Edexcel C4 — Question 4 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeVolume with trigonometric functions
DifficultyChallenging +1.2 This is a volume of revolution problem requiring integration of (2sin x + cosec x)² from π/6 to π/2. While it involves 8 marks and requires expanding the square, using trigonometric identities (double angle formulas), and careful integration of multiple terms, the techniques are all standard C4 content. The 'show that' format provides a target to work towards, reducing problem-solving demand. The algebraic manipulation is moderately involved but follows predictable patterns for this topic.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^24.08d Volumes of revolution: about x and y axes

\includegraphics{figure_1} Figure 1 shows the curve with equation \(y = 2\sin x + \cosec x\), \(0 < x < \pi\). The shaded region bounded by the curve, the \(x\)-axis and the lines \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\) is rotated through \(360°\) about the \(x\)-axis. Show that the volume of the solid formed is \(\frac{1}{2}\pi(4\pi + 3\sqrt{3})\). [8]

AnswerMarks Guidance
\(= \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (2\sin x + \cosec x)^2 \, dx\)M1
\(= \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (4\sin^2 x + 4 + \cosec^2 x) \, dx\)A1
\(= \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (2 - 2\cos 2x + 4 + \cosec^2 x) \, dx\)M1
\(= \pi[6x - \sin 2x - \cot x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\)M1 A2
\(= \pi\left[(3\pi + 0) - \left(\pi - \frac{\sqrt{3}}{2} - \sqrt{3}\right)\right]\)M1
\(= \pi\left(2\pi + \frac{3\sqrt{3}}{2}\right) = \frac{1}{2}\pi(4\pi + 3\sqrt{3})\)A1 (8)
$= \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (2\sin x + \cosec x)^2 \, dx$ | M1 |
$= \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (4\sin^2 x + 4 + \cosec^2 x) \, dx$ | A1 |
$= \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (2 - 2\cos 2x + 4 + \cosec^2 x) \, dx$ | M1 |
$= \pi[6x - \sin 2x - \cot x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$ | M1 A2 |
$= \pi\left[(3\pi + 0) - \left(\pi - \frac{\sqrt{3}}{2} - \sqrt{3}\right)\right]$ | M1 |
$= \pi\left(2\pi + \frac{3\sqrt{3}}{2}\right) = \frac{1}{2}\pi(4\pi + 3\sqrt{3})$ | A1 | (8)
\includegraphics{figure_1}

Figure 1 shows the curve with equation $y = 2\sin x + \cosec x$, $0 < x < \pi$.

The shaded region bounded by the curve, the $x$-axis and the lines $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$ is rotated through $360°$ about the $x$-axis.

Show that the volume of the solid formed is $\frac{1}{2}\pi(4\pi + 3\sqrt{3})$. [8]

\hfill \mbox{\textit{Edexcel C4  Q4 [8]}}