The region enclosed between the curves \(y = \sqrt x {{\text{e}}^x}\) and \(y = {\text{e}}\sqrt x \) is rotated through \(2\pi \) about the x-axis. Find the volume of the solid obtained.

\(\sqrt x {{\text{e}}^x} = {\text{e}}\sqrt x  \Rightarrow x = 0{\text{ or 1}}\)     (A1)

attempt to find \(\int {{y^2}{\text{d}}x} \)     M1

\({V_1} = \pi \int_0^1 {{{\text{e}}^2}x{\text{d}}x} \)

\( = \pi \left[ {\frac{1}{2}{{\text{e}}^2}{x^2}} \right]_0^1\)

\( = \frac{{\pi {{\text{e}}^2}}}{2}\)     A1

\({V_2} = \pi \int_0^1 {x{{\text{e}}^{2x}}{\text{d}}x} \)

\( = \pi \left( {\left[ {\frac{1}{2}x{{\text{e}}^{2x}}} \right]_0^1 - \int_0^1 {\frac{1}{2}{{\text{e}}^{2x}}{\text{d}}x} } \right)\)     M1A1

Note: Award M1 for attempt to integrate by parts.

\( = \frac{{\pi {{\text{e}}^2}}}{2} - \pi \left[ {\frac{1}{4}{{\text{e}}^{2x}}} \right]_0^1\)

finding difference of volumes     M1

volume\( = {V_1} - {V_2}\)

\( = \pi \left[ {\frac{1}{4}{{\text{e}}^{2x}}} \right]_0^1\)

\( = \frac{1}{4}\pi ({{\text{e}}^2} - 1)\)     A1

[7 marks]

While only a minority of candidates achieved full marks in this question, many candidates made good attempts. Quite a few candidates obtained the limits correctly and many realized a square was needed in the integral, though a number of them subtracted then squared rather than squaring and then subtracting. There was evidence that quite a few knew about integration by parts. One common mistake was to have \(2\pi \), rather than \(\pi \) in the integral.