OCR MEI C4 — Question 1 6 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about y-axis, standard curve
DifficultyStandard +0.3 This is a straightforward volumes of revolution question requiring students to rearrange y in terms of x, set up the integral V = π∫x²dy with correct limits (found by substituting y=2), and integrate a simple polynomial. While it involves multiple steps, each is routine for C4 level with no conceptual challenges beyond standard technique application.
Spec4.08d Volumes of revolution: about x and y axes
1 Fig. 6 shows the region enclosed by the curve \(y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }\) and the line \(y = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-1_426_664_277_714} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\).
Question 1:
\(y = (1 + 2x^2)^{1/3}\) \(\Rightarrow\) \(y^3 = 1 + 2x^2\) \(\Rightarrow\) \(x^2 = \frac{1}{2}(y^3 - 1)\)
\(V = \pi \int_1^2 x^2 \, dy = \pi \int_1^2 (y^3 - 1) \, dy\)
\(= \pi \left[\frac{1}{2} \cdot \frac{y^4}{4} - y\right]_1^2\)
\(= \pi\left(2 - \frac{1}{4}\right) = \frac{11\pi}{8}\)
M1: Finding \(x^2\) (or \(x\)) correctly in terms of \(y\)
M1: For M1 need \(\pi x^2 \, dy\) with substitution for their \(x^2\) (in terms of \(y\) only). Condone absence of \(dy\) throughout if intentions clear (need \(\pi\)).
A1: For A1 it must be correct with correct limits 1 and 2, but they may appear later.
B1: \(\frac{1}{2}\left[\frac{y^4}{4} - y\right]\) independent of \(\pi\) and limits
M1: Substituting both their limits in correct order in correct expression. Condone a minor slip for M1. (If using \(y = 0\) as lower limit then '\(-0\)' is enough.) Condone absence of \(\pi\) for M1.
A1: oe exact only (e.g. \(1\frac{3}{8}\pi\) or \(1.375\pi\))
[6]
Question 1:

$y = (1 + 2x^2)^{1/3}$ $\Rightarrow$ $y^3 = 1 + 2x^2$ $\Rightarrow$ $x^2 = \frac{1}{2}(y^3 - 1)$

$V = \pi \int_1^2 x^2 \, dy = \pi \int_1^2 (y^3 - 1) \, dy$

$= \pi \left[\frac{1}{2} \cdot \frac{y^4}{4} - y\right]_1^2$

$= \pi\left(2 - \frac{1}{4}\right) = \frac{11\pi}{8}$

M1: Finding $x^2$ (or $x$) correctly in terms of $y$

M1: For M1 need $\pi x^2 \, dy$ with substitution for their $x^2$ (in terms of $y$ only). Condone absence of $dy$ throughout if intentions clear (need $\pi$).

A1: For A1 it must be correct with correct limits 1 and 2, but they may appear later.

B1: $\frac{1}{2}\left[\frac{y^4}{4} - y\right]$ independent of $\pi$ and limits

M1: Substituting both their limits in correct order in correct expression. Condone a minor slip for M1. (If using $y = 0$ as lower limit then '$-0$' is enough.) Condone absence of $\pi$ for M1.

A1: oe exact only (e.g. $1\frac{3}{8}\pi$ or $1.375\pi$)

[6]
1 Fig. 6 shows the region enclosed by the curve $y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }$ and the line $y = 2$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-1_426_664_277_714}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}

This region is rotated about the $y$-axis. Find the volume of revolution formed, giving your answer as a multiple of $\pi$.

\hfill \mbox{\textit{OCR MEI C4  Q1 [6]}}
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