Answer
$$V = \pi \left( {25\ln 5 - 12} \right)$$
Work Step by Step
$$\eqalign{
& y = \ln x,{\text{ }}y = 0,{\text{ }}x = 5 \cr
& {\text{Let }}y = 0 \cr
& 0 = \ln x \Rightarrow x = 1 \cr
& {\text{The limits of integration are }}\left[ {1,5} \right] \cr
& {\text{Calculate the Volume by cylindrical shells about the }}y{\text{ - axis}} \cr
& V = 2\pi \int_a^b {xf\left( x \right)} dx,{\text{ then}} \cr
& {\text{The Volume of the solid is given by}} \cr
& V = 2\pi \int_1^5 {x\ln x} dx \cr
& {\text{Integrating by parts}} \cr
& u = \ln x,{\text{ }}dv = xdx \cr
& du = \frac{1}{x}dx{\text{ }}v = \frac{{{x^2}}}{2} \cr
& \int {x\ln xdx = \frac{{{x^2}}}{2}\ln x - \int {\left( {\frac{{{x^2}}}{2}} \right)\left( {\frac{1}{x}} \right)dx} } \cr
& {\text{ }} = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\int {xdx} \cr
& {\text{ }} = \frac{{{x^2}}}{2}\ln x - \frac{1}{4}{x^2} + C \cr
& V = 2\pi \int_1^5 {x\ln x} dx = 2\pi \left[ {\frac{{{x^2}}}{2}\ln x - \frac{1}{4}{x^2}} \right]_1^5 \cr
& V = 2\pi \left[ {\frac{{{{\left( 5 \right)}^2}}}{2}\ln \left( 5 \right) - \frac{1}{4}{{\left( 5 \right)}^2}} \right] - 2\pi \left[ {\frac{{{{\left( 1 \right)}^2}}}{2}\ln \left( 1 \right) - \frac{1}{4}{{\left( 1 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left[ {\frac{{25}}{2}\ln \left( 5 \right) - \frac{{25}}{4}} \right] - 2\pi \left[ {0 - \frac{1}{4}{{\left( 1 \right)}^2}} \right] \cr
& V = 25\pi \ln 5 - \frac{{25}}{2}\pi + \frac{\pi }{2} \cr
& V = 25\pi \ln 5 - 12\pi \cr
& V = \pi \left( {25\ln 5 - 12} \right) \cr} $$
