Question Number 190052 by Mastermind last updated on 26/Mar/23
$$\mathrm{Evaluate}\:\int\int_{\mathrm{A}} \left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \mathrm{dxdy}\:\mathrm{over}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ellipse}\: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Anybody}? \\ $$
Answered by CElcedricjunior last updated on 26/Mar/23
![โซโซ_A (x+y)^2 dxdy A={(x;y);(x^2 /a^2 )+(y^2 /b^2 )=1} En changeant les coordonnees de cartesienne a^ polaire on a : { ((x=arcos๐)),((y=brsin๐)) :}1>r>0 ๐โ[0;2๐] et dxdy=โฃ determinant (((acos๐ โarsin๐)),((bsin๐ brcos๐)))โฃdrd๐ dxdy=abrdrd๐ A={(r;๐) 0<r<1;0โค๐โค2๐} โ Moivre โซโซ_A abr^3 (acos๐+bsin๐)^2 drd๐=k k=โซ_0 ^(2๐) [((ab)/4)r^4 (acos๐+bsin๐)^2 ]_0 ^1 d๐ =((ab)/4)โซ_0 ^(2๐) (a^2 cos๐^2 +2abcos๐sin๐+b^2 sin^2 ๐)d๐ ((ab)/4)[a^2 ((1/2)๐+(1/4)sin2๐)โ((abcos2๐)/2)+b^2 ((1/2)๐โ(1/4)sin2๐)]_0 ^(2๐) =((ab)/4)(๐(a^2 +b)) โ
Cedric junior =(๐/4)ab(a^2 +b^2 )](https://www.tinkutara.com/question/Q190071.png)
$$\int\int_{\boldsymbol{{A}}} \left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\mathrm{2}} \boldsymbol{\mathrm{dxdy}} \\ $$$$\boldsymbol{\mathrm{A}}=\left\{\left(\boldsymbol{{x}};\boldsymbol{{y}}\right);\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }+\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }=\mathrm{1}\right\} \\ $$$$\boldsymbol{{En}}\:\boldsymbol{{changeant}}\:\boldsymbol{{les}}\:\boldsymbol{{coordonnees}}\:\boldsymbol{{de}}\:\boldsymbol{{cartesienne}}\:\grave {\boldsymbol{{a}}}\:\boldsymbol{{polaire}} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\::\begin{cases}{\boldsymbol{{x}}=\boldsymbol{{arcos}\theta}}\\{\boldsymbol{{y}}=\boldsymbol{{brsin}\theta}}\end{cases}\mathrm{1}>\boldsymbol{{r}}>\mathrm{0}\:\boldsymbol{\theta}\in\left[\mathrm{0};\mathrm{2}\boldsymbol{\pi}\right] \\ $$$$\boldsymbol{{et}}\:\boldsymbol{{dxdy}}=\mid\begin{vmatrix}{\boldsymbol{{acos}\theta}\:\:\:\:\:\:\:โ\boldsymbol{{arsin}\theta}}\\{\boldsymbol{{bsin}\theta}\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{brcos}\theta}}\end{vmatrix}\mid\boldsymbol{{drd}\theta} \\ $$$$\boldsymbol{{dxdy}}=\boldsymbol{{abrdrd}\theta} \\ $$$${A}=\left\{\left(\boldsymbol{{r}};\boldsymbol{\theta}\right)\:\mathrm{0}<\boldsymbol{{r}}<\mathrm{1};\mathrm{0}\leqslant\boldsymbol{\theta}\leqslant\mathrm{2}\boldsymbol{\pi}\right\}\:\blacksquare\boldsymbol{{M}}{oivre} \\ $$$$\int\int_{\boldsymbol{{A}}} \boldsymbol{{abr}}^{\mathrm{3}} \left(\boldsymbol{{acos}\theta}+\boldsymbol{{bsin}\theta}\right)^{\mathrm{2}} \boldsymbol{{drd}\theta}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\mathrm{2}\boldsymbol{\pi}} \left[\frac{\boldsymbol{{ab}}}{\mathrm{4}}\boldsymbol{{r}}^{\mathrm{4}} \left(\boldsymbol{{acos}\theta}+\boldsymbol{{bsin}\theta}\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{d}\theta} \\ $$$$=\frac{\boldsymbol{{ab}}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{2}\boldsymbol{\pi}} \left(\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{cos}}\overset{\mathrm{2}} {\boldsymbol{\theta}}+\mathrm{2}\boldsymbol{{abcos}\theta{sin}\theta}+\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{si}}\overset{\mathrm{2}} {\boldsymbol{{n}}\theta}\right)\boldsymbol{{d}\theta} \\ $$$$\frac{\boldsymbol{{ab}}}{\mathrm{4}}\left[\boldsymbol{{a}}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\theta}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta}\right)โ\frac{\boldsymbol{{abcos}}\mathrm{2}\boldsymbol{\theta}}{\mathrm{2}}+\boldsymbol{{b}}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\theta}โ\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta}\right)\right]_{\mathrm{0}} ^{\mathrm{2}\boldsymbol{\pi}} \\ $$$$=\frac{\boldsymbol{{ab}}}{\mathrm{4}}\left(\boldsymbol{\pi}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)\right)\:\bigstar\mathscr{C}{edric}\:{junior} \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{4}}\boldsymbol{{ab}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \right) \\ $$