Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.3 Volumes By Cylindrical Shells - Exercises Set 5.3 - Page 370: 4

$$V = \frac{{16}}{3}\pi $$

$$\eqalign{ & {\text{From the graph we have }}x = y,\,\,\,y = \sqrt {x + 2} ,\,\,\,0 \leqslant y \leqslant 2 \cr & y = \sqrt {x + 2} \Rightarrow x = {y^2} - 2 \cr & y \geqslant {y^2} - 2{\text{ on the interval }}\,0 \leqslant y \leqslant 2 \cr & {\text{The volume of the solid can be calculated using cylindrical shells}} \cr & V = \int_c^d {2\pi y\left[ {f\left( y \right) - g\left( y \right)} \right]dy} \cr & {\text{Let }}f\left( y \right) = y{\text{ and }}g\left( y \right) = {y^2} - 2 \cr & V = \int_0^2 {2\pi y\left[ {y - \left( {{y^2} - 2} \right)} \right]dy} \cr & V = 2\pi \int_0^2 {y\left( {y - {y^2} + 2} \right)dy} \cr & V = 2\pi \int_0^2 {\left( {{y^2} - {y^3} + 2y} \right)dy} \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{1}{3}{y^3} - \frac{1}{4}{y^4} + {y^2}} \right]_0^2 \cr & V = 2\pi \left[ {\frac{1}{3}{{\left( 2 \right)}^3} - \frac{1}{4}{{\left( 2 \right)}^4} + {{\left( 2 \right)}^2}} \right] - 2\pi \left[ 0 \right] \cr & V = 2\pi \left( {\frac{8}{3}} \right) \cr & V = \frac{{16}}{3}\pi \cr} $$