| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with trigonometric functions |
| Difficulty | Standard +0.8 This is a multi-step volumes of revolution question requiring algebraic manipulation of trigonometric identities (sin x + cos x)² = 1 + sin 2x, followed by integration by parts twice for the x² sin 2x term. While the techniques are standard C3/C4 content, the combination of trig manipulation and repeated integration by parts makes it moderately challenging, above average difficulty but not requiring novel insight. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y^2 = (x(\sin x + \cos x))^2 = x^2(\sin^2 x + \cos^2 x + 2\sin x\cos x)\) | M1 | Squaring \(y\) AND attempting to expand bracket. Minimum: \(y^2=x^2(\sin^2 x+\cos^2 x+\ldots)\) |
| \(= x^2(1+\sin 2x)\) | A1 | Using \(\sin^2 x+\cos^2 x=1\) and \(2\sin x\cos x = \sin 2x\) |
| \(V = \int_0^{\pi/4}\pi y^2\,dx = \int_0^{\pi/4}\pi x^2(1+\sin 2x)\,dx\) | A1* | Must be stated or implied that \(V=\int_0^{\pi/4}\pi y^2\,dx\). Correct proof with limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^{\pi/4}\pi x^2\,dx = \left[\pi\frac{x^3}{3}\right]_0^{\pi/4} = \frac{\pi(\pi/4)^3}{3}\) | M1A1 | M1: splits integral and integrates \(x^2\) or \(\pi x^2\) to \(Ax^3\). A1: correct value \(\frac{(\pi/4)^3}{3}\) |
| \(\int x^2\sin 2x\,dx = \pm Bx^2\cos 2x \pm C\int x\cos 2x\,dx\) | M1 | Integration by parts on \(\int \pi x^2\sin 2x\,dx\) the correct way around |
| \(= -x^2\frac{\cos 2x}{2} + \int x\cos 2x\,dx\) | A1 | Correct first application |
| \(= \pm Bx^2\cos 2x \pm Cx\sin 2x \pm \int D\sin 2x\,dx\) | dM1 | Second application of integration by parts, correct way. Dependent on previous M1 |
| \(= -x^2\frac{\cos 2x}{2} + x\frac{\sin 2x}{2} + \frac{\cos 2x}{4}\) | A1 | Fully correct integral of \(\int x^2\sin 2x\,dx\) |
| \(\int_0^{\pi/4} x^2\sin 2x\,dx = \left(\frac{\pi}{8}-\frac{1}{4}\right)\) | ddM1 | Substitutes both limits and subtracts. Both M1s for by parts must have been scored |
| \(V = \left(\frac{\pi^4}{192}\right) + \left(\frac{\pi^2}{8}-\frac{\pi}{4}\right)\) | A1, A1 | A1: either \(\frac{\pi^4}{192}\) or \(\frac{\pi^2}{8}-\frac{\pi}{4}\). A1: fully correct. Accept \(V=\pi\left(\frac{\pi^3}{192}+\frac{\pi}{8}-\frac{1}{4}\right)\) |
## Question 12:
### Part 12(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^2 = (x(\sin x + \cos x))^2 = x^2(\sin^2 x + \cos^2 x + 2\sin x\cos x)$ | M1 | Squaring $y$ AND attempting to expand bracket. Minimum: $y^2=x^2(\sin^2 x+\cos^2 x+\ldots)$ |
| $= x^2(1+\sin 2x)$ | A1 | Using $\sin^2 x+\cos^2 x=1$ and $2\sin x\cos x = \sin 2x$ |
| $V = \int_0^{\pi/4}\pi y^2\,dx = \int_0^{\pi/4}\pi x^2(1+\sin 2x)\,dx$ | A1* | Must be stated or implied that $V=\int_0^{\pi/4}\pi y^2\,dx$. Correct proof with limits |
### Part 12(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{\pi/4}\pi x^2\,dx = \left[\pi\frac{x^3}{3}\right]_0^{\pi/4} = \frac{\pi(\pi/4)^3}{3}$ | M1A1 | M1: splits integral and integrates $x^2$ or $\pi x^2$ to $Ax^3$. A1: correct value $\frac{(\pi/4)^3}{3}$ |
| $\int x^2\sin 2x\,dx = \pm Bx^2\cos 2x \pm C\int x\cos 2x\,dx$ | M1 | Integration by parts on $\int \pi x^2\sin 2x\,dx$ the correct way around |
| $= -x^2\frac{\cos 2x}{2} + \int x\cos 2x\,dx$ | A1 | Correct first application |
| $= \pm Bx^2\cos 2x \pm Cx\sin 2x \pm \int D\sin 2x\,dx$ | dM1 | Second application of integration by parts, correct way. Dependent on previous M1 |
| $= -x^2\frac{\cos 2x}{2} + x\frac{\sin 2x}{2} + \frac{\cos 2x}{4}$ | A1 | Fully correct integral of $\int x^2\sin 2x\,dx$ |
| $\int_0^{\pi/4} x^2\sin 2x\,dx = \left(\frac{\pi}{8}-\frac{1}{4}\right)$ | ddM1 | Substitutes both limits and subtracts. Both M1s for by parts must have been scored |
| $V = \left(\frac{\pi^4}{192}\right) + \left(\frac{\pi^2}{8}-\frac{\pi}{4}\right)$ | A1, A1 | A1: either $\frac{\pi^4}{192}$ or $\frac{\pi^2}{8}-\frac{\pi}{4}$. A1: fully correct. Accept $V=\pi\left(\frac{\pi^3}{192}+\frac{\pi}{8}-\frac{1}{4}\right)$ |12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5b698944-41ac-4072-b5e1-c580b7752c39-40_695_1212_276_420}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve with equation
$$y = x ( \sin x + \cos x ) , \quad 0 \leqslant x \leqslant \frac { \pi } { 4 }$$
The finite region $R$, shown shaded in Figure 4, is bounded by the curve, the $x$-axis and the line $x = \frac { \pi } { 4 }$. This shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution, with volume $V$.
\begin{enumerate}[label=(\alph*)]
\item Assuming the formula for volume of revolution show that $V = \int _ { 0 } ^ { \frac { \pi } { 4 } } \pi x ^ { 2 } ( 1 + \sin 2 x ) \mathrm { d } x$
\item Hence using calculus find the exact value of $V$.
You must show your working.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q12 [12]}}