Question Number 42395 by abdo.msup.com last updated on 24/Aug/18
$${calculate}\:\int\int_{{D}} \:\:\:\:\:\:\frac{{xy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}{dxdy}\:{with} \\ $$$${D}\:=\left\{\left({x},{y}\right)\in\:{R}^{\mathrm{2}} \:\:/\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:,\mathrm{0}\leqslant{y}\leqslant\mathrm{1},\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{1}\right\} \\ $$
Commented by maxmathsup by imad last updated on 25/Aug/18
![let consider the diffeomorphisme (r,θ) →(x,y) =(rcosθ,rsinθ) ∫∫_D f(x,y)dxdy = ∫∫_W foϕ(t) ∣J_ϕ ∣ dr dθ = ∫∫_(0≤θ≤(π/(2 )) and 0≤r≤1) ((r^2 cosθ sinθ)/((1+r^2 ))) rdr dθ =∫_0 ^(π/2) ( ∫_0 ^1 (r^3 /(1+r^2 ))dr)cosθ sinθ dθ = ∫_0 ^1 (r^3 /(1+r^2 ))dr .((1/2) ∫_0 ^(π/2) sin(2θ)dθ) but ∫_0 ^(π/2) sin(2θ)dθ =[(1/2)cos(2θ)]_0 ^(π/2) =(1/2)(−2) =−1 and changement r =tanα give ∫_0 ^1 (r^3 /(1+r^2 )) dr = ∫_0 ^(π/4) ((tan^3 α)/((1+tan^2 α))) (1+tan^2 α)dα =∫_0 ^(π/4) tan^3 α dα let A =∫_0 ^(π/4) tan^3 t dt A=∫_0 ^(π/4) (1+tan^2 t −1)tant dt =∫_0 ^(π/4) (1+tan^2 t)tantdt −∫_0 ^(π/4) tant dt =[tant tant]_0 ^(π/4) −∫_0 ^(π/4) tant(1+tan^2 t)dt −∫_0 ^(π/4) tant dt =1−2 ∫_0 ^(π/4) tant dt −∫_0 ^(π/4) tan^3 t dt ⇒2A =1−2 ∫_0 ^(π/4) tant dt ⇒ A =(1/2) −∫_0 ^(π/4) ((sint)/(cost)) dt =(1/2) +[ln∣cost∣]_0 ^(π/4) =(1/2) +ln((1/(√2))) =(1/2) −(1/2)ln(2) ⇒ ∫∫_D ((xy)/((1+x^2 +y^2 )))dx dy =((ln(2))/2) −(1/2) . error at the final line ∫∫ ((xy)/(1+x^2 +y^2 ))dxdy =(1/2)(((ln(2))/2) −(1/2)) =((ln(2))/4) −(1/4) .](Q42444.png)
$${let}\:{consider}\:{the}\:{diffeomorphisme}\:\left({r},\theta\right)\:\rightarrow\left({x},{y}\right)\:=\left({rcos}\theta,{rsin}\theta\right) \\ $$$$\int\int_{{D}} {f}\left({x},{y}\right){dxdy}\:=\:\int\int_{{W}} {fo}\varphi\left({t}\right)\:\mid{J}_{\varphi} \mid\:{dr}\:{d}\theta \\ $$$$=\:\int\int_{\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}\:}\:\:\:{and}\:\:\mathrm{0}\leqslant{r}\leqslant\mathrm{1}} \:\:\frac{{r}^{\mathrm{2}} \:{cos}\theta\:{sin}\theta}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)}\:{rdr}\:{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left(\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{r}^{\mathrm{3}} }{\mathrm{1}+{r}^{\mathrm{2}} }{dr}\right){cos}\theta\:{sin}\theta\:{d}\theta\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{r}^{\mathrm{3}} }{\mathrm{1}+{r}^{\mathrm{2}} }{dr}\:.\left(\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left(\mathrm{2}\theta\right){d}\theta\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}\left(\mathrm{2}\theta\right){d}\theta\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}\right)\:=−\mathrm{1}\:{and}\:\:{changement}\:{r}\:={tan}\alpha\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{r}^{\mathrm{3}} }{\mathrm{1}+{r}^{\mathrm{2}} }\:{dr}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{tan}^{\mathrm{3}} \alpha}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right){d}\alpha \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{\mathrm{3}} \alpha\:{d}\alpha\:\:\:\:{let}\:\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{\mathrm{3}} {t}\:{dt} \\ $$$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\:−\mathrm{1}\right){tant}\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){tantdt}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tant}\:{dt} \\ $$$$=\left[{tant}\:{tant}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tant}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}\:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tant}\:{dt} \\ $$$$=\mathrm{1}−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tant}\:{dt}\:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{\mathrm{3}} {t}\:{dt}\:\Rightarrow\mathrm{2}{A}\:=\mathrm{1}−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tant}\:{dt}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sint}}{{cost}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+\left[{ln}\mid{cost}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:+{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\int\int_{{D}} \:\:\:\frac{{xy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}{dx}\:{dy}\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$${error}\:{at}\:{the}\:{final}\:{line}\:\:\:\int\int\:\:\:\:\frac{{xy}}{\mathrm{1}+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$