Answer
$$V = \frac{\pi }{3}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{ - 2}},\,\,\,{\text{Interval }}\left[ {1,\infty } \right) \cr
& {\text{Calculate the volume using the disk method }}V = \int_a^b {\pi f{{\left( x \right)}^2}dx} \cr
& V = \int_1^\infty {\pi {x^{ - 2}}dx} \cr
& {\text{Then}}{\text{,}} \cr
& V = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\pi {{\left( {{x^{ - 2}}} \right)}^2}dx} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \int_1^b {{x^{ - 4}}dx} \cr
& {\text{Evaluate the integral}} \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{{3{x^3}}}} \right]_1^b \cr
& V = \pi \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{3} - \frac{1}{{3{b^3}}}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& V = \pi \left[ {\frac{1}{3} - \frac{1}{\infty }} \right] \cr
& V = \frac{\pi }{3} \cr} $$
