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: 14

$$V = \frac{1}{2}\pi $$

$$\eqalign{ & {\text{We have that }}f\left( x \right) = \cos x,\,\,\,\,g\left( x \right) = \sin x,\,\,\,\,\,x = 0,\,\,\,x = \pi /4 \cr & {\text{Where }} \cr & \cos x \geqslant \sin x{\text{ on the interval }}\left( {0,\frac{\pi }{4}} \right) \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_a^b {\pi \left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_0^{\pi /4} {\left( {{{\left[ {\cos x} \right]}^2} - {{\left[ {\sin x} \right]}^2}} \right)dx} \cr & V = \pi \int_0^{\pi /4} {\left( {{{\cos }^2}x - {{\sin }^2}x} \right)dx} \cr & {\text{Using the trig}}{\text{. identity cos2}}\theta = {\cos ^2}\theta - {\sin ^2}\theta \cr & V = \pi \int_0^{\pi /4} {\cos 2xdx} \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{1}{2}\sin 2x} \right]_0^{\pi /4} \cr & V = \frac{\pi }{2}\left[ {\sin 2\left( {\frac{\pi }{4}} \right) - \sin 2\left( 0 \right)} \right] \cr & V = \frac{\pi }{2}\left[ {1 - 0} \right] \cr & V = \frac{1}{2}\pi \cr} $$