The region bounded by the curve \(y = \frac{{\ln (x)}}{x}\) and the lines x = 1, x = e, y = 0 is rotated through \(2\pi \) radians about the x-axis.

Find the volume of the solid generated.

METHOD 1

\(V = \pi \int_1^{\text{e}} {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x} \)     M1

Integrating by parts:

\(u = {(\ln x)^2},{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{1}{{{x^2}}}\)     (M1)

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{{2\ln x}}{x},{\text{ }}v = - \frac{1}{x}\)

\( \Rightarrow V = \pi \left( { - \frac{{{{(\ln x)}^2}}}{x} + 2\int {\frac{{\ln x}}{{{x^2}}}{\text{d}}x} } \right)\)     A1

\(u = \ln x,{\text{ }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{1}{{{x^2}}}\)     (M1)

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x},{\text{ }}v = - \frac{1}{x}\)

\(\therefore \int {\frac{{\ln x}}{{{x^2}}}{\text{d}}x = - \frac{{\ln x}}{x} + \int {\frac{1}{{{x^2}}}{\text{d}}x = - \frac{{\ln x}}{x} - \frac{1}{x}} } \)     A1

\(\therefore V = \pi \left[ { - \frac{{{{(\ln x)}^2}}}{x} + 2\left( { - \frac{{\ln x}}{x} - \frac{1}{x}} \right)} \right]_1^{\text{e}}\)

\( = 2\pi - \frac{{5\pi }}{e}\)     A1

[6 marks]

METHOD 2

\(V = \pi \int_1^{\text{e}} {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x} \)     M1

Let \(\ln x = u \Rightarrow x = {{\text{e}}^u},{\text{ }}\frac{{{\text{d}}x}}{x} = {\text{d}}u\)     (M1)

\(\int {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x = \int {\frac{{{u^2}}}{{{{\text{e}}^u}}}{\text{d}}u = \int {{{\text{e}}^{ - u}}} {u^2}{\text{d}}u = - {{\text{e}}^{ - u}}{u^2} + 2\int {{{\text{e}}^{ - u}}u{\text{d}}u} } } \)     A1

\( = - {{\text{e}}^{ - u}}{u^2} + 2\left( { - {{\text{e}}^{ - u}}u + \int {{{\text{e}}^{ - u}}{\text{d}}u} } \right) = - {{\text{e}}^{ - u}}{u^2} - 2{{\text{e}}^{ - u}}u - 2{{\text{e}}^{ - u}}\)

\( = - {{\text{e}}^{ - u}}({u^2} + 2u + 2)\)     A1

When x = e, u = 1. When x = 1, u = 0 .

\(\therefore {\text{ Volume}} = \pi \left[ { - {{\text{e}}^{ - u}}({u^2} + 2u + 2)} \right]_0^1\)     M1

\( = \pi ( - 5{{\text{e}}^{ - 1}} + 2){\text{ }}\left( { = 2\pi  - \frac{{5\pi }}{{\text{e}}}} \right)\)     A1

[6 marks]

Only the best candidates were able to make significant progress with this question. It was disappointing to see that many candidates could not state that the formula for the required volume was \(\pi \int_1^e {{{\left( {\frac{{\ln x}}{x}} \right)}^2}{\text{d}}x} \) . Of those who could, very few either attempted integration by parts or used an appropriate substitution.