Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 362: 4

$$\,V = \frac{9}{2}\pi $$

$$\eqalign{ & {\text{From the graph we have }} \cr & y = \frac{1}{x}\,\,\, \Rightarrow \,\,\,\,x = \frac{1}{y}{\text{ and }}x = 2{\text{ on the interval }}\left[ {\frac{1}{2},2} \right] \cr & {\text{let }}w\left( y \right) = 2{\text{ and }}v\left( y \right) = \frac{1}{y}{\text{ with }}w\left( y \right) \geqslant v\left( y \right) \cr & {\text{The volume of the solid can be calculated using}} \cr & \,\,\,\,V = \int_c^d {\pi \left( {{{\left[ {w\left( y \right)} \right]}^2} - {{\left[ {v\left( y \right)} \right]}^2}} \right)dy} \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,V = \int_{1/2}^2 {\pi \left[ {{{\left( 2 \right)}^2} - {{\left( {\frac{1}{y}} \right)}^2}} \right]dy} \cr & \,\,\,\,V = \pi \int_{1/2}^2 {\left( {4 - \frac{1}{{{y^2}}}} \right)dy} \cr & \,\,\,\,V = \pi \left[ {4y + \frac{1}{y}} \right]_{1/2}^2 \cr & {\text{Evaluating}} \cr & \,\,\,\,V = \pi \left[ {4\left( 2 \right) + \frac{1}{2}} \right] - \pi \left[ {4\left( {\frac{1}{2}} \right) + 2} \right] \cr & \,\,\,\,V = \pi \left[ {\frac{{17}}{2}} \right] - \pi \left[ 4 \right] \cr & \,\,\,\,V = \frac{9}{2}\pi \cr} $$