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: 8

$$V = 8\pi $$

$$\eqalign{ & {\text{From the graph we have }} \cr & y = {x^2} - 1\,\,\, \Rightarrow \,\,\,\,x = \sqrt {y + 1} {\text{ and }}x = 2{\text{ on the interval }}\left[ { - 1,3} \right] \cr & {\text{let }}w\left( y \right) = 2{\text{ and }}v\left( y \right) = \sqrt {y + 1} {\text{ with }}w\left( y \right) \geqslant v\left( y \right) \cr & {\text{The volume of the solid can be calculated using}} \cr & \,\,\,\,V = \int_c^d {\pi \left( {{{\left[ {w\left( y \right)} \right]}^2} - {{\left[ {v\left( y \right)} \right]}^2}} \right)dy} \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,V = \int_{ - 1}^3 {\pi \left( {{{\left[ 2 \right]}^2} - {{\left[ {\sqrt {y + 1} } \right]}^2}} \right)dy} \cr & \,\,\,\,V = \int_{ - 1}^3 {\pi \left( {4 - y - 1} \right)dy} \cr & \,\,\,\,V = \pi \int_{ - 1}^3 {\left( {3 - y} \right)dy} \cr & \,\,\,\,V = \pi \left[ {3y - \frac{1}{2}{y^2}} \right]_{ - 1}^3 \cr & {\text{Evaluating}} \cr & \,\,\,\,V = \pi \left[ {3\left( 3 \right) - \frac{1}{2}{{\left( 3 \right)}^2}} \right] - \pi \left[ {3\left( { - 1} \right) - \frac{1}{2}{{\left( { - 1} \right)}^2}} \right] \cr & \,\,\,\,V = \pi \left[ {9 - \frac{9}{2}} \right] - \pi \left[ { - 3 - \frac{1}{2}} \right] \cr & \,\,\,\,V = \pi \left[ {9 - \frac{9}{2}} \right] - \pi \left[ { - 3 - \frac{1}{2}} \right] \cr & \,\,\,\,V = \frac{9}{2}\pi + \frac{7}{2}\pi \cr & \,\,\,\,V = 8\pi \cr} $$