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: 5

$$V = \left( {1 - \frac{{\sqrt 2 }}{2}} \right)\pi $$

$$\eqalign{ & {\text{From the graph we have }} \cr & y = f\left( x \right) = \sqrt {\cos x} {\text{ on the interval }}\left[ {\frac{\pi }{4},\frac{\pi }{2}} \right]{\text{ and the }}x{\text{ axis}} \cr & {\text{The volume of the solid can be calculated using}} \cr & \,\,\,\,V = \int_a^b {\pi {{\left[ {f\left( x \right)} \right]}^2}dx} \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,V = \int_{\pi /4}^{\pi /2} {\pi {{\left[ {\sqrt {\cos x} } \right]}^2}dx} \cr & \,\,\,\,V = \pi \int_{\pi /4}^{\pi /2} {\cos xdx} \cr & \,\,\,\,V = \pi \left[ {\sin x} \right]_{\pi /4}^{\pi /2} \cr & {\text{Evaluating}} \cr & \,\,\,\,V = \pi \left[ {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( {\frac{\pi }{4}} \right)} \right] \cr & \,\,\,\,V = \pi \left[ {1 - \frac{{\sqrt 2 }}{2}} \right] \cr & \,\,\,\,V = \left( {1 - \frac{{\sqrt 2 }}{2}} \right)\pi \cr} $$