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Question Number 34714 by abdo mathsup 649 cc last updated on 10/May/18

calculate ∫∫_(x^2 +2y^2 ≤1) (x^2 −y^2 )dxdy

$${calculate}\:\int\int_{{x}^{\mathrm{2}} \:+\mathrm{2}{y}^{\mathrm{2}} \:\leqslant\mathrm{1}} \left({x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} \right){dxdy} \\ $$

Commented by math khazana by abdo last updated on 10/May/18

let consider the diffeomorphism (r,θ)→(x,y)=(rcosθ,(r/(√2))sinθ)=(ϕ_1 (r,θ),ϕ_2 (r,θ)) M_j = ((((∂ϕ_1 /∂r) (∂ϕ_1 /∂θ))),(((∂ϕ_2 /∂r) (∂ϕ_2 /∂θ))) ) = (((cosθ −rsinθ)),((((sinθ)/(√2)) (r/(√2))cosθ)) ) and det(M_j )= (r/(√2)) cos^2 θ +(r/(√2)) sin^2 θ = (1/(√2)) r I = ∫∫_(0≤r≤1 , −π≤θ≤π) ( r^2 cos^2 θ −(r^2 /2) sin^2 θ)(r/(√2))drdθ I = (1/(√2)) ∫_0 ^1 r^3 dr ∫_(−π) ^π cos^2 θ dθ −(1/(2(√2))) ∫_0 ^1 r^3 dr ∫_(−π) ^π sin^2 θdθ = (1/(4(√2))) ∫_(−π) ^π cos^2 θ dθ −(1/(8(√2))) ∫_(−π) ^π sin^2 θ dθ but ∫_(−π) ^π cos^2 θ dθ = 2 ∫_0 ^π ((1+cos(2θ))/2)dθ = ∫_0 ^π ( 1+cos(2θ))dθ = π ∫_(−π) ^π sin^2 θdθ = 2 ∫_0 ^π ((1−cos(2θ))/2)dθ = ∫_0 ^π (1−cos(2θ))dθ = π ⇒ I = (π/(4(√2))) −(π/(8(√2))) = ((2π−π)/(8(√2))) = (π/(8(√2))) I = (π/(8(√2))) .

$${let}\:{consider}\:{the}\:{diffeomorphism} \\ $$$$\left({r},\theta\right)\rightarrow\left({x},{y}\right)=\left({rcos}\theta,\frac{{r}}{\sqrt{\mathrm{2}}}{sin}\theta\right)=\left(\varphi_{\mathrm{1}} \left({r},\theta\right),\varphi_{\mathrm{2}} \left({r},\theta\right)\right) \\ $$$${M}_{{j}} =\:\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial\theta}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial\theta}}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{{cos}\theta\:\:\:\:\:\:\:\:\:\:\:−{rsin}\theta}\\{\frac{{sin}\theta}{\sqrt{\mathrm{2}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{\sqrt{\mathrm{2}}}{cos}\theta}\end{pmatrix}\:\:\:\:\:\:{and}\:{det}\left({M}_{{j}} \right)= \\ $$$$\frac{{r}}{\sqrt{\mathrm{2}}}\:{cos}^{\mathrm{2}} \theta\:+\frac{{r}}{\sqrt{\mathrm{2}}}\:{sin}^{\mathrm{2}} \theta\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{r} \\ $$$${I}\:\:=\:\int\int_{\mathrm{0}\leqslant{r}\leqslant\mathrm{1}\:\:,\:\:\:−\pi\leqslant\theta\leqslant\pi} \left(\:{r}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\:−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\:{sin}^{\mathrm{2}} \theta\right)\frac{{r}}{\sqrt{\mathrm{2}}}{drd}\theta \\ $$$${I}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{r}^{\mathrm{3}} {dr}\:\int_{−\pi} ^{\pi} \:{cos}^{\mathrm{2}} \theta\:{d}\theta\:\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{r}^{\mathrm{3}} {dr}\:\int_{−\pi} ^{\pi} \:{sin}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\int_{−\pi} ^{\pi} \:{cos}^{\mathrm{2}} \theta\:{d}\theta\:\:−\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{2}}}\:\int_{−\pi} ^{\pi} \:\:{sin}^{\mathrm{2}} \theta\:{d}\theta\:\:{but} \\ $$$$\int_{−\pi} ^{\pi} \:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta\: \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \left(\:\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\:=\:\pi \\ $$$$\int_{−\pi} ^{\pi} \:{sin}^{\mathrm{2}} \theta{d}\theta\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta\: \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\left(\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\:=\:\pi\:\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:\:−\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:\:=\:\frac{\mathrm{2}\pi−\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}} \\ $$$${I}\:\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:. \\ $$

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