Question Number 61533 by maxmathsup by imad last updated on 04/Jun/19
![∫∫_([0,1]^2 ) ((x−y)/((x^2 +3y^(2 ) +1)^2 )) dxdy](https://www.tinkutara.com/question/Q61533.png)
$$\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\:\:\frac{{x}−{y}}{\left({x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}\:} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dxdy}\: \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
![let use the diffeomorphism x=rcosθ and y =(r/( (√3))) sinθ 0≤x≤1 and 0≤y≤1 ⇒0≤x^2 +y^2 ≤2 ⇒0≤r^2 ≤2 ⇒ 0≤r≤(√2) (r,θ)→ϕ(r,θ) =(x,y)=(ϕ_1 (r,θ),ϕ_2 (r,θ))=(rcosθ,(r/( (√3)))sinθ) M_j (ϕ) = ((((∂ϕ_1 /∂r) (∂ϕ_1 /∂θ))),(((∂ϕ_2 /∂r) (∂ϕ_2 /∂θ))) ) = (((cosθ −rsinθ)),((((sinθ)/( (√3))) (r/( (√3))) cosθ)) ) ⇒detM_j = (r/( (√3))) ⇒A=∫∫_([0,1]^2 ) ((x−y)/((x^2 +3y^2 +1)^2 ))dxdy =∫∫_(0≤r≤(√2) and 0≤θ ≤(π/2)) ((rcosθ −(r/( (√3)))sinθ)/((r^2 +1)^2 )) (r/( (√3)))dr dθ =(1/( (√3))) ∫_0 ^(√2) (r^2 /((r^2 +1)^2 ))dr ∫_0 ^(π/2) (cosθ −sinθ)dθ ∫_0 ^(√2) (r^2 /((r^2 +1)^2 )) dr =∫_0 ^(√2) ((r^2 +1−1)/((r^2 +1)^2 )) dr =∫_0 ^(√2) (dr/(1+r^2 )) −∫_0 ^(√2) (dr/((1+r^2 )^2 )) =arctan((√2))−∫_0 ^(√2) (dr/((1+r^2 )^2 )) ∫_0 ^(√2) (dr/((1+r^2 )^2 )) =_(r=tanu) ∫_0 ^(arctan((√2))) ((1+tan^2 u)/((1+tan^2 u)^2 ))du =∫_0 ^(arctan((√2))) (du/(1+tan^2 u)) =∫_0 ^(arctan((√2))) cos^2 u du =(1/2) ∫_0 ^(arctan((√2))) (1+cos(2u))du =((arctan((√2)))/2) +(1/4)[sin(2u)]_0 ^(arctan((√2))) =((arctan((√2)))/2) +(1/4)sin(2arctan((√2)) ∫_0 ^(π/2) (cosθ −sinθ)dθ =[sinθ +cosθ]_0 ^(π/2) =1−1 =0 ⇒ A =0 .](https://www.tinkutara.com/question/Q61588.png)
$${let}\:{use}\:{the}\:{diffeomorphism}\:\:{x}={rcos}\theta\:{and}\:{y}\:=\frac{{r}}{\:\sqrt{\mathrm{3}}}\:{sin}\theta\:\: \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\mathrm{0}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}} \\ $$$$\left({r},\theta\right)\rightarrow\varphi\left({r},\theta\right)\:=\left({x},{y}\right)=\left(\varphi_{\mathrm{1}} \left({r},\theta\right),\varphi_{\mathrm{2}} \left({r},\theta\right)\right)=\left({rcos}\theta,\frac{{r}}{\:\sqrt{\mathrm{3}}}{sin}\theta\right) \\ $$$${M}_{{j}} \left(\varphi\right)\:=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial\theta}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial\theta}}\end{pmatrix}\:\:\:\:\:\:=\begin{pmatrix}{{cos}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:−{rsin}\theta}\\{\frac{{sin}\theta}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{\:\sqrt{\mathrm{3}}}\:{cos}\theta}\end{pmatrix} \\ $$$$\Rightarrow{detM}_{{j}} =\:\frac{{r}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow{A}=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\:\:\frac{{x}−{y}}{\left({x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dxdy}\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:{and}\:\:\mathrm{0}\leqslant\theta\:\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{rcos}\theta\:−\frac{{r}}{\:\sqrt{\mathrm{3}}}{sin}\theta}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\frac{{r}}{\:\sqrt{\mathrm{3}}}{dr}\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{r}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\left({cos}\theta\:−{sin}\theta\right){d}\theta \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{r}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dr}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{r}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dr}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{dr}}{\mathrm{1}+{r}^{\mathrm{2}} }\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$={arctan}\left(\sqrt{\mathrm{2}}\right)−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{{r}={tanu}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} \:\:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {u}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right)^{\mathrm{2}} }{du}\:=\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} \:\:\frac{{du}}{\mathrm{1}+{tan}^{\mathrm{2}} {u}} \\ $$$$=\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} {cos}^{\mathrm{2}} {u}\:{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} \left(\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)\right){du} \\ $$$$=\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} =\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\mathrm{2}}\right)\right. \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}\theta\:−{sin}\theta\right){d}\theta\:=\left[{sin}\theta\:+{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{1}−\mathrm{1}\:=\mathrm{0}\:\Rightarrow\:{A}\:=\mathrm{0}\:. \\ $$
Commented by maxmathsup by imad last updated on 05/Jun/19
![error from line 6 I =(1/3) ∫_0 ^(√2) (r^2 /((1+r^2 )^2 ))dr ∫_0 ^(π/2) ((√3)cosθ −sinθ)dθ ∫_0 ^((πu)/2) ( (√3)cosθ −sinθ)dθ =[(√3)sinθ +cosθ]_0 ^(π/2) =(√3)−1 ⇒ I =(((√3)−1)/3) ∫_0 ^(√2) ((r^2 dr)/((1+r^2 )^2 )) =(((√3)−1)/3) { arctan((√2))−(1/2) arctan((√2))−(1/4)sin(2arctan((√2)))} ⇒ I =(((√3)−1)/3){ ((arctan((√2)))/2) −(1/4)sin(2arctan((√2))} .](https://www.tinkutara.com/question/Q61598.png)
$${error}\:{from}\:{line}\:\mathrm{6}\: \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\sqrt{\mathrm{3}}{cos}\theta\:−{sin}\theta\right){d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi{u}}{\mathrm{2}}} \left(\:\sqrt{\mathrm{3}}{cos}\theta\:−{sin}\theta\right){d}\theta\:=\left[\sqrt{\mathrm{3}}{sin}\theta\:+{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\sqrt{\mathrm{3}}−\mathrm{1}\:\Rightarrow \\ $$$${I}\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{r}^{\mathrm{2}} {dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}\:\left\{\:{arctan}\left(\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\mathrm{2}}\right)\right)\right\} \\ $$$$\Rightarrow\:{I}\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}\left\{\:\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\mathrm{2}}\right)\right\}\:.\right. \\ $$