Question Number 46846 by maxmathsup by imad last updated on 01/Nov/18
$${calculate}\:\int\int_{{D}} \:\:\:\:\frac{{x}+{y}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}{dxdy}\:{with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}\geqslant\mathrm{0},{y}\geqslant\mathrm{0},{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} <\mathrm{1}\right\} \\ $$
Commented by maxmathsup by imad last updated on 03/Nov/18
![let consider the diffeomorphisme (r,θ)→ϕ(r,θ)=(x,y) with x =rcosθ and y =rsinθ ⇒ ∫∫_D f(x,y)dxdy =∫∫_w foϕ(r,θ)rdrdθ =∫∫_(0≤r≤1 and 0≤θ≤(π/2)) ((rcosθ +rsinθ)/( (√(1−r^2 )))) rdrdθ = ∫_0 ^1 (r^2 /( (√(1−r^2 )))) dr∫_0 ^(π/2) (cosθ +sinθ)dθ but ∫_0 ^(π/2) (cosθ +sinθ)dθ =[sinθ −cosθ]_0 ^(π/2) =1 +1 =2 also changement r =sint give ∫_0 ^1 (r^2 /( (√(1−r^2 ))))dr =∫_0 ^(π/2) ((sin^2 t)/(cost)) cost dt =∫_0 ^(π/2) ((1−cos(2t))/2)dt =(π/4) −(1/4)[sin(2t)]_0 ^(π/2) =(π/4) ⇒∫∫_D ((x+y)/( (√(1−x^2 −y^2 ))))dxdy =(π/2) .](https://www.tinkutara.com/question/Q46982.png)
$${let}\:{consider}\:{the}\:{diffeomorphisme}\:\left({r},\theta\right)\rightarrow\varphi\left({r},\theta\right)=\left({x},{y}\right)\:{with} \\ $$$${x}\:={rcos}\theta\:{and}\:{y}\:={rsin}\theta\:\Rightarrow \\ $$$$\int\int_{{D}} \:{f}\left({x},{y}\right){dxdy}\:=\int\int_{{w}} {fo}\varphi\left({r},\theta\right){rdrd}\theta\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{rcos}\theta\:+{rsin}\theta}{\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}\:{rdrd}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{r}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}\:{dr}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}\theta\:+{sin}\theta\right){d}\theta\:\:{but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}\theta\:+{sin}\theta\right){d}\theta \\ $$$$=\left[{sin}\theta\:−{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{1}\:+\mathrm{1}\:=\mathrm{2}\:\:\:{also}\:{changement}\:{r}\:={sint}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{r}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}{dr}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}^{\mathrm{2}} {t}}{{cost}}\:{cost}\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{4}}\:\Rightarrow\int\int_{{D}} \frac{{x}+{y}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}{dxdy}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$