C4 June 2006 Q3
3.
The curve with equation \(y = 3\sin\dfrac{x}{2}\), \(0 \leqslant x \leqslant 2\pi\), is shown in Figure 1. The finite region enclosed by the curve and the \(x\)-axis is shaded.
This region is rotated through \(2\pi\) radians about the \(x\)-axis.
| Scheme | Marks |
|---|---|
| Area Shaded \(= \displaystyle\int_0^{2\pi} 3\sin\left(\tfrac{x}{2}\right)\mathrm{d}x\) | |
| \(= \left[\dfrac{-3\cos\left(\frac{x}{2}\right)}{\frac{1}{2}}\right]_0^{2\pi}\) | M1 |
| \(= \left[-6\cos\left(\tfrac{x}{2}\right)\right]_0^{2\pi}\) | A1 oe. |
| \(= [-6(-1)] - [-6(1)] = 6 + 6 = \underline{12}\) | A1 cao |
| (3) |
Notes
M1 Integrating \(3\sin\left(\frac{x}{2}\right)\) to give \(k\cos\left(\frac{x}{2}\right)\) with \(k \neq 1\). Ignore limits.
A1 \(-6\cos\left(\frac{x}{2}\right)\) or \(\frac{-3}{\frac{1}{2}}\cos\left(\frac{x}{2}\right)\)
A1 \(\underline{12}\)
(Answer of 12 with no working scores M0A0A0.)
Beware: Owing to the symmetry of the curve between \(x = 0\) and \(x = 2\pi\) candidates can find: Area \(= 2\displaystyle\int_0^{\pi} 3\sin\left(\tfrac{x}{2}\right)\mathrm{d}x\) in part (a).
| Scheme | Marks |
|---|---|
| Volume \(= \pi\displaystyle\int_0^{2\pi} \left(3\sin\left(\tfrac{x}{2}\right)\right)^2\mathrm{d}x = 9\pi\int_0^{2\pi} \sin^2\left(\tfrac{x}{2}\right)\mathrm{d}x\) | M1 |
| \(\left[\text{NB: } \underline{\cos 2x = \pm 1 \pm 2\sin^2 x} \text{ gives } \sin^2 x = \tfrac{1 - \cos 2x}{2}\right]\) \(\left[\text{NB: } \underline{\cos x = \pm 1 \pm 2\sin^2\left(\tfrac{x}{2}\right)} \text{ gives } \sin^2\left(\tfrac{x}{2}\right) = \tfrac{1 - \cos x}{2}\right]\) | M1* |
| \(\therefore\) Volume \(= 9(\pi)\displaystyle\int_0^{2\pi} \left(\frac{1 - \cos x}{2}\right)\mathrm{d}x\) | A1 |
| \(= \dfrac{9(\pi)}{2}\displaystyle\int_0^{2\pi} (1 - \cos x)\,\mathrm{d}x\) | |
| \(= \dfrac{9(\pi)}{2}\left[x - \sin x\right]_0^{2\pi}\) | depM1*; A1 |
| \(= \dfrac{9\pi}{2}\left[(2\pi - 0) - (0 - 0)\right]\) | |
| \(= \dfrac{9\pi}{2}(2\pi) = \underline{9\pi^2}\) or \(\underline{88.8264\ldots}\) | A1 cso |
| (6) | |
| (9 marks) |
Notes
M1 Use of \(\underline{V = \pi\int y^2\,\mathrm{d}x}\). Can be implied. Ignore limits.
M1* Consideration of the Half Angle Formula for \(\sin^2\left(\frac{x}{2}\right)\) or the Double Angle Formula for \(\sin^2 x\)
A1 Correct expression for Volume. Ignore limits and \(\pi\).
depM1* Integrating to give \(\pm ax \pm b\sin x\);
A1 Correct integration \(\underline{k - k\cos x \to kx - k\sin x}\)
A1 cso Use of limits to give either \(9\pi^2\) or awrt 88.8. Solution must be completely correct. No flukes allowed.
Note: \(\pi\) is not needed for the middle four marks of question 3(b).
Beware: Owing to the symmetry of the curve between \(x = 0\) and \(x = 2\pi\) candidates can find: Volume \(= 2\pi\displaystyle\int_0^{\pi} \left(3\sin\left(\tfrac{x}{2}\right)\right)^2\mathrm{d}x\)