Chapter 14 - Multiple Integrals - 14.3 Double Integrals In Polar Coordinates - Exercises Set 14.3 - Page 1024: 6

$$\frac{3}{{64}}\pi $$

$$\eqalign{ & \int_0^{\pi /2} {\int_0^{\cos \theta } {{r^3}} } drd\theta \cr & = \int_0^{\pi /2} {\left[ {\int_0^{\cos \theta } {{r^3}} dr} \right]} d\theta \cr & {\text{solve the inner integral and treat }}\theta {\text{ as a constant}} \cr & = \int_0^{\cos \theta } {{r^3}} dr \cr & {\text{then}} \cr & = \left[ {\frac{{{r^4}}}{4}} \right]_0^{\cos \theta } \cr & {\text{evaluating the limits in the variable }}r \cr & = \frac{1}{4}\left[ {{{\cos }^4}\theta - {0^4}} \right] \cr & = \frac{1}{4}{\cos ^4}\theta \cr & {\text{then}} \cr & \int_0^{\pi /2} {\left[ {\int_0^{\cos \theta } {{r^3}} dr} \right]} d\theta = \int_0^{\pi /2} {\frac{1}{4}{{\cos }^4}\theta } d\theta \cr & = \frac{1}{4}\int_0^{\pi /2} {{{\cos }^4}\theta } d\theta \cr & = \frac{1}{4}\int_0^{\pi /2} {{{\left( {{{\cos }^2}\theta } \right)}^2}} d\theta \cr & {\text{use the identity co}}{{\text{s}}^2}x = \frac{{1 + \cos 2x}}{2} \cr & = \frac{1}{4}\int_0^{\pi /2} {{{\left( {\frac{{1 + \cos 2\theta }}{2}} \right)}^2}} d\theta \cr & = \frac{1}{4}\int_0^{\pi /2} {\left( {\frac{{1 + 2\cos 2\theta + {{\cos }^2}2\theta }}{4}} \right)} d\theta \cr & = \frac{1}{{16}}\int_0^{\pi /2} {\left( {1 + 2\cos 2\theta + {{\cos }^2}2\theta } \right)} d\theta \cr & = \frac{1}{{16}}\int_0^{\pi /2} {\left( {1 + 2\cos 2\theta + \frac{1}{2} + \frac{{\cos 4\theta }}{2}} \right)} d\theta \cr & = \frac{1}{{16}}\int_0^{\pi /2} {\left( {\frac{3}{2} + 2\cos 2\theta + \frac{{\cos 4\theta }}{2}} \right)} d\theta \cr & {\text{integrating }} \cr & = \frac{1}{{16}}\left( {\frac{3}{2}\theta + \sin 2\theta + \frac{1}{8}\sin 4\theta } \right)_0^{\pi /2} \cr & {\text{evaluate}} \cr & = \frac{1}{{16}}\left( {\frac{3}{2}\left( {\frac{\pi }{2}} \right) + \sin 2\left( {\frac{\pi }{2}} \right) + \frac{1}{8}\sin 4\left( {\frac{\pi }{2}} \right)} \right) - \frac{1}{{16}}\left( {\frac{3}{2}\left( 0 \right) + \sin 2\left( 0 \right) + \frac{1}{8}\sin 4\left( 0 \right)} \right) \cr & = \frac{1}{{16}}\left( {\frac{3}{2}\left( {\frac{\pi }{2}} \right) + 0} \right) - \frac{1}{{16}}\left( 0 \right) \cr & = \frac{3}{{64}}\pi \cr} $$