Answer
$$V = \pi \ln \left( 3 \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}}{\text{, }}y = \frac{1}{{\sqrt x }},{\text{ }}y = 0,{\text{ }}x = 2{\text{ and }}x = 6 \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}\underbrace {\left[ {2,6} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Revolving the region about the }}x{\text{ - axis}}{\text{, using the Disk Method }} \cr
& {\text{about the }}x{\text{ - axis }} \cr
& V = \int_a^b {\pi f{{\left( x \right)}^2}} dx \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_2^6 {\pi {{\left( {\frac{1}{{\sqrt x }}} \right)}^2}} dx \cr
& V = \int_2^6 {\pi \left( {\frac{1}{x}} \right)} dx \cr
& V = \pi \int_2^6 {\frac{1}{x}} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\ln x} \right]_2^6 \cr
& V = \pi \left( {\ln 6 - \ln 2} \right) \cr
& V = \pi \ln \left( {\frac{6}{2}} \right) \cr
& V = \pi \ln \left( 3 \right) \cr} $$
