| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about y-axis, region between two curves |
| Difficulty | Standard +0.3 This is a straightforward integration question requiring standard techniques: finding k from a given area using ∫(k/x)dx = k ln x, then computing a volume of revolution. The multi-step nature and volume calculation add slight complexity, but both parts follow routine C3 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Either: Integrate to obtain \(k \ln x\) | B1 | — |
| Use at least one relevant logarithm property | M1 | — |
| Obtain \(k \ln 3 = \ln 81\) and hence \(k = 4\) | A1 3 AG | accurate work required |
| Answer | Marks | Guidance |
|---|---|---|
| Integrate to obtain \(k \ln x\) | B1 | — |
| Obtain correct explicit expression for \(k\) and conclude \(k = 4\) with no error seen | B2 3 AG | e.g. \(k = \frac{\ln 81}{\ln 6 - \ln 2} = 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Integrate to obtain \(4 \ln x\) | B1 | — |
| Use at least one relevant logarithm property | M1 | — |
| Obtain \(\ln 81\) legitimately with no error seen | A1 3 AG | accurate work required |
| Answer | Marks | Guidance |
|---|---|---|
| State volume involves \(\int \pi \left(\frac{\pi}{x}\right)^2 dx\) | B1 | possibly implied |
| Obtain integral of form \(k_1 x^{-1}\) | M1 | any constant \(k_1\) including \(\pi\) or not |
| Use correct process for finding volume produced from \(S\) | M1 | \(\int (k_2 2^x - k_3 y^x) dx\), including \(\pi\) or not with correct limits indicated; or equiv |
| Obtain \(16\pi - \frac{16}{3}\pi\) and hence \(\frac{32}{3}\pi\) | A1 4 | or exact equiv |
## (i)
**Either:** Integrate to obtain $k \ln x$ | B1 | —
Use at least one relevant logarithm property | M1 | —
Obtain $k \ln 3 = \ln 81$ and hence $k = 4$ | A1 3 AG | accurate work required
**Or 1:** (where solution involves no use of a logarithm property)
Integrate to obtain $k \ln x$ | B1 | —
Obtain correct explicit expression for $k$ and conclude $k = 4$ with no error seen | B2 3 AG | e.g. $k = \frac{\ln 81}{\ln 6 - \ln 2} = 4$
**Or 2:** (where solution involves verification of result by initial substitution of 4 for $k$)
Integrate to obtain $4 \ln x$ | B1 | —
Use at least one relevant logarithm property | M1 | —
Obtain $\ln 81$ legitimately with no error seen | A1 3 AG | accurate work required
## (ii)
State volume involves $\int \pi \left(\frac{\pi}{x}\right)^2 dx$ | B1 | possibly implied
Obtain integral of form $k_1 x^{-1}$ | M1 | any constant $k_1$ including $\pi$ or not
Use correct process for finding volume produced from $S$ | M1 | $\int (k_2 2^x - k_3 y^x) dx$, including $\pi$ or not with correct limits indicated; or equiv
Obtain $16\pi - \frac{16}{3}\pi$ and hence $\frac{32}{3}\pi$ | A1 4 | or exact equiv
---\includegraphics{figure_4}
The diagram shows part of the curve $y = \frac{k}{x}$, where $k$ is a positive constant. The points A and B on the curve have $x$-coordinates 2 and 6 respectively. Lines through A and B parallel to the axes as shown meet at the point C. The region R is bounded by the curve and the lines $x = 2$, $x = 6$ and $y = 0$. The region S is bounded by the curve and the lines AC and BC. It is given that the area of the region R is $\ln 81$.
\begin{enumerate}[label=(\roman*)]
\item Show that $k = 4$. [3]
\item Find the exact volume of the solid produced when the region S is rotated completely about the $x$-axis. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2010 Q4 [7]}}