The graph of \(y = \sqrt x \) between \(x = 0\) and \(x = a\) is rotated \(360^\circ \) about the x-axis. The volume of the solid formed is \(32\pi \) . Find the value of a.

attempt to substitute into formula \(V = \int {\pi {y^2}{\rm{d}}x} \)     (M1) 

integral expression     A1

e.g. \(\pi \int_0^a {(\sqrt x } {)^2}{\rm{d}}x\) ,  \(\pi \int x \)

correct integration     (A1)

e.g. \(\int x{{\rm{d}}x = \frac{1}{2}{x^2}}\)

correct substitution \(V = \pi \left[ {\frac{1}{2}{a^2}} \right]\)     (A1)

equating their expression to \(32\pi \)     M1

e.g. \(\pi \left[ {\frac{1}{2}{a^2}} \right] = 32\pi \)

\({a^2} = 64\)

\(a = 8\)     A2     N2

[7 marks]

Despite the “reverse” nature of this question, many candidates performed well with the integration. Some forgot to square the function, while others did not discard the negative value of a. Some attempted to equate \(32\pi \) to the formula for volume of a sphere, which suggests this topic was not fully covered in some centres.