**(a)**

$\text{Vol} = \pi \int_0^{\pi/2} y^2 dx = \pi \int_0^{\pi/2} \cos^2 x \, dx$ | M1 |

$= \pi \int_0^{\pi/2} \frac{1}{2}(\cos 2x + 1) dx$ | M1 |

$= \pi \left[\frac{1}{2}\sin 2x + x\right]_0^{\pi/2} = \frac{\pi^2}{4}$ | DM1A1 | (4)

**(b)**

$\pi \int_0^{\pi/2} y^2 x \, dx = \pi \int_0^{\pi/2} x \cos^2 x \, dx$ | M1 |

$= \pi \int_0^{\pi/2} \frac{1}{2}x(\cos 2x + 1) dx$ | M1 |

$= \frac{\pi}{2} \int_0^{\pi/2} x \cos 2x \, dx + \frac{\pi}{2}\left[\frac{x^2}{2}\right]_0^{\pi/2}$ |  |

$\frac{\pi}{2}\left[x \cdot \frac{1}{2}\sin 2x\right]_0^{\pi/2} - \frac{\pi}{2}\int_0^{\pi/2} \frac{1}{2}\sin 2x \, dx + \frac{\pi^3}{16}$ | M1,B1 |

$= 0 + \frac{\pi}{2}\left[\frac{1}{4}\cos 2x\right]_0^{\pi/2} + \frac{\pi^3}{16}$ | DM1 |

$= \frac{\pi}{8}[-1-1] + \frac{\pi^3}{16}$ | A1ft |

$\overline{x} = \frac{\pi^3 - 4\pi}{16} \div \frac{\pi^2}{4} = \frac{\pi^2 - 4}{4\pi}$ or $0.467088....$ | M1A1 | (7) [11] |

**Notes for Question 5(a):**

M1 | for using Vol $= \pi \int_0^{\pi} \cos^2 x \, dx$. If $\pi$ is missing here it must be included later to earn this mark. Limits not needed. |

M1 | for using the double angle formula (correct) to prepare for integration. Formula must be correct. $\pi$ and limits not needed for this mark. |

M1 dep | for attempting to integrate and substitute the correct limits (only sub of non-zero limit needed to be seen) dependent on both M marks. |

A1 cso | for $\frac{\pi^2}{4}$ * (check integration is correct, answer can be obtained by luck due to the limits) |

**(b)**

NB: The first 5 marks can be earned with or without $\pi$ |

M1 | for using $\pi \int_0^{\pi/2} x \cos^2 x \, dx$. $\pi$ not needed; limits not needed. |

M1 | for using the double angle formula (correct) and attempting the first stage of integration by parts |

B1 | for $\frac{\pi^2}{16}$ or $\frac{\pi^2}{16}$ if $\pi$ not included. NB integration by parts not needed for this mark |

M1 dep | for completing the integration by parts, limits not needed yet |

A1 ft | for $= \frac{\pi}{8}[-1-1] + \frac{\pi^3}{16} - \frac{1}{16}$ ft on $\frac{\pi^3}{16}$ |

$\quad$ or $= \frac{8}[-1-1] + \frac{\pi^3}{16} - \frac{\pi}{4}$ |

M1 | for using $\overline{x} = \frac{\int \pi y^2 x \, dx}{\int \pi y^2 \, dx}$. The numerator integral need not be correct. |

M1 | $\pi$ should be seen in both or neither integral |

for $\overline{x} = \frac{\pi^3 - 4}{4\pi}$ oe eg $\frac{\pi}{4} - \frac{1}{\pi}$ or $0.467088....$ |

A1 cso | Accept 0.47 or better but no fractions within fractions |

(a) has a given answer, so the cso applies to the solution of (b) only. |

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