Standard +0.8 This question requires finding the volume of revolution for a region bounded by two curves (sin x and cos x) and the x-axis. Students must identify the intersection point (x = π/4), set up two separate integrals (one for each curve dominating different intervals), apply the formula V = π∫y² dx correctly to each part, and evaluate trigonometric integrals using double angle identities. The multi-step nature, need to split the region, and integration of sin²x and cos²x make this moderately challenging but still within standard C4 scope.
6
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The diagram shows the curves \(y = \cos x\) and \(y = \sin x\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The region \(R\) is bounded by the curves and the \(x\)-axis. Find the volume of the solid of revolution formed when \(R\) is rotated completely about the \(x\)-axis, giving your answer in terms of \(\pi\).
State or imply that graphs cross at \(x = \frac{1}{4}\pi\)
B1
Limits on integrals may clarify. Be lenient here
\(\pi\int y^2\,dx\) used with either \(y = \sin x\) or \(y = \cos x\)
*M1
The \(\pi\) element(s) may not appear until later in working
\(\pi\int_0^{\frac{1}{4}\pi}\sin^2 x\,(dx) + \pi\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\cos^2 x\,(dx)\) or \(2\pi\int_0^{\frac{1}{4}\pi}\sin^2 x\,(dx)\)
A1
Changing \(\sin^2 x\) or \(\cos^2 x\) into \(f(\cos 2x)\)
dep*M1
\(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\) or \(\cos^2 x = \frac{1}{2}(1+\cos 2x)\)
A1
\(\int\cos 2x\,(dx) = \frac{1}{2}\sin 2x\) anywhere in this part
B1
\(\frac{1}{4}\pi^2 - \frac{1}{2}\pi\)
A1 ISW
[7]
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply that graphs cross at $x = \frac{1}{4}\pi$ | B1 | Limits on integrals may clarify. Be lenient here |
| $\pi\int y^2\,dx$ used with either $y = \sin x$ or $y = \cos x$ | *M1 | The $\pi$ element(s) may not appear until later in working |
| $\pi\int_0^{\frac{1}{4}\pi}\sin^2 x\,(dx) + \pi\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\cos^2 x\,(dx)$ or $2\pi\int_0^{\frac{1}{4}\pi}\sin^2 x\,(dx)$ | A1 | |
| Changing $\sin^2 x$ or $\cos^2 x$ into $f(\cos 2x)$ | dep*M1 | |
| $\sin^2 x = \frac{1}{2}(1 - \cos 2x)$ or $\cos^2 x = \frac{1}{2}(1+\cos 2x)$ | A1 | |
| $\int\cos 2x\,(dx) = \frac{1}{2}\sin 2x$ anywhere in this part | B1 | |
| $\frac{1}{4}\pi^2 - \frac{1}{2}\pi$ | A1 ISW | |
| **[7]** | | |
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6\\
\includegraphics[max width=\textwidth, alt={}, center]{cf154c94-6248-4dda-91e8-61349cc10482-3_606_846_251_614}
The diagram shows the curves $y = \cos x$ and $y = \sin x$, for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$. The region $R$ is bounded by the curves and the $x$-axis. Find the volume of the solid of revolution formed when $R$ is rotated completely about the $x$-axis, giving your answer in terms of $\pi$.
\hfill \mbox{\textit{OCR C4 2012 Q6 [7]}}