A2 June 2025 Paper 2 Q14

AQACurrent spec8 marksPolar Coordinates

14 A children’s play area in a park contains a paddling pool.

The outline of the paddling pool is modelled by the polar curve

\[r = 6 + 2\cos(2\theta)\]

where \(r\) is measured in metres.

Figure 2 shows the outline of the paddling pool.

Figure 2: the closed curve r = 6 + 2cos(2θ), an oval that is slightly pinched in at the top and bottom, with the initial line drawn from the pole to the right — Figure 2

There is a solid concrete island inside the paddling pool.

The boundary of the island is modelled by the polar curve

\[r = 2 + \cos\theta\]

where \(r\) is measured in metres.

(a) Sketch the boundary of the island on Figure 2 [2 marks]

(b) The paddling pool has a constant depth of 0.3 metres.

Find the volume of water in the paddling pool.

Give your answer to four significant figures.

Fully justify your answer. [6 marks]

Figure 2 with the island r = 2 + cos θ drawn as a small closed curve around the pole, extending further to the right than to the left, inside the pool outline — Mark scheme (a)

Scheme	Marks	AO
Draws closed curve, approximately symmetrical about the initial line, completely inside the existing curve.	B1	1.1b
Draws closed curve with pole inside their shape, with more of the curve to the right.	B1	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
Deduces a correct expression for the total area (PI). Condone omission of ½	M1	2.2a
Uses a trig identity to correctly transform \(4\cos^2 2\theta\) or \(\cos^2\theta\)	M1	3.1a
Integrates \((6 + 2\cos 2\theta)^2\) or \((2 + \cos\theta)^2\) correctly.	A1	1.1b
Substitutes limits correctly for at least one integral.	M1	1.1a
Obtains \(\dfrac{67\pi}{2}\) from correct working.	A1	1.1b
Obtains AWRT 31.57 m³ Must include units.	A1	3.2a
	(6)
	(8 marks)

Typical solution

\[\begin{aligned} A &= 2 \times \frac{1}{2} \times \int_0^{\pi} (6 + 2\cos 2\theta)^2\,\mathrm{d}\theta - 2 \times \frac{1}{2} \times \int_0^{\pi} (2 + \cos\theta)^2\,\mathrm{d}\theta \\ &= \int_0^{\pi} \left(36 + 24\cos 2\theta + 4\cos^2 2\theta\right)\mathrm{d}\theta - \int_0^{\pi} (4 + 4\cos\theta + \cos^2\theta)\,\mathrm{d}\theta \\ &= \int_0^{\pi} \left(36 + 24\cos 2\theta + 2 + 2\cos 4\theta\right)\mathrm{d}\theta - \int_0^{\pi} \left(4 + 4\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right)\mathrm{d}\theta \\ &= \left[38\theta + 12\sin 2\theta + \frac{1}{2}\sin 4\theta\right]_0^{\pi} - \left[\frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\pi} \\ &= (38\pi - 0) - \left(\frac{9}{2}\pi - 0\right) = \frac{67\pi}{2} \end{aligned}\]\[V = 0.3 \times \frac{67\pi}{2} = \frac{201\pi}{20}\]\[V = 31.57 \text{ m}^3\]