Chapter 14 - Multiple Integration - 14.3 Exercises - Page 991: 20

$$\frac{3}{{32}}\pi $$

$$\eqalign{ & \int_0^1 {\int_{ - \sqrt {x - {x^2}} }^{\sqrt {x - {x^2}} } {\left( {{x^2} + {y^2}} \right)} dy} dx \cr & R{\text{ has the bounds}} \cr & - \sqrt {x - {x^2}} \leqslant y \leqslant \sqrt {x - {x^2}} ,{\text{ }}0 \leqslant x \leqslant 1 \cr & y = \sqrt {x - {x^2}} \cr & {y^2} = x - {x^2} \cr & {x^2} + {y^2} = x \cr & {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = r\cos \theta \cr & {r^2} = r\cos \theta \cr & r = \cos \theta \cr & r = \pm \cos \theta \cr & {\text{The region }}R{\text{ in polar coordinates is}} \cr & R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant \cos \theta ,{\text{ 0}} \leqslant \theta \leqslant \pi } \right\} \cr & {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr & {\text{So}},{\text{ we have}} \cr & \int_0^1 {\int_{ - \sqrt {x - {x^2}} }^{\sqrt {x - {x^2}} } {\left( {{x^2} + {y^2}} \right)} dy} dx = \int_0^\pi {\int_0^{\cos \theta } {\left( {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } \right)} rdr} d\theta \cr & = \int_0^\pi {\int_0^{\cos \theta } {{r^3}} dr} d\theta \cr & {\text{Integrate with respect to }}r \cr & = \int_0^\pi {\left[ {\frac{{{r^4}}}{4}} \right]_0^{\cos \theta }} d\theta \cr & = \int_0^\pi {\left[ {\frac{{{{\left( {\cos \theta } \right)}^4}}}{4}} \right]} d\theta \cr & = \frac{1}{4}\int_0^\pi {{{\cos }^4}\theta } d\theta \cr & {\text{Integrating we obtain}} \cr & = \frac{1}{4}\left( {\frac{3}{8}\pi } \right) \cr & = \frac{3}{{32}}\pi \cr} $$