| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Applied context: real-world solid |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring standard application of the formula V = π∫y² dx with a simple rational function, followed by a similarity scaling calculation. The integration is routine (substitution or recognition of standard form), and part (b) is a direct application of the volume scale factor k³. Slightly easier than average due to the simple function and standard technique. |
| Spec | 4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((V)=\pi\int_{-1}^{\frac{2}{3}}\frac{4}{(4+3x)^2}\,dx\) | B1 | Need correct statement with \(\pi\), correct limits, and \(dx\); allow \(\left(\frac{2}{4+3x}\right)^2\) form |
| \((\pi)\int\frac{4}{(4+3x)^2}\,dx = (\pi)\left(-\frac{4}{3}(4+3x)^{-1}\right)\) | M1A1 | M1: substitution or reverse chain rule achieving \(k(4+3x)^{-1}\); A1: \(-\frac{4}{3}(4+3x)^{-1}\); no need for \(\pi\) or limits |
| \(=(\pi)\left[-\frac{4}{3}(4+3x)^{-1}\right]_{-1}^{\frac{2}{3}} = (\pi)\left[-\frac{4}{3}(4+2)^{-1} - -\frac{4}{3}(4-3)^{-1}\right]\) | M1 | Substitutes correct limits into changed/integrated function and subtracts |
| \(=\frac{10}{9}\pi\) | A1 | Accept \(1.1\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Length scale factor \(9\), so volume scale factor \(9^3\); Volume \(= 9^3 \times \frac{10}{9}\pi = 810\pi\) or \(2545\text{ cm}^3\) | M1 A1 | M1: attempts to multiply answer in (a) by \(729\); may be implied by \(\frac{10}{9}\pi\to 810\); A1: accept awrt \(2540\), \(2550\), or \(810\pi\) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(V)=\pi\int_{-1}^{\frac{2}{3}}\frac{4}{(4+3x)^2}\,dx$ | B1 | Need correct statement with $\pi$, correct limits, and $dx$; allow $\left(\frac{2}{4+3x}\right)^2$ form |
| $(\pi)\int\frac{4}{(4+3x)^2}\,dx = (\pi)\left(-\frac{4}{3}(4+3x)^{-1}\right)$ | M1A1 | M1: substitution or reverse chain rule achieving $k(4+3x)^{-1}$; A1: $-\frac{4}{3}(4+3x)^{-1}$; no need for $\pi$ or limits |
| $=(\pi)\left[-\frac{4}{3}(4+3x)^{-1}\right]_{-1}^{\frac{2}{3}} = (\pi)\left[-\frac{4}{3}(4+2)^{-1} - -\frac{4}{3}(4-3)^{-1}\right]$ | M1 | Substitutes correct limits into changed/integrated function and subtracts |
| $=\frac{10}{9}\pi$ | A1 | Accept $1.1\pi$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Length scale factor $9$, so volume scale factor $9^3$; Volume $= 9^3 \times \frac{10}{9}\pi = 810\pi$ or $2545\text{ cm}^3$ | M1 A1 | M1: attempts to multiply answer in (a) by $729$; may be implied by $\frac{10}{9}\pi\to 810$; A1: accept awrt $2540$, $2550$, or $810\pi$ |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-05_504_844_255_543}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$ with equation $y = \frac { 2 } { ( 4 + 3 x ) } , x > - \frac { 4 } { 3 }$ is shown in Figure 1\\
The region bounded by the curve, the $x$-axis and the lines $x = - 1$ and $x = \frac { 2 } { 3 }$, is shown shaded in Figure 1
This region is rotated through 360 degrees about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the exact value of the volume of the solid generated.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-05_583_433_1398_753}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a candle with axis of symmetry $A B$ where $A B = 15 \mathrm {~cm}$. $A$ is a point at the centre of the top surface of the candle and $B$ is a point at the centre of the base of the candle. The candle is geometrically similar to the solid generated in part (a).
\item Find the volume of this candle.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q4 [7]}}