Answer
$$V = \frac{{2\pi }}{{27}}\left( {13{e^6} - 1} \right)$$
Work Step by Step
$$\eqalign{ & {\text{Using the disk method, the Volume of the solid can be}} \cr & {\text{calculated by:}} \cr & V\int_a^b {\pi f{{\left( x \right)}^2}dx} \cr & {\text{With }}a = 1,\,\,\,\,b = {e^2},\,\,\,\,f\left( x \right) = x\ln x \cr & {\text{Then,}} \cr & V = \int_1^{{e^2}} {\pi {{\left( {x\ln x} \right)}^2}dx} \cr & V = \pi \int_1^{{e^2}} {{x^2}{{\ln }^2}xdx} \cr & {\text{Integrating }}\int {{x^2}{{\ln }^2}xdx} {\text{ by parts }} \cr & \left( {{\text{See the solution of exercise 26 in this section}}} \right){\text{ we obtain:}} \cr & \int {{x^2}{{\ln }^2}xdx} = \frac{{{x^3}{{\ln }^2}x}}{3} - \frac{{2{x^3}\ln x}}{9} + \frac{2}{{27}}{x^3} + C \cr & {\text{Then,}} \cr & \,\,V = \pi \left[ {\frac{{{x^3}{{\ln }^2}x}}{3} - \frac{{2{x^3}\ln x}}{9} + \frac{2}{{27}}{x^3}} \right]_1^{{e^2}} \cr & \,\,V = \pi \left[ {\frac{{{e^6}{{\ln }^2}{e^2}}}{3} - \frac{{2{e^6}\ln {e^2}}}{9} + \frac{2}{{27}}{e^6}} \right] - \pi \left[ {\frac{{{1^3}{{\ln }^2}1}}{3} - \frac{{2{{\left( 1 \right)}^3}\ln 1}}{9} + \frac{2}{{27}}} \right] \cr & \,\,V = \pi \left( {\frac{{{e^6}\left( 4 \right)}}{3} - \frac{{2{e^6}\left( 2 \right)}}{9} + \frac{{2{e^6}}}{{27}}} \right) - \frac{{2\pi }}{{27}} \cr & \,\,V = \pi \left( {\frac{{4{e^6}}}{3} - \frac{{4{e^6}}}{9} + \frac{{2{e^6}}}{{27}}} \right) - \frac{{2\pi }}{{27}} \cr & V = \frac{{26{e^6}}}{{27}}\pi - \frac{2}{{27}}\pi \cr & V = \frac{{2\pi }}{{27}}\left( {13{e^6} - 1} \right) \cr} $$
