| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about y-axis, standard curve |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring students to rearrange y = 1 + x² to get x² in terms of y, then apply the standard formula V = π∫x²dy with limits y = 1 to y = 2. The integration is simple (integrating y - 1) and requires no special techniques. Slightly above average difficulty only because rotation about the y-axis requires the rearrangement step, but this is a standard C4 exercise. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 1 + x^2 \Rightarrow x^2 = y - 1\) | B1 | |
| \(V = \int_1^2 \pi(y-1)dy\) | M1 | |
| \(= \pi\left[\frac{1}{2}y^2 - y\right]_1^2\) | B1 | \(\left[\frac{1}{2}y^2 - y\right]\) |
| \(= \pi(2 - 2 - \frac{1}{2} + 1)\) | M1 | Substituting limits into integrand |
| \(= \frac{1}{2}\pi\) | A1 | |
| [5] |
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 1 + x^2 \Rightarrow x^2 = y - 1$ | B1 | |
| $V = \int_1^2 \pi(y-1)dy$ | M1 | |
| $= \pi\left[\frac{1}{2}y^2 - y\right]_1^2$ | B1 | $\left[\frac{1}{2}y^2 - y\right]$ |
| $= \pi(2 - 2 - \frac{1}{2} + 1)$ | M1 | Substituting limits into integrand |
| $= \frac{1}{2}\pi$ | A1 | |
| **[5]** | | |
---2 Fig. 3 shows part of the curve $y = 1 + x ^ { 2 }$, together with the line $y = 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-2_558_716_302_687}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
The region enclosed by the curve, the $y$-axis and the line $y = 2$ is rotated through $360 ^ { \circ }$ about the $y$-axis. Find the volume of the solid generated, giving your answer in terms of $\pi$.
\hfill \mbox{\textit{OCR MEI C4 Q2 [5]}}