Question Number 51998 by maxmathsup by imad last updated on 01/Jan/19
$${let}\:{U}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{1}\leqslant{x}^{\mathrm{2}} \:+\mathrm{2}{y}^{\mathrm{2}} \leqslant\mathrm{3}\right\} \\ $$$${calculate}\:\int\int_{{U}} \:\:\:\:\frac{{x}−{y}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdxy} \\ $$
Commented by Abdo msup. last updated on 05/Jan/19
![let consider the diffeomorphism (r,θ)→ϕ(r,θ)=(x,y) with x=rcosθ and y =(r/( (√2)))sinθ 1≤x^2 +2y^2 ≤3 ⇒1 ≤r^2 ≤3 ⇒1≤r≤(√3) and 0≤θ≤2π ⇒ ∫∫_U ((x−y)/(x^2 +y^2 ))dxdy =∫∫_(1≤r≤(√3)and 0≤θ≤2π) ((r(cosθ−((sinθ)/( (√2)))))/(r^2 cos^2 θ +(r^2 /2)sin^2 θ))rdrdθ ∫_1 ^(√3) dr ∫_0 ^(2π) (((√2)cosθ −sinθ)/(2cos^2 θ +sin^2 θ))2dθ =2((√3)−1) ∫_0 ^(2π) (((√2)cosθ −sinθ)/(1+cos^2 θ)) dθ but ∫_0 ^(2π) (((√2)cosθ −sinθ)/(1+cos^2 θ)) dθ =(√2)∫_0 ^(2π) ((cosθ)/(1+cos^2 θ))dθ +∫_0 ^(2π) ((−sinθ)/(1+cos^2 θ)) dθ ∫_0 ^(2π) ((−sinθ)/(1+cos^2 θ)) dθ =[arctan(cosθ)]_0 ^(2π) =0 also ∫_0 ^(2π) ((cosθ)/(1+cos^2 θ)) dθ =∫_0 ^π ((cosθ)/(1+cos^2 θ)) +∫_π ^(2π) ((cosθ)/(1+cos^2 θ))dθ ∫_π ^(2π) ((cosθ)/(1+cos^2 θ)) dθ =_(θ =π +t) ∫_0 ^π ((−cost)/(1+cos^2 t)) dt ⇒ ∫_0 ^(2π) ((cosθ)/(1+cos^2 θ)) dθ =0 ⇒ ∫∫_U ((x−y)/(x^2 +y^2 )) dxdy =0](https://www.tinkutara.com/question/Q52241.png)
$${let}\:{consider}\:{the}\:{diffeomorphism} \\ $$$$\left({r},\theta\right)\rightarrow\varphi\left({r},\theta\right)=\left({x},{y}\right)\:{with}\:{x}={rcos}\theta\:{and}\:{y}\:=\frac{{r}}{\:\sqrt{\mathrm{2}}}{sin}\theta \\ $$$$\mathrm{1}\leqslant{x}^{\mathrm{2}} \:+\mathrm{2}{y}^{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow\mathrm{1}\:\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{3}}\:{and}\: \\ $$$$\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi\:\Rightarrow \\ $$$$\int\int_{{U}} \frac{{x}−{y}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy}\:=\int\int_{\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{3}}{and}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi} \frac{{r}\left({cos}\theta−\frac{{sin}\theta}{\:\sqrt{\mathrm{2}}}\right)}{{r}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\:+\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{sin}^{\mathrm{2}} \theta}{rdrd}\theta \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {dr}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\sqrt{\mathrm{2}}{cos}\theta\:−{sin}\theta}{\mathrm{2}{cos}^{\mathrm{2}} \theta\:+{sin}^{\mathrm{2}} \theta}\mathrm{2}{d}\theta \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\sqrt{\mathrm{2}}{cos}\theta\:−{sin}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\sqrt{\mathrm{2}}{cos}\theta\:−{sin}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta\:\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{cos}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}{d}\theta \\ $$$$+\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{−{sin}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{−{sin}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta\:=\left[{arctan}\left({cos}\theta\right)\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$=\mathrm{0}\:\:{also} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cos}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cos}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{cos}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}{d}\theta \\ $$$$\int_{\pi} ^{\mathrm{2}\pi} \:\:\frac{{cos}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta\:=_{\theta\:=\pi\:+{t}} \:\:\int_{\mathrm{0}} ^{\pi} \:\frac{−{cost}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cos}\theta}{\mathrm{1}+{cos}^{\mathrm{2}} \theta}\:{d}\theta\:=\mathrm{0}\:\Rightarrow \\ $$$$\int\int_{{U}} \:\frac{{x}−{y}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:{dxdy}\:=\mathrm{0} \\ $$$$ \\ $$