Answer
$$V = 90\pi $$
Work Step by Step
$$\eqalign{
& y = 2{x^{ - 3/2}},{\text{ }}y = 2,{\text{ }}y = 16,{\text{ and }}x = 0 \cr
& y = 2{x^{ - 3/2}} \to x = {\left( {\frac{2}{y}} \right)^{2/3}} \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& {\left( {\frac{2}{y}} \right)^{2/3}} > 0{\text{ on the interval }}\left[ {2,16} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_2^{16} {2\pi y\left( {{{\left( {\frac{2}{y}} \right)}^{2/3}} - 0} \right)} dy \cr
& V = 2\pi \int_2^{16} {{2^{2/3}}{y^{1/3}}} dy \cr
& V = {2^{5/3}}\pi \int_2^{16} {{y^{1/3}}} dy \cr
& {\text{Integrating}} \cr
& V = {2^{5/3}}\pi \left[ {\frac{{{y^{4/3}}}}{{4/3}}} \right]_2^{16} \cr
& {\text{Simplifying}} \cr
& V = \frac{{{2^{5/3}}\left( 3 \right)\pi }}{4}\left[ {{{\left( {16} \right)}^{4/3}} - {{\left( 2 \right)}^{4/3}}} \right] \cr
& V = \frac{{3\pi }}{{{2^{1/3}}}}\left[ {{{\left( {{2^4}} \right)}^{4/3}} - {{\left( 2 \right)}^{4/3}}} \right] \cr
& V = \frac{{3\pi }}{{{2^{1/3}}}}\left( {{2^{16/3}} - {2^{4/3}}} \right) \cr
& V = 3\pi \left( {{2^5} - 2} \right) \cr
& V = 3\pi \left( {30} \right) \cr
& V = 90\pi \cr} $$
