Chapter 14 - Multiple Integration - 14.3 Exercises - Page 991: 23

$$\frac{2}{3}$$

$$\eqalign{ & \int_0^2 {\int_0^{\sqrt {2x - {x^2}} } {xy} dydx} \cr & R{\text{ has the bounds}} \cr & 0 \leqslant y \leqslant \sqrt {2x - {x^2}} ,{\text{ }}0 \leqslant x \leqslant 2 \cr & y = \sqrt {2x - {x^2}} \cr & {y^2} = 2x - {x^2} \cr & {x^2} + {y^2} = 2x \cr & {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 2r\cos \theta \cr & {r^2} = 2r\cos \theta \cr & r = 2\cos \theta \cr & {\text{The region }}R{\text{ in polar coordinates is}} \cr & R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 2\cos \theta ,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2}} \right\} \cr & {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr & {\text{So}},{\text{ we have}} \cr & \int_0^2 {\int_0^{\sqrt {2x - {x^2}} } {xy} dydx} = \int_0^{\pi /2} {\int_0^{2\cos \theta } {\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)rdrd\theta } } \cr & = \int_0^{\pi /2} {\int_0^{2\cos \theta } {{r^3}\cos \theta \sin \theta drd\theta } } \cr & {\text{Integrate with respect to }}r \cr & = \int_0^{\pi /2} {\left[ {\frac{{{r^4}}}{4}\cos \theta \sin \theta } \right]_0^{2\cos \theta }d\theta } \cr & = \int_0^{\pi /2} {\left[ {\frac{{{{\left( {2\cos \theta } \right)}^4}}}{4}\cos \theta \sin \theta - \frac{{{{\left( 0 \right)}^4}}}{4}\cos \theta \sin \theta } \right]d\theta } \cr & = \int_0^{\pi /2} {\frac{{16{{\cos }^5}\theta }}{4}\sin \theta d\theta } \cr & = - 4\int_0^{\pi /2} {{{\cos }^5}\theta \left( { - \sin \theta } \right)d\theta } \cr & {\text{Integrate}} \cr & = - 4\left[ {\frac{{{{\cos }^6}\theta }}{6}} \right]_0^{\pi /2} \cr & = - \frac{2}{3}\left[ {{{\cos }^6}\left( {\frac{\pi }{2}} \right) - {{\cos }^6}\left( 0 \right)} \right] \cr & = - \frac{2}{3}\left( { - 1} \right) \cr & = \frac{2}{3} \cr} $$