Calculate the volume of the observatory.

The hemispherical roof is to be painted.

Calculate the area that is to be painted.

\(V = \pi {(15)^2}(12) + 0.5 \times \frac{{4\pi {{(15)}^3}}}{3}\)     (M1)(M1)(M1)

Note: Award (M1) for correctly substituted cylinder formula, (M1) for correctly substituted sphere formula, (M1) for dividing the sphere formula by 2.

\( = 15550.8 \ldots \)

\( = 15600{\text{ }}{{\text{m}}^3}{\text{ }}(4950\pi {\text{ }}{{\text{m}}^3})\)     (A1)     (C4)

Notes: The final answer is \(15600{\text{ }}{{\text{m}}^3}\); the units are required. The use of \(\pi  = 3.14\) which gives a final answer of \(15 500\) (\(15 543\)) is premature rounding; the final (A1) is not awarded.

[4 marks]

\(SA = 0.5 \times 4\pi {\left( {15} \right)^2}\)     (M1)

\( = 1413.71 \ldots \)

\( = 1410{\text{ }}{{\text{m}}^2}{\text{ }}(450\pi {\text{ }}{{\text{m}}^2})\)     (A1)     (C2)

Notes: The final answer is \(1410{\text{ }}{{\text{m}}^2}\) ; do not penalize lack of units if this has been penalized in part (a).

[2 marks]

This question was only done well by the more able candidates. For the lower quartile of candidates, about a half of these scored no more than one mark in total for this question. In many cases, formulae were either misquoted or misused. Indeed, in part (a) many ignored the hemisphere, choosing to use the formula for the volume of a sphere instead. Some candidates who correctly used a hemisphere in part (a) then treated the surface area as a sphere in part (b). A minority of candidates thought the area to be painted in the last part of the question was a circle rather than a hemisphere.

This question was only done well by the more able candidates. For the lower quartile of candidates, about a half of these scored no more than one mark in total for this question. In many cases, formulae were either misquoted or misused. Indeed, in part (a) many ignored the hemisphere, choosing to use the formula for the volume of a sphere instead. Some candidates who correctly used a hemisphere in part (a) then treated the surface area as a sphere in part (b). A minority of candidates thought the area to be painted in the last part of the question was a circle rather than a hemisphere.