Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.3 Volumes By Cylindrical Shells - Exercises Set 5.3 - Page 370: 2

$$V = \frac{8}{3}\pi \left( {2 - \sqrt 2 } \right)$$

$$\eqalign{ & {\text{From the graph we have }}y = x,\,\,\,\,\,y = \sqrt {4 - {x^2}} ,\,\,\,x = 0 \cr & {\text{Let }}y = y \cr & x = \sqrt {4 - {x^2}} \cr & {x^2} = 4 - {x^2} \cr & 2{x^2} = 4 \cr & x = \pm \sqrt 2 ,\,x > 0 \cr & x = \sqrt 2 \cr & {\text{The volume of the solid can be calculated using cylindrical shells}} \cr & V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]dx} \cr & {\text{Let }}f\left( x \right) = \sqrt {4 - {x^2}} {\text{ and }}g\left( x \right) = x \cr & V = 2\pi \int_0^{\sqrt 2 } {x\left( {\sqrt {4 - {x^2}} - x} \right)dx} \cr & V = 2\pi \int_0^{\sqrt 2 } {x\sqrt {4 - {x^2}} dx} - 2\pi \int_0^{\sqrt 2 } {{x^2}dx} \cr & V = - \pi \int_0^{\sqrt 2 } {\sqrt {4 - {x^2}} \left( { - 2x} \right)dx} - 2\pi \int_0^{\sqrt 2 } {{x^2}dx} \cr & {\text{Integrating}} \cr & V = - \pi \left[ {\frac{2}{3}{{\left( {4 - {x^2}} \right)}^{3/2}}} \right]_0^{\sqrt 2 } - \frac{{2\pi }}{3}\left[ {{x^3}} \right]_0^{\sqrt 2 } \cr & V = - \frac{2}{3}\pi \left[ {{{\left( {4 - 2} \right)}^{3/2}} - {{\left( {4 - 0} \right)}^{3/2}}} \right] - \frac{{2\pi }}{3}\left[ {{{\left( {\sqrt 2 } \right)}^3} - {{\left( 0 \right)}^3}} \right] \cr & V = - \frac{2}{3}\pi \left[ {{2^{3/2}} - 8} \right] - \frac{{2\pi }}{3}\left[ {{2^{3/2}}} \right] \cr & V = - \frac{{2{{\left( 2 \right)}^{3/2}}}}{3}\pi + \frac{{16}}{3}\pi - \frac{{2{{\left( 2 \right)}^{3/2}}}}{3}\pi \cr & V = \frac{{16}}{3}\pi - \frac{4}{3}{\left( 2 \right)^{3/2}}\pi \cr & V = \frac{{16}}{3}\pi - \frac{8}{3}\sqrt 2 \pi \cr & V = \frac{8}{3}\pi \left( {2 - \sqrt 2 } \right) \cr} $$