AS June 2020 Paper 1 Q15

AQACurrent spec4 marksIntegration

15 A segment of the line \(y = kx\) is rotated about the \(x\)-axis to generate a cone with vertex \(O\).

The distance of \(O\) from the centre of the base of the cone is \(h\).

The radius of the base of the cone is \(r\).

The line y = kx through O and its reflection in the x-axis, forming a cone with vertex O and a circular base drawn as an ellipse to the right

(a) Find \(k\) in terms of \(r\) and \(h\). [1 mark]

(b) Use calculus to prove that the volume of the cone is\[\frac{1}{3}\pi r^2h\]

[3 marks]

Scheme	Marks	AO
Obtains the correct expression \(k = \frac{r}{h}\).	B1	1.1b

Mark scheme (b)

Scheme	Marks	AO
Uses the formula for volume of revolution \(V = \pi\int mx^2\,dx\). Condone missing \(\pi\), \(dx\) and missing or incorrect limits.	M1	1.1a
Correctly integrates their \((kx)^2\), with an expression for \(k\) in terms of \(r\) and \(h\).	M1	1.1a
Completes a rigorous proof to show that \(V = \frac{1}{3}\pi r^2h\).	R1	2.1
	(4 marks)

Typical solution

\[\begin{aligned}\text{Volume} &= \pi\int_0^h \left(\frac{rx}{h}\right)^2 dx \\ &= \pi\int_0^h \frac{r^2x^2}{h^2}\,dx \\ &= \pi\left[\frac{r^2x^3}{3h^2}\right]_0^h \\ &= \pi\left(\frac{r^2h^3}{3h^2} - 0\right) = \frac{1}{3}\pi r^2h\end{aligned}\]