Answer
$$V = 2\pi \ln \left( {\frac{5}{2}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^2}}},{\text{ }}y = 0,{\text{ }}x = 2,{\text{ }}x = 5 \cr
& {\text{From the graph shown below, we can apply the shell method}} \cr
& V = 2\pi \int_a^b {p\left( x \right)h\left( x \right)} dx \cr
& {\text{Let }}p\left( x \right) = x{\text{ and }}h\left( x \right) = \frac{1}{{{x^2}}},{\text{ }}a = 2{\text{ and }}b = 5 \cr
& V = 2\pi \int_2^5 {x\left( {\frac{1}{{{x^2}}}} \right)} dx \cr
& {\text{Simplify and integrate}} \cr
& V = 2\pi \int_2^5 {\frac{1}{x}} dx \cr
& V = 2\pi \left[ {\ln \left| x \right|} \right]_2^5 \cr
& V = 2\pi \left[ {\ln \left| 5 \right| - \ln \left| 2 \right|} \right] \cr
& V = 2\pi \ln \left( {\frac{5}{2}} \right) \cr} $$
