Answer
$$V = \frac{\pi }{3}{\ln ^3}\left( 2 \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}}{\text{, }}y = \frac{{\ln x}}{{\sqrt x }},{\text{ }}y = 0{\text{ and }}x = 2 \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}\underbrace {\left[ {1,2} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Revolving the region about the }}x{\text{ - axis}}{\text{, using the Disk Method }} \cr
& {\text{about the }}x{\text{ - axis }} \cr
& V = \int_a^b {\pi f{{\left( x \right)}^2}} dx \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_1^2 {\pi {{\left( {\frac{{\ln x}}{{\sqrt x }}} \right)}^2}} dx \cr
& V = \pi \int_1^2 {\frac{{{{\ln }^2}x}}{x}} dx \cr
& V = \pi \int_1^2 {{{\ln }^2}x\left( {\frac{1}{x}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\frac{1}{3}{{\ln }^3}x} \right]_1^2 \cr
& V = \pi \left[ {\frac{1}{3}{{\ln }^3}\left( 2 \right) - \frac{1}{3}{{\ln }^3}\left( 1 \right)} \right] \cr
& V = \frac{\pi }{3}{\ln ^3}\left( 2 \right) \cr} $$
