Higher June 2018 Paper 2 Q19
19 Shape S is one quarter of a solid sphere, centre \(O\).
| Volume of sphere \(= \dfrac{4}{3}\pi r^3\) Surface area of sphere \(= 4\pi r^2\) |  |
The volume of S is \(576\pi\) cm3
Find the surface area of S.
Give your answer correct to 3 significant figures.
You must show your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 905 | P1 | for correct use of formula for the volume of a sphere eg \(\dfrac{1}{4} \times \dfrac{4}{3} \times \pi \times r^3\) (= \(576\pi\) or 1809…) OR \(576\pi \times 4\) or \(2304\pi\) or 7238…\(\left(= \dfrac{4}{3} \times \pi \times r^3\right)\) |
| P1 | for a complete correct process to find \(r\), eg \(r = \sqrt[3]{\dfrac{576 \times 4 \times 3}{4}}\) or \(r = 12\) | |
| P1 | for a process to find the curved surface area eg \(\dfrac{4 \times \pi \times [\text{radius}]^2}{4}\) (\(=144\pi\) or 452…) OR the surface area of both flat surfaces eg \(\left(2 \times \dfrac{\pi \times [\text{radius}]^2}{2}\right)\) OR complete expression for the total surface area eg \(\dfrac{4\pi r^2}{4} + \dfrac{\pi r^2}{2} \times 2\) oe | |
| P1 | for process to find the complete surface area eg \(\dfrac{4 \times \pi \times [\text{radius}]^2}{4} + \left(2 \times \dfrac{\pi \times [\text{radius}]^2}{2}\right)\) | |
| A1 | answer in the range 904.7 – 905 or \(288\pi\) (SCB2 for an answer in the range 358.1 – 359.2) |
Additional guidance
P1 (volume): We do not need to see what is in the brackets to award this mark.
The contents of the bracket alone would score P0
P1 (find \(r\)): Could be shown in several stages
\(\sqrt[3]{\dfrac{576 \times 4 \times 3}{4}} = \sqrt[3]{1728}\)
P1 (curved surface area): Radius used must be clearly identified as their radius of the solid
If an answer is given in the range but then incorrectly rounded, award full marks.