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.