The region \(R\) is enclosed by the graph of \(y = {e^{ - {x^2}}}\), the \(x\)-axis and the lines \(x =  - 1\) and \(x = 1\).

Find the volume of the solid of revolution that is formed when \(R\) is rotated through \(2\pi \) about the \(x\)-axis.

\(\int_{ - 1}^1 {\pi {{\left( {{{\text{e}}^{ - {x^2}}}} \right)}^2}{\text{d}}x} \;\;\;\left( {\int_{ - 1}^1 {\pi {{\text{e}}^{ - 2{x^2}}}{\text{d}}x} \;\;\;{\text{or}}\;\;\;\int_0^1 {2\pi {{\text{e}}^{ - 2{x^2}}}{\text{d}}x} } \right)\)     (M1)(A1)(A1)

Note:     Award M1 for integral involving the function given; A1 for correct limits; A1 for \(\pi \) and \({{{\left( {{{\text{e}}^{ - {x^2}}}} \right)}^2}}\)

\( = 3.758249 \ldots  = 3.76\)     A1

[4 marks]

Most candidates answered this question correctly. Those candidates who attempted to manipulate the function or attempt an integration wasted time and obtained 3/4 marks. The most common errors were an extra factor ‘2’ and a fourth power when attempting to square the function. Many candidates wrote down the correct expression but not all were able to use their calculator correctly.