2022 H2 Maths Paper 2 Q5 — Worked Solution

Part (a)

5 marks

Rotating the shaded region between the circle $x^{2} + y^{2} = r^{2}$ and the line $y = r - h$ about the $y$ -axis, the cross-section at height $y$ is a disc of radius $x$ , where $x^{2} = r^{2} - y^{2}$ . The spherical cap is generated for $y$ from $r - h$ to $r$ :

V = \int_{r - h}^{r}\pi x^{2}\, dy

= \pi\int_{r - h}^{r}\left( r^{2} - y^{2} \right)dy

= \pi\left\lbrack r^{2}y - \frac{y^{3}}{3} \right\rbrack_{r - h}^{r}

Evaluating the limits, and expanding $(r - h)^{3} = r^{3} - 3r^{2}h + 3rh^{2} - h^{3}$ ,

V = \pi\left( r^{3} - \frac{r^{3}}{3} \right) - \pi\left( r^{2}(r - h) - \frac{(r - h)^{3}}{3} \right)

= \frac{2\pi r^{3}}{3} - \pi r^{3} + \pi r^{2}h + \frac{\pi(r - h)^{3}}{3}

= - \frac{\pi r^{3}}{3} + \pi r^{2}h + \frac{\pi r^{3}}{3} - \pi r^{2}h + \pi rh^{2} - \frac{\pi h^{3}}{3}

= \pi rh^{2} - \frac{\pi h^{3}}{3}

= \frac{1}{3}\pi h^{2}(3r - h)

To check your answer: at

h = r

(the cap is a full hemisphere),

V = \frac{1}{3}\pi r^{2}(3r - r) = \frac{2\pi r^{3}}{3}

, matching the known hemisphere volume ✓

$\therefore\ V = \frac{1}{3}\pi h^{2}(3r - h)$ (shown)

Part (b)

4 marks

Approach

Apply the part (a) formula twice, once to the top cap (height $p$ ) and once to the bottom cap (height $3p$ ), with $r = 15$ throughout, and set (sphere) $\text{-}$ (top cap) $\text{-}$ (bottom cap) equal to the given ornament volume.

The sphere has volume $\frac{4}{3}\pi(15)^{3} = 4500\pi$ . The top cap has height $p$ , so its volume is $\frac{1}{3}\pi p^{2}(45 - p)$ ; the bottom cap has height $3p$ , so its volume is $\frac{1}{3}\pi(3p)^{2}(45 - 3p) = 3\pi p^{2}(45 - 3p)$ . The ornament volume is

4500\pi - \frac{1}{3}\pi p^{2}(45 - p) - 3\pi p^{2}(45 - 3p) = 3402\pi

Dividing by $\pi$ and multiplying by $3$ ,

13500 - p^{2}(45 - p) - 9p^{2}(45 - 3p) = 10206

\Rightarrow 13500 - 45p^{2} + p^{3} - 405p^{2} + 27p^{3} = 10206

\Rightarrow 28p^{3} - 450p^{2} + 3294 = 0

Dividing by $2$ ,

14p^{3} - 225p^{2} + 1647 = 0

Testing $p = 3$ : $14(27) - 225(9) + 1647 = 378 - 2025 + 1647 = 0$ ✓, so $(p - 3)$ is a factor:

14p^{3} - 225p^{2} + 1647 = (p - 3)\left( 14p^{2} - 183p - 549 \right)

The remaining quadratic has discriminant $183^{2} + 4(14)(549) = 64233$ , which is not a perfect square, giving $p = \frac{183 \pm \sqrt{64233}}{28} \approx - 2.52$ or $15.59$ . Both are rejected: we need $0 < p$ and $3p < 15$ (so that the bottom cap does not exceed the sphere), i.e. $0 < p < 5$ .

To check your answer: with

p = 3

, top cap

\text{=}\frac{1}{3}\pi(9)(42) = 126\pi

, bottom cap

\text{=}\frac{1}{3}\pi(81)(18) = 486\pi

; ornament

\text{=}4500\pi - 126\pi - 486\pi = 3402\pi

✓

$\therefore\ 14p^{3} - 225p^{2} + 1647 = 0$ , and $p = 3$ .

Part (c)

2 marks

Approach

The two ornaments share both face radii but not the same volume. Each face radius corresponds to two possible cutting heights ( $y = \pm \text{value}$ ), so the same pair of radii can be produced by cutting planes on either the same or opposite sides of the centre. The part (b) ornament uses opposite sides (spanning most of the sphere); the smaller one must use the same side instead.

With $p = 3$ : the top face of the part (b) ornament is at $y = 15 - 3 = 12$ (radius $\sqrt{15^{2} - 12^{2}} = 9$ ). The bottom cap has height $3p = 9$ , so the bottom face is at $y = - (15 - 9) = - 6$ (radius $\sqrt{15^{2} - 6^{2}} = 3\sqrt{21}$ ). This ornament spans $y = - 6$ to $y = 12$ , a height of $18$ .

A face of radius $9$ occurs only at $y = 12$ or $y = - 12$ ; a face of radius $3\sqrt{21}$ occurs only at $y = 6$ or $y = - 6$ . For a smaller solid with a top above a bottom, the only remaining valid pairing is $y = 12$ (top) and $y = 6$ (bottom), giving a height of just $6$ :

V = \int_{6}^{12}\pi\left( 15^{2} - y^{2} \right)dy

= \pi\left\lbrack 225y - \frac{y^{3}}{3} \right\rbrack_{6}^{12}

= \pi\left\lbrack (2700 - 576) - (1350 - 72) \right\rbrack

= \pi(2124 - 1278)

= 846\pi

To check your answer:

846\pi < 3402\pi

✓, confirming a smaller volume, as required.

$\therefore\ V = 846\pi\ cm^{3}$ .

2022 H2 Maths Paper 2, Question 5

Part (a)

Part (b)

Part (c)

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