Question Number 57323 by turbo msup by abdo last updated on 02/Apr/19
$${calculate}\:\int\int_{{D}} \:\:\frac{{x}+{y}}{\mathrm{3}+\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\right. \\ $$$$\left.{and}\:{x}\geqslant\mathrm{0}\:,{y}\geqslant\mathrm{0}\right\} \\ $$
Commented by maxmathsup by imad last updated on 03/Apr/19
![let use the diffeomorphism x=rcosθ and y=rsinθ we have x^2 +y^2 ≤2 ⇒ r^2 ≤2 ⇒0≤r≤(√2) also x≥0 and y ≥0 ⇒0≤θ ≤(π/2) ⇒ ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy =∫∫_(0≤r≤(√2) and 0≤θ≤(π/2)) ((r(cosθ +sinθ))/(3+r)) rdrdθ =∫_0 ^(√2) (r^2 /(3+r))dr .∫_0 ^(π/2) (cosθ +sinθ)dθ but ∫_0 ^(√2) (r^2 /(r+3)) dr =∫_0 ^(√2) ((r^2 −9 +9)/(r+3))dr =∫_0 ^(√2) (r−3)dr +9 ∫_1 ^(√2) (dr/(r+3)) =[(r^2 /2)−3r]_0 ^(√2) +9[ln∣r+3∣]_1 ^(√2) =1−3(√2) +9{ ln(3+(√2))−2ln(2)} =1−3ln(2)+9ln(3+(√2))−18ln(2) =1+9ln(3+(√2))−21ln(2) ∫_0 ^(π/2) (cosθ +sinθ)dθ =[sinθ −cosθ]_0 ^(π/2) =1−(−1) =2 ⇒ ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy =2 +18ln(3+(√2))−42ln(2).](https://www.tinkutara.com/question/Q57346.png)
$${let}\:{use}\:\:{the}\:{diffeomorphism}\:\:{x}={rcos}\theta\:{and}\:{y}={rsin}\theta\:\:{we}\:{have}\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{2}\:\Rightarrow \\ $$$${r}^{\mathrm{2}} \:\leqslant\mathrm{2}\:\Rightarrow\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:\:\:{also}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\:\geqslant\mathrm{0}\:\Rightarrow\mathrm{0}\leqslant\theta\:\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$$\int\int_{{D}} \:\:\frac{{x}+{y}}{\mathrm{3}+\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}{dxdy}\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\frac{{r}\left({cos}\theta\:+{sin}\theta\right)}{\mathrm{3}+{r}}\:{rdrd}\theta \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{{r}^{\mathrm{2}} }{\mathrm{3}+{r}}{dr}\:.\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({cos}\theta\:+{sin}\theta\right){d}\theta\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{{r}^{\mathrm{2}} }{{r}+\mathrm{3}}\:{dr}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{r}^{\mathrm{2}} −\mathrm{9}\:+\mathrm{9}}{{r}+\mathrm{3}}{dr}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \left({r}−\mathrm{3}\right){dr}\:+\mathrm{9}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\frac{{dr}}{{r}+\mathrm{3}} \\ $$$$=\left[\frac{{r}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}{r}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:+\mathrm{9}\left[{ln}\mid{r}+\mathrm{3}\mid\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:=\mathrm{1}−\mathrm{3}\sqrt{\mathrm{2}}\:+\mathrm{9}\left\{\:{ln}\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)\right\} \\ $$$$=\mathrm{1}−\mathrm{3}{ln}\left(\mathrm{2}\right)+\mathrm{9}{ln}\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)−\mathrm{18}{ln}\left(\mathrm{2}\right)\:=\mathrm{1}+\mathrm{9}{ln}\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)−\mathrm{21}{ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({cos}\theta\:+{sin}\theta\right){d}\theta\:=\left[{sin}\theta\:−{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{1}−\left(−\mathrm{1}\right)\:=\mathrm{2}\:\Rightarrow \\ $$$$\int\int_{{D}} \frac{{x}+{y}}{\mathrm{3}+\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}{dxdy}\:=\mathrm{2}\:+\mathrm{18}{ln}\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)−\mathrm{42}{ln}\left(\mathrm{2}\right). \\ $$