A cone has height h and base radius r . Deduce the formula for the volume of this cone by rotating the triangular region, enclosed by the line \(y = h - \frac{h}{r}x\) and the coordinate axes, through \(2\pi \) about the y-axis.

\(x = r - \frac{r}{h}y{\text{ or }}x = \frac{r}{h}(h - y){\text{ (or equivalent)}}\)     (A1)

\(\int {\pi {x^2}{\text{d}}y} \)

\( = \pi \int_0^h {{{\left( {r - \frac{r}{h}y} \right)}^2}{\text{d}}y} \)     M1A1 

Note: Award M1 for \(\int {{x^2}{\text{d}}y} \) and A1 for correct expression.

Accept \(\pi \int_0^h {{{\left( {\frac{r}{h}y - r} \right)}^2}{\text{d}}y{\text{ and }}\pi \int_0^h {{{\left( { \pm \left( {r - \frac{r}{h}x} \right)} \right)}^2}{\text{d}}x} } \)

\( = \pi \int_0^h {\left( {{r^2} - \frac{{2{r^2}}}{h}y + \frac{{{r^2}}}{{{h^2}}}{y^2}} \right){\text{d}}y} \)     A1

Note: Accept substitution method and apply markscheme to corresponding steps.

\( = \pi \left[ {{r^2}y - \frac{{{r^2}{y^2}}}{h} + \frac{{{r^2}{y^3}}}{{3{h^2}}}} \right]_0^h\)     M1A1 

Note: Award M1 for attempted integration of any quadratic trinomial.

\( = \pi \left( {{r^2}h - {r^2}h + \frac{1}{3}{r^2}h} \right)\)     M1A1 

Note: Award M1 for attempted substitution of limits in a trinomial.

\( = \frac{1}{3}\pi {r^2}h\)     A1 

Note: Throughout the question do not penalize missing dx/dy as long as the integrations are done with respect to correct variable.

[9 marks]

Most candidates attempted this question using either the formula given in the information booklet or the disk method. However, many were not successful, either because they started off with the incorrect expression or incorrect integration limits or even attempted to integrate the correct expression with respect to the incorrect variable.