OCR MEI C4 2016 June — Question 3 4 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2016
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about y-axis, standard curve
DifficultyModerate -0.3 This is a straightforward volumes of revolution question requiring students to set up and evaluate a standard integral about the y-axis. The main steps are expressing x in terms of y (x = y^(1/4)), applying the formula V = π∫x²dy with limits 0 to 4, and integrating y^(1/2). While it requires careful algebraic manipulation and knowledge of the rotation formula, it's a routine textbook exercise with no conceptual surprises, making it slightly easier than average.
Spec4.08d Volumes of revolution: about x and y axes

3 Fig. 3 shows the curve \(y = x ^ { 4 }\) and the line \(y = 4\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8b807b2e-777b-4c9a-b3dd-890d21d33174-2_509_510_778_774} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} The finite region enclosed by the curve and the line is rotated through \(180 ^ { \circ }\) about the \(y\)-axis. Find the exact volume of revolution generated.

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(V = \pi\int_0^4 x^2 \, \mathrm{d}y\)M1 M1 for \(k(\pi)\int_0^4 x^2(\mathrm{d}y)\) with correct limits and \(k=1\) or \(\frac{1}{2}\); condone lack of \(\pi\) for M mark and \(\mathrm{d}y\) throughout
\(V = \pi\int_0^4 y^{\frac{1}{2}} \, \mathrm{d}y\)A1 Correct (or with a \(1/2\)) – limits may be seen or implied through later working
\(\frac{2}{3}y^{\frac{3}{2}}\)B1 \(\frac{2}{3}y^{\frac{3}{2}}\) or \(\frac{1}{3}y^{\frac{3}{2}}\) (but only if \(k=\frac{1}{2}\)), condone \(\frac{y^{1.5}}{1.5}\) (oe)
\(= \pi\left[\frac{2}{3}y^{\frac{3}{2}}\right]_0^4 = \frac{16\pi}{3}\)A1 Exact – mark final answer (so no isw if correct answer subsequently halved); but if exact value seen then followed by 16.755… then isw
# Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \pi\int_0^4 x^2 \, \mathrm{d}y$ | M1 | M1 for $k(\pi)\int_0^4 x^2(\mathrm{d}y)$ with correct limits and $k=1$ or $\frac{1}{2}$; condone lack of $\pi$ for M mark and $\mathrm{d}y$ throughout |
| $V = \pi\int_0^4 y^{\frac{1}{2}} \, \mathrm{d}y$ | A1 | Correct (or with a $1/2$) – limits may be seen or implied through later working |
| $\frac{2}{3}y^{\frac{3}{2}}$ | B1 | $\frac{2}{3}y^{\frac{3}{2}}$ or $\frac{1}{3}y^{\frac{3}{2}}$ (but only if $k=\frac{1}{2}$), condone $\frac{y^{1.5}}{1.5}$ (oe) |
| $= \pi\left[\frac{2}{3}y^{\frac{3}{2}}\right]_0^4 = \frac{16\pi}{3}$ | A1 | Exact – mark **final** answer (so no isw if correct answer subsequently halved); but if exact value seen then followed by 16.755… then isw |

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3 Fig. 3 shows the curve $y = x ^ { 4 }$ and the line $y = 4$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8b807b2e-777b-4c9a-b3dd-890d21d33174-2_509_510_778_774}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

The finite region enclosed by the curve and the line is rotated through $180 ^ { \circ }$ about the $y$-axis. Find the exact volume of revolution generated.

\hfill \mbox{\textit{OCR MEI C4 2016 Q3 [4]}}