| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate volume |
| Difficulty | Standard +0.8 This is a multi-step parametric volume of revolution question requiring: (1) deriving the volume integral formula using dx/dt, (2) finding the correct limits by solving x = 3/2 = 3sin(t), and (3) evaluating a trigonometric integral using substitution. While the techniques are standard C4 content, the combination of parametric equations, volume of revolution, and trigonometric integration with careful limit handling makes this moderately challenging but still within typical A-level scope. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \int y^2\,dx = \int y^2\frac{dx}{dt}\,dt = \int(2\sin 2t)^2 3\cos t\,dt\) | M1A1 | M1: Attempts \(\int y^2\,dx = \int y^2\frac{dx}{dt}\,dt\) where \(\frac{dx}{dt} = \pm k\cos t\). A1: \(= \int(2\sin 2t)^2 3\cos t\,(dt)\) (\(dt\) can be missing as long as M is scored) |
| \(= \int(4\sin t\cos t)^2\cdot 3\cos t\,dt\) | M1 | Uses \(\sin 2t = 2\sin t\cos t\) |
| \(x = \frac{3}{2} \Rightarrow t = \frac{\pi}{6}\) or \(k = 48\) | B1 | Correct value for \(a\) (must be exact) or correct value for \(k\) |
| \(V = \int \pi y^2\,dx = 48\pi\int_0^{\pi/6}\sin^2 t\cos^3 t\,dt\)* | A1* | Achieves printed answer including "\(dt\)" with correct limits and \(48\pi\) in place with no errors. Or achieves printed answer with letters \(a\) and \(k\) and states correct values of \(a\) and \(k\). (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = \sin t \Rightarrow \frac{du}{dt} = \cos t\) | B1 | States \(\frac{du}{dt} = \cos t\) or equivalent. May be implied. |
| \(V = k\int\sin^2 t\cos^3 t\,dt = k\int u^2\cos^2 t\,du = k\int u^2(1-\sin^2 t)\,du = k\int u^2(1-u^2)\,du\) | M1A1ft | M1: Substitutes fully including for \(dt\) using \(u=\sin t\) and \(\cos^2 t = \pm 1\pm\sin^2 t\). A1ft: Fully correct integral in terms of \(u\), follow through on incorrect \(k\)'s |
| \(= k\left[\frac{u^3}{3} - \frac{u^5}{5}\right]\) | M1 | Multiplies out to form polynomial in \(u\) and integrates with \(u^n \to u^{n+1}\) for at least one power |
| Volume \(= 48\pi\left[\frac{u^3}{3}-\frac{u^5}{5}\right]_0^{\frac{1}{2}} = \frac{17\pi}{10}\) | dM1A1 | dM1: All methods must have been scored. Uses limits 0 and \(\frac{1}{2}\), or limits 0 and \(\frac{\pi}{6}\) if returning to \(\sin t\). A1: \(V = \frac{17\pi}{10}\) oe such as \(\frac{51\pi}{30}\) (6) (11 marks) |
| If \(\frac{du}{dt} = -\cos t\) is used, maximum B0M1A0M1M1A0 is possible |
# Question 12(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \int y^2\,dx = \int y^2\frac{dx}{dt}\,dt = \int(2\sin 2t)^2 3\cos t\,dt$ | M1A1 | M1: Attempts $\int y^2\,dx = \int y^2\frac{dx}{dt}\,dt$ where $\frac{dx}{dt} = \pm k\cos t$. A1: $= \int(2\sin 2t)^2 3\cos t\,(dt)$ ($dt$ can be missing as long as M is scored) |
| $= \int(4\sin t\cos t)^2\cdot 3\cos t\,dt$ | M1 | Uses $\sin 2t = 2\sin t\cos t$ |
| $x = \frac{3}{2} \Rightarrow t = \frac{\pi}{6}$ or $k = 48$ | B1 | Correct value for $a$ (must be exact) or correct value for $k$ |
| $V = \int \pi y^2\,dx = 48\pi\int_0^{\pi/6}\sin^2 t\cos^3 t\,dt$* | A1* | Achieves printed answer including "$dt$" with correct limits and $48\pi$ in place with no errors. Or achieves printed answer with letters $a$ and $k$ and states correct values of $a$ and $k$. **(5)** |
# Question 12(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \sin t \Rightarrow \frac{du}{dt} = \cos t$ | B1 | States $\frac{du}{dt} = \cos t$ or equivalent. May be implied. |
| $V = k\int\sin^2 t\cos^3 t\,dt = k\int u^2\cos^2 t\,du = k\int u^2(1-\sin^2 t)\,du = k\int u^2(1-u^2)\,du$ | M1A1ft | M1: Substitutes **fully** including for $dt$ using $u=\sin t$ and $\cos^2 t = \pm 1\pm\sin^2 t$. A1ft: Fully correct integral in terms of $u$, follow through on incorrect $k$'s |
| $= k\left[\frac{u^3}{3} - \frac{u^5}{5}\right]$ | M1 | Multiplies out to form polynomial in $u$ and integrates with $u^n \to u^{n+1}$ for at least one power |
| Volume $= 48\pi\left[\frac{u^3}{3}-\frac{u^5}{5}\right]_0^{\frac{1}{2}} = \frac{17\pi}{10}$ | dM1A1 | dM1: All methods must have been scored. Uses limits 0 and $\frac{1}{2}$, or limits 0 and $\frac{\pi}{6}$ if returning to $\sin t$. A1: $V = \frac{17\pi}{10}$ oe such as $\frac{51\pi}{30}$ **(6) (11 marks)** |
| If $\frac{du}{dt} = -\cos t$ is used, maximum B0M1A0M1M1A0 is possible | | |
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d67f716-c8af-4460-8a6b-62073ba9b825-23_503_1333_267_301}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve with parametric equations
$$x = 3 \sin t , \quad y = 2 \sin 2 t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
The finite region $S$, shown shaded in Figure 3, is bounded by the curve, the $x$-axis and the line with equation $x = \frac { 3 } { 2 }$
The shaded region $S$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume of the solid of revolution is given by
$$k \int _ { 0 } ^ { a } \sin ^ { 2 } t \cos ^ { 3 } t \mathrm {~d} t$$
where $k$ and $a$ are constants to be given in terms of $\pi$.
\item Use the substitution $u = \sin t$, or otherwise, to find the exact value of this volume, giving your answer in the form $\frac { p \pi } { q }$ where $p$ and $q$ are integers. (Solutions based entirely on graphical or numerical methods are not acceptable.)
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q12 [11]}}