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: 19

$$V = \frac{{72}}{5}\pi $$

$$\eqalign{ & {\text{We have }}x = {y^2},\,\,\,\,x = y + 2 \cr & {\text{Let }}x = x \cr & {y^2} = y + 2 \cr & {y^2} - y - 2 = 0 \cr & \left( {y - 2} \right)\left( {y + 1} \right) = 0 \cr & {y_1} = - 1{\text{ and }}{y_2} = 2 \cr & \cr & {\text{Let }}w\left( y \right) = {y^2}{\text{ and }}v\left( y \right) = y + 2 \cr & w\left( 0 \right) = 0{\text{ and }}v\left( 0 \right) = 2 \cr & v\left( y \right) \geqslant w\left( y \right){\text{ on the interval }} - 1 \leqslant y \leqslant 2 \cr & \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_c^d {\pi \left( {{{\left[ {v\left( y \right)} \right]}^2} - {{\left[ {w\left( y \right)} \right]}^2}} \right)dy} \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_{ - 1}^2 {\left( {{{\left[ {y + 2} \right]}^2} - {{\left[ {{y^2}} \right]}^2}} \right)dy} \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{1}{3}{{\left( {y + 2} \right)}^3} - \frac{1}{5}{y^5}} \right]_{ - 1}^2 \cr & V = \pi \left[ {\frac{1}{3}{{\left( {2 + 2} \right)}^3} - \frac{1}{5}{{\left( 2 \right)}^5}} \right] - \pi \left[ {\frac{1}{3}{{\left( { - 1 + 2} \right)}^3} - \frac{1}{5}{{\left( { - 1} \right)}^5}} \right] \cr & V = \pi \left( {\frac{{224}}{{15}}} \right) - \pi \left( {\frac{8}{{15}}} \right) \cr & V = \frac{{72}}{5}\pi \cr} $$