Question Number 78703 by abdomathmax last updated on 20/Jan/20
$${let}\:{a}>\mathrm{0}\:\:{calculate}\:\int\int_{{D}_{{a}} } \:\:\:\:\frac{{xdxdy}}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} } \\ $$$${and}\:{D}_{{a}} =\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} \:\:{and}\:{x}>\mathrm{0}\right\} \\ $$
Answered by mind is power last updated on 20/Jan/20
![x=rcos(θ),y=rsin(θ) D_a ⇔{(r,θ)∈R^2 /r^2 <a^2 cos(θ)>0} r∈[0,a],θ∈[−(π/2),(π/2)[ ∫∫_D_a ((xdxdy)/(a^2 +x^2 +y^2 ))=∫_((−π)/2) ^(π/2) ∫_0 ^a ((r^2 cos(θ)drdθ)/(a^2 +r^2 )) =∫_(−(π/2)) ^(π/2) cos(θ)dθ.∫_0 ^a (r^2 /(a^2 +r^2 ))dr =[sin(θ)]_(−(π/2)) ^(π/2) .[1−(a^2 /(a^2 +r^2 ))]_0 ^a 2[x−aarctan((r/a))]_0 ^a 2(a−a(π/4))=a(((4−π)/2))](https://www.tinkutara.com/question/Q78742.png)
$$\mathrm{x}=\mathrm{rcos}\left(\theta\right),\mathrm{y}=\mathrm{rsin}\left(\theta\right) \\ $$$$\mathrm{D}_{\mathrm{a}} \Leftrightarrow\left\{\left(\mathrm{r},\theta\right)\in\mathbb{R}^{\mathrm{2}} /\mathrm{r}^{\mathrm{2}} <\mathrm{a}^{\mathrm{2}} \mathrm{cos}\left(\theta\right)>\mathrm{0}\right\} \\ $$$$\mathrm{r}\in\left[\mathrm{0},\mathrm{a}\right],\theta\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$$\int\int_{\mathrm{D}_{\mathrm{a}} } \frac{\mathrm{xdxdy}}{\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{a}} \frac{\mathrm{r}^{\mathrm{2}} \mathrm{cos}\left(\theta\right)\mathrm{drd}\theta}{\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} } \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\theta\right)\mathrm{d}\theta.\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\mathrm{dr} \\ $$$$=\left[\mathrm{sin}\left(\theta\right)\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} .\left[\mathrm{1}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{a}} \\ $$$$\mathrm{2}\left[\mathrm{x}−\mathrm{aarctan}\left(\frac{\mathrm{r}}{\mathrm{a}}\right)\right]_{\mathrm{0}} ^{\mathrm{a}} \mathrm{2}\left(\mathrm{a}−\mathrm{a}\frac{\pi}{\mathrm{4}}\right)=\mathrm{a}\left(\frac{\mathrm{4}−\pi}{\mathrm{2}}\right) \\ $$