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Question Number 43538 by abdo.msup.com last updated on 11/Sep/18

calculate ∫∫_((x^2 /a^2 ) +(y^2 /b^2 ) ≤1) (x^2 −y^2 )dxdy whit a>0 and b>0 .

$${calculate}\:\int\int_{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\leqslant\mathrm{1}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}\:{whit} \\ $$ $${a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}\:. \\ $$

Commented bymaxmathsup by imad last updated on 16/Sep/18

let consider the diffeomorphism (r,θ) →(x,y)=(arcosθ,brsinθ) I = ∫∫_(0≤r≤1 and 0≤θ≤2π) (a^2 cos^2 θ −b^2 sin^2 θ)ab r dr dθ =a^3 b∫_0 ^1 rdr ∫_0 ^(2π) cos^2 θ dθ −ab^3 ∫_0 ^1 rdr ∫_0 ^(2π) sin^2 θ dθ =((a^3 b)/2) ∫_0 ^(2π) ((1+cos(2θ))/2)dθ −((ab^3 )/2) ∫_0 ^(2π) ((1−cos(2θ))/2) dθ A=((π a^3 b)/2) +((a^3 b)/4) ∫_0 ^(2π) cos(2θ)dθ −((πab^3 )/2) + ((ab^3 )/4) ∫_0 ^(2π) cos(2θ)dθ =((πab)/2)(a^2 −b^2 ) +0+0 ⇒A =((πab)/2){a^2 −b^2 } .

$${let}\:{consider}\:{the}\:{diffeomorphism}\:\left({r},\theta\right)\:\rightarrow\left({x},{y}\right)=\left({arcos}\theta,{brsin}\theta\right) \\ $$ $${I}\:=\:\int\int_{\mathrm{0}\leqslant{r}\leqslant\mathrm{1}\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi} \left({a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\:−{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right){ab}\:{r}\:{dr}\:{d}\theta \\ $$ $$={a}^{\mathrm{3}} {b}\int_{\mathrm{0}} ^{\mathrm{1}} {rdr}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}^{\mathrm{2}} \theta\:{d}\theta\:\:−{ab}^{\mathrm{3}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {rdr}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}^{\mathrm{2}} \theta\:{d}\theta \\ $$ $$=\frac{{a}^{\mathrm{3}} {b}}{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta\:−\frac{{ab}^{\mathrm{3}} }{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:{d}\theta \\ $$ $${A}=\frac{\pi\:{a}^{\mathrm{3}} {b}}{\mathrm{2}}\:\:+\frac{{a}^{\mathrm{3}} {b}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}\left(\mathrm{2}\theta\right){d}\theta\:−\frac{\pi{ab}^{\mathrm{3}} }{\mathrm{2}}\:+\:\frac{{ab}^{\mathrm{3}} }{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{cos}\left(\mathrm{2}\theta\right){d}\theta \\ $$ $$=\frac{\pi{ab}}{\mathrm{2}}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:+\mathrm{0}+\mathrm{0}\:\Rightarrow{A}\:=\frac{\pi{ab}}{\mathrm{2}}\left\{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \right\}\:. \\ $$

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