Answer
$$V = \pi {\ln ^2}\left( 3 \right)$$
Work Step by Step
$$\eqalign{
& {\text{From the graph of the region shown below}} \cr
& f\left( x \right) = \frac{{\ln x}}{{{x^2}}}{\text{, }}g\left( x \right) = 0 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}\left[ {1,3} \right] \cr
& {\text{Use the Shell method about the }}y{\text{ - axis}} \cr
& V = \int_c^d {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_1^3 {2\pi x\left( {\frac{{\ln x}}{{{x^2}}} - 0} \right)} dx \cr
& V = 2\pi \int_1^3 {\frac{{\ln x}}{x}} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{{{{\ln }^2}x}}{2}} \right]_1^3 \cr
& V = 2\pi \left[ {\frac{{{{\ln }^2}\left( 3 \right)}}{2} - \frac{{{{\ln }^2}\left( 1 \right)}}{2}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left[ {\frac{{{{\ln }^2}\left( 3 \right)}}{2} - \frac{0}{2}} \right] \cr
& V = \pi {\ln ^2}\left( 3 \right) \cr} $$
