A2 June 2021 Paper 2 Q7

AQACurrent spec7 marksIntegration

Astroid: a four-cusped curve symmetrical about both axes, with cusps on the axes; the cusp on the positive x-axis is marked t = 0 and the cusp on the positive y-axis is marked t = π/2

The diagram shows a curve known as an astroid.

The curve has parametric equations

\[\begin{gathered} x = 4\cos^3 t \\ y = 4\sin^3 t \\ (0 \leqslant t \lt 2\pi) \end{gathered}\]

The section of the curve from \(t = 0\) to \(t = \dfrac{\pi}{2}\) is rotated through \(2\pi\) radians about the \(x\)-axis.

Show that the curved surface area of the shape formed is equal to \(\dfrac{b\pi}{c}\), where \(b\) and \(c\) are integers. [7 marks]

Mark scheme

Scheme	Marks	AO
Obtains derivatives of \(x\) and \(y\)	M1	1.1a
Obtains correct expression for \(\dot{x}^2 + \dot{y}^2\)	A1	1.1b
Uses trig identity to simplify their expression for \(\dot{x}^2 + \dot{y}^2\)	B1	2.2a
Substitutes their expression for \(\dot{x}^2 + \dot{y}^2\) into the formula for surface area Condone missing limits of integration and missing “\(2\pi\)”	M1	1.1a
Obtains correct expression for the surface area including correct limits of integration	A1	1.1b
Obtains \(k\sin^5 t\) by integration Condone missing limits of integration	A1	1.1b
Completes a rigorous argument to show the required result	R1	2.1
	(7 marks)

Typical solution

\[\dot{x} = -12\cos^2 t\sin t\]\[\dot{y} = 12\sin^2 t\cos t\]\[\begin{aligned} \dot{x}^2 + \dot{y}^2 &= 144\cos^4 t\sin^2 t + 144\sin^4 t\cos^2 t \\ &= 144\cos^2 t\sin^2 t\,(\cos^2 t + \sin^2 t) \\ &= 144\cos^2 t\sin^2 t \end{aligned}\]\[\sqrt{\dot{x}^2 + \dot{y}^2} = 12\cos t\sin t\]\[\begin{aligned} S &= 2\pi\int_0^{\frac{\pi}{2}} y\sqrt{\dot{x}^2 + \dot{y}^2}\,\mathrm{d}t \\ &= 2\pi\int_0^{\frac{\pi}{2}} 4\sin^3 t\,(12\cos t\sin t)\,\mathrm{d}t \\ &= 96\pi\int_0^{\frac{\pi}{2}} \sin^4 t\cos t\,\mathrm{d}t \\ &= 96\pi\left[\frac{\sin^5 t}{5}\right]_0^{\frac{\pi}{2}} \\ &= \frac{96\pi}{5} \end{aligned}\]