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: 3

$$\,\,V = \frac{{13}}{6}\pi $$

$$\eqalign{ & {\text{From the graph we have }} \cr & y = 3 - 2x\,\,\, \Rightarrow \,\,\,\,x = - \frac{1}{2}y + \frac{3}{2}{\text{ on the interval }}\left[ {0,2} \right]{\text{ and the }}y{\text{ axis}} \cr & {\text{The volume of the solid can be calculated using}} \cr & \,\,\,\,V = \int_c^d {\pi {{\left[ {u\left( y \right)} \right]}^2}dy} \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,V = \int_0^2 {\pi {{\left[ { - \frac{1}{2}y + \frac{3}{2}} \right]}^2}dy} \cr & \,\,\,\,V = \pi \int_0^2 {\left( {\frac{9}{4} - \frac{3}{2}y + \frac{1}{4}{y^2}} \right)dy} \cr & \,\,\,\,V = \pi \left[ {\frac{1}{{12}}{y^3} - \frac{3}{4}{y^2} + \frac{9}{4}y} \right]_0^2 \cr & {\text{Evaluating}} \cr & \,\,\,\,V = \pi \left[ {\frac{1}{{12}}{{\left( 2 \right)}^3} - \frac{3}{4}{{\left( 2 \right)}^2} + \frac{9}{4}\left( 2 \right)} \right] - \pi \left[ 0 \right] \cr & \,\,\,\,V = \pi \left[ {\frac{2}{3} - 3 + \frac{9}{2}} \right] \cr & \,\,\,\,V = \frac{{13}}{6}\pi \cr} $$