \frac{32}{5}\pi$ | M1 A1 | (13 marks)6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0f2d48ab-1f61-4fb9-b35a-25d684dbd50f-10_454_602_255_479}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve with parametric equations
$$x = 3 \sin t , \quad y = 2 \sin 2 t , \quad 0 \leq t < \pi .$$
The curve meets the $x$-axis at the origin, $O$, and at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ at $O$ and the value of $t$ at $A$.
The region enclosed by the curve is rotated through $\pi$ radians about the $x$-axis.
\item Show that the volume of the solid formed is given by
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } 12 \pi \sin ^ { 2 } 2 t \cos t \mathrm {~d} t$$
\item Using the substitution $u = \sin t$, or otherwise, evaluate this integral, giving your answer as an exact multiple of $\pi$.\\
6. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [13]}}