Answer
$$V = \frac{1}{{\ln 2}}\pi $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{\sqrt x \ln x}},\,\,\,{\text{Interval }}\left[ {2,\infty } \right) \cr
& {\text{Calculate the volume using the disk method }}V = \int_a^b {\pi f{{\left( x \right)}^2}dx} \cr
& V = \int_2^\infty {\pi {{\left( {\frac{1}{{\sqrt x \ln x}}} \right)}^2}dx} \cr
& {\text{Then}}{\text{,}} \cr
& V = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{\pi }{{x{{\ln }^2}x}}dx} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \int_2^b {{{\left( {\ln x} \right)}^{ - 2}}\frac{1}{x}dx} \cr
& {\text{Evaluate the integral}} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{{\ln x}}} \right]_2^b \cr
& V = - \pi \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{\ln b}} - \frac{1}{{\ln 2}}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& V = - \pi \left[ {\frac{1}{{\ln \infty }} - \frac{1}{{\ln 2}}} \right] \cr
& V = \frac{1}{{\ln 2}}\pi \cr} $$
