| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about y-axis, region between two curves |
| Difficulty | Challenging +1.2 This is a volumes of revolution question requiring rotation about the y-axis with two curves, but the hint about the cone formula significantly reduces the difficulty. Students must express x in terms of y for both curves, set up the integral V = π∫(x_line² - x_curve²)dy from y=0 to y=2, and recognize that the cone formula handles the linear part. While multi-step, this is a fairly standard C4 question with helpful scaffolding, making it moderately above average difficulty. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vol = vol of rev of curve + vol of rev of line | M1 | (soi) at any stage |
| vol of rev of curve \(= \int_0^2 \pi x^2\,dy\) | ||
| \(= \int_0^2 \pi\frac{y}{2}\,dy\) | M1 | substituting \(x^2 = y/2\); for M1 need \(\pi\), substitution for \(x^2\), (\(dy\) soi), intention to integrate and correct limits |
| \(= \pi\left[\frac{y^2}{4}\right]_0^2\) | B1 | \(\left[\frac{y^2}{4}\right]\) even if \(\pi\) missing or limits incorrect or missing |
| \(= \pi\) | A1 | cao |
| height of cone \(= 3 - 2 = 1\) | B1 | h=1 soi; OR \(\pi\int_2^3(3-y)^2\,dy\) M1 (even if expanded incorrectly) |
| so vol of cone \(= \frac{1}{3}\pi\ 1^2 \times 1 = \pi/3\) | B1 | \(= \pi/3\) A1 www |
| so total vol \(= 4\pi/3\) | A1 | www cao |
| [7] |
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vol = vol of rev of curve + vol of rev of line | M1 | (soi) at any stage |
| vol of rev of curve $= \int_0^2 \pi x^2\,dy$ | | |
| $= \int_0^2 \pi\frac{y}{2}\,dy$ | M1 | substituting $x^2 = y/2$; for M1 need $\pi$, substitution for $x^2$, ($dy$ soi), intention to integrate and correct limits |
| $= \pi\left[\frac{y^2}{4}\right]_0^2$ | B1 | $\left[\frac{y^2}{4}\right]$ even if $\pi$ missing or limits incorrect or missing |
| $= \pi$ | A1 | cao |
| height of cone $= 3 - 2 = 1$ | B1 | h=1 soi; OR $\pi\int_2^3(3-y)^2\,dy$ M1 (even if expanded incorrectly) |
| so vol of cone $= \frac{1}{3}\pi\ 1^2 \times 1 = \pi/3$ | B1 | $= \pi/3$ A1 www |
| so total vol $= 4\pi/3$ | A1 | www cao |
| **[7]** | | |
---3 Fig. 6 shows the region enclosed by part of the curve $y = 2 x ^ { 2 }$, the straight line $x + y = 3$, and the $y$-axis. The curve and the straight line meet at $\mathrm { P } ( 1,2 )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-3_640_923_399_613}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
The shaded region is rotated through $360 ^ { \circ }$ about the $y$-axis. Find, in terms of $\pi$, the volume of the solid of revolution formed.\\[0pt]
[You may use the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$ for the volume of a cone.]
\hfill \mbox{\textit{OCR MEI C4 Q3 [7]}}