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Date	May 2022	Marks available	5	Reference code	22M.2.AHL.TZ2.6
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Determine	Question number	6	Adapted from	N/A

Question

The following diagram shows the curve $\frac{x^{2}}{36} + \frac{{(y - 4)}^{2}}{16} = 1$ , where $h \leq y \leq 4$ .

The curve from point $Q$ to point $B$ is rotated $360 °$ about the $y$ -axis to form the interior surface of a bowl. The rectangle $OPQR$ , of height $h cm$ , is rotated $360 °$ about the $y$ -axis to form a solid base.

The bowl is assumed to have negligible thickness.

Given that the interior volume of the bowl is to be $285 {cm}^{3}$ , determine the height of the base.

Markscheme

attempts to express $x^{2}$ in terms of $y$ (M1)

$V = π \int_{h}^{4} 36 (1 - \frac{{(y - 4)}^{2}}{16}) d y$ A1

Note: Correct limits are required.

Attempts to solve $π \int_{h}^{4} 36 (1 - \frac{{(y - 4)}^{2}}{16}) d y = 285$ for $h$ (M1)

Note: Award M1 for attempting to solve $36 π (\frac{h^{3}}{48} - \frac{h^{2}}{4} + \frac{8}{3}) = 285$ or equivalent for $h$ .

$h = 0.7926 \dots$

$h = 0.793$ (cm) A2

[5 marks]

Examiners report

This question was a struggle for many candidates. To start with, many candidates had difficulty understanding the diagram. Some candidates tried to include the base in their equation.

Because of this confusion, the question was poorly attempted. Some only received one mark for rearranging the equation to make $x^{2}$ the subject but were unable to set the correct definite integral with correct terminals. Again, many candidates tried to solve by hand instead of using their GDC. The correct answer was not seen that often.

Those candidates who recognised that the volume was around the y -axis and used their GDC to solve, usually achieved full marks for this question.

Syllabus sections

Topic 5 —Calculus » AHL 5.17—Areas under curve onto y-axis, volume of revolution (about x and y axes)

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19N.2.SL.TZ0.S_8b.i:
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