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: 2

$$\,\,\,\,V = \frac{{38}}{{15}}\pi $$

$$\eqalign{ & {\text{From the graph we have }} \cr & f\left( x \right) = 2 - {x^2}{\text{ and }}g\left( x \right) = x{\text{ with }}f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}\left[ {0,1} \right] \cr & {\text{The volume of the solid can be calculated using}} \cr & \,\,\,\,V = \int_a^b {\pi \left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,V = \int_0^1 {\pi \left( {{{\left[ {2 - {x^2}} \right]}^2} - {{\left[ x \right]}^2}} \right)dx} \cr & \,\,\,\,V = \pi \int_0^1 {\left( {4 - 4{x^2} + {x^4} - {x^2}} \right)dx} \cr & \,\,\,\,V = \pi \int_0^1 {\left( {{x^4} - 5{x^2} + 4} \right)dx} \cr & \,\,\,\,V = \pi \left[ {\frac{1}{5}{x^5} - \frac{5}{3}{x^3} + 4x} \right]_0^1 \cr & {\text{Evaluating}} \cr & \,\,\,\,V = \pi \left[ {\frac{1}{5}{{\left( 1 \right)}^5} - \frac{5}{3}{{\left( 1 \right)}^3} + 4\left( 1 \right)} \right] - \pi \left[ 0 \right] \cr & \,\,\,\,V = \pi \left[ {\frac{{38}}{{15}}} \right] \cr & \,\,\,\,V = \frac{{38}}{{15}}\pi \cr} $$