The region enclosed by the curve of f and the x-axis is rotated \(360^\circ \) about the x-axis.

Find the volume of the solid formed.

attempt to set up integral expression     M1

e.g. \(\pi {\int {\sqrt {16 - 4{x^2}} } ^2}{\rm{d}}x\) , \(2\pi \int_0^2 {(16 - 4{x^2})} \) ,  \({\int {\sqrt {16 - 4{x^2}} } ^2}{\rm{d}}x\)

\(\int {16} {\rm{d}}x = 16x\) , \(\int {4{x^2}{\rm{d}}x = } \frac{{4{x^3}}}{3}\) (seen anywhere)     A1A1

evidence of substituting limits into the integrand     (M1)

e.g. \(\left( {32 - \frac{{32}}{3}} \right) - \left( { - 32 + \frac{{32}}{3}} \right)\) , \(64 - \frac{{64}}{3}\)

volume \(= \frac{{128\pi }}{3}\)     A2     N3

[6 marks]

Many candidates correctly integrated using \(f(x)\) , although some neglected to square the function and mired themselves in awkward integration attempts. Upon substituting the limits, many were unable to carry the calculation to completion. Occasionally the \(\pi \) was neglected in a final answer. Weaker candidates considered the solid formed to be a sphere and did not use integration.