Answer
$$V = 4\pi \ln 2$$
Work Step by Step
$$\eqalign{
& {\text{From the graph of the region shown below}} \cr
& f\left( x \right) = \frac{1}{{{x^2}}}{\text{, }}g\left( x \right) = 0 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}\left[ {2,8} \right] \cr
& {\text{Use the Shell method about the }}y{\text{ - axis}} \cr
& V = \int_c^d {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_2^8 {2\pi x\left( {\frac{1}{{{x^2}}} - 0} \right)} dx \cr
& V = 2\pi \int_2^8 {\frac{1}{x}} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\ln \left| x \right|} \right]_2^8 \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left( {\ln 8 - \ln 2} \right) \cr
& V = 2\pi \ln \left( 4 \right) \cr
& V = 2\pi \ln \left( {{2^2}} \right) \cr
& V = 4\pi \ln 2 \cr} $$
