A2 June 2025 Paper 1 Q16

AQACurrent spec7 marksIntegration

16 The picture shows a solar cooker. Part of the solar cooker is a parabolic dish.

Picture of a solar cooker: a parabolic dish on legs with a kettle held at the centre above the dish

The shape of the internal surface of the dish is formed by rotating the part of the parabola \(y^2 = 0.8x\) between \(x = 0\) and \(x = 0.25\) through \(2\pi\) radians about the \(x\)-axis, where \(x\) and \(y\) are measured in metres.

Use integration to show that the internal surface area of the dish, to three decimal places, is 0.796 square metres.

Fully justify your answer. [7 marks]

Mark scheme

Scheme	Marks	AO
Differentiates in order to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)	M1	3.1a
Obtains a correct expression for \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) or \(y\)	A1	1.1b
Substitutes their \(y\) and their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) into the formula for surface area in terms of one variable. Condone missing or incorrect limits	M1	1.1a
Writes integrand in the form \(k(x + 0.2)^{1/2}\) or \(k(5x + 1)^{1/2}\) or \(k(0.8x + 0.16)^{1/2}\) or \(k\cosh^2 u\,\sinh u\) or \(k\,y(y^2 + 0.16)^{1/2}\) OE	M1	2.2a
Obtains \(k(x + 0.2)^{3/2}\) or \(k(5x + 1)^{3/2}\) or \(k(0.8x + 0.16)^{3/2}\) or \(k\cosh^3 u\) or \(k(y^2 + 0.16)^{3/2}\) OE	A1	1.1b
Substitutes correct upper and lower limits into an integrated expression of the form \(k(ax + b)^{3/2}\) or \(k\cosh^3 u\) or \(k(ay^2 + b)^{3/2}\) and subtracts.	M1	1.1a
Completes a reasoned argument to obtain 0.796 Must see 0.7958… or \(\dfrac{19\pi}{75}\) Condone omission of units. AG	R1	2.1
	(7 marks)

Typical solution

\[y^2 = 0.8x\]\[2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0.8 \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{0.4}{y}\]\[\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{0.16}{y^2} = \frac{0.16}{0.8x} = \frac{0.2}{x}\]\[\begin{aligned} S &= 2\pi\int_0^{0.25} y\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x \\ &= 2\pi\int_0^{0.25} \sqrt{0.8x}\sqrt{1 + \frac{0.2}{x}}\,\mathrm{d}x \\ &= 2\pi\sqrt{0.8}\int_0^{0.25} (x + 0.2)^{\frac{1}{2}}\,\mathrm{d}x \\ &= 2\pi\sqrt{0.8}\left[\frac{2}{3}(x + 0.2)^{\frac{3}{2}}\right]_0^{0.25} \\ &= \frac{4}{3}\pi\sqrt{0.8}\left(0.45^{\frac{3}{2}} - 0.2^{\frac{3}{2}}\right) \\ &= 0.79587\ldots \\ &= 0.796 \text{ m}^2 \text{ to 3 decimal places} \end{aligned}\]