Question Number 43538 by abdo.msup.com last updated on 11/Sep/18
$${calculate}\:\int\int_{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\leqslant\mathrm{1}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}\:{whit} \\ $$$${a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}\:. \\ $$
Commented by maxmathsup by imad last updated on 16/Sep/18
$${let}\:{consider}\:{the}\:{diffeomorphism}\:\left({r},\theta\right)\:\rightarrow\left({x},{y}\right)=\left({arcos}\theta,{brsin}\theta\right) \\ $$$${I}\:=\:\int\int_{\mathrm{0}\leqslant{r}\leqslant\mathrm{1}\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi} \left({a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\:−{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right){ab}\:{r}\:{dr}\:{d}\theta \\ $$$$={a}^{\mathrm{3}} {b}\int_{\mathrm{0}} ^{\mathrm{1}} {rdr}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}^{\mathrm{2}} \theta\:{d}\theta\:\:−{ab}^{\mathrm{3}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {rdr}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$=\frac{{a}^{\mathrm{3}} {b}}{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta\:−\frac{{ab}^{\mathrm{3}} }{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:{d}\theta \\ $$$${A}=\frac{\pi\:{a}^{\mathrm{3}} {b}}{\mathrm{2}}\:\:+\frac{{a}^{\mathrm{3}} {b}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}\left(\mathrm{2}\theta\right){d}\theta\:−\frac{\pi{ab}^{\mathrm{3}} }{\mathrm{2}}\:+\:\frac{{ab}^{\mathrm{3}} }{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{cos}\left(\mathrm{2}\theta\right){d}\theta \\ $$$$=\frac{\pi{ab}}{\mathrm{2}}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:+\mathrm{0}+\mathrm{0}\:\Rightarrow{A}\:=\frac{\pi{ab}}{\mathrm{2}}\left\{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \right\}\:. \\ $$