Answer
$$V = \frac{3}{2}\pi $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {\frac{{x + 1}}{{{x^3}}}} ,\,\,\,{\text{Interval }}\left[ {1,\infty } \right) \cr
& {\text{Calculate the volume using the disk method }}V = \int_a^b {\pi f{{\left( x \right)}^2}dx} \cr
& V = \int_1^\infty {\sqrt {\frac{{x + 1}}{{{x^3}}}} dx} \cr
& {\text{Then}}{\text{,}} \cr
& V = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\pi {{\left( {\sqrt {\frac{{x + 1}}{{{x^3}}}} } \right)}^2}dx} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{x + 1}}{{{x^3}}}dx} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \int_1^b {\left( {{x^{ - 2}} + {x^{ - 3}}} \right)dx} \cr
& {\text{Evaluate the integral}} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{x} - \frac{1}{{2{x^2}}}} \right]_1^b \cr
& V = - \pi \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{x} + \frac{1}{{2{x^2}}}} \right]_1^b \cr
& V = - \pi \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{b} + \frac{1}{{2{b^2}}} - \frac{1}{1} - \frac{1}{2}} \right] \cr
& V = - \pi \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{b} + \frac{1}{{2{b^2}}} - \frac{3}{2}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& V = - \pi \left[ {\frac{1}{\infty } + \frac{1}{\infty } - \frac{3}{2}} \right] \cr
& V = \frac{3}{2}\pi \cr} $$
