calculate ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy with D={(x,y)∈R^2 /x^2 +y^2 ≤2 and x≥0 ,y≥0}


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Question Number 57323 by turbo msup by abdo last updated on 02/Apr/19
$calculate ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy with D={(x,y)∈R^2 /x^2 +y^2 ≤2 and x≥0 ,y≥0}$
$c a l c u l a t e \int \int_{D} \frac{x + y}{3 + \sqrt{x^{2} + y^{2}}} d x d y$ $w i t h D = {(x, y) \in R^{2} / x^{2} + y^{2} ⩽ 2$ $a n d x ⩾ 0, y ⩾ 0}$
Commented by maxmathsup by imad last updated on 03/Apr/19
$let use the diffeomorphism x=rcosθ and y=rsinθ we have x^2 +y^2 ≤2 ⇒ r^2 ≤2 ⇒0≤r≤(√2) also x≥0 and y ≥0 ⇒0≤θ ≤(π/2) ⇒ ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy =∫∫_(0≤r≤(√2) and 0≤θ≤(π/2)) ((r(cosθ +sinθ))/(3+r)) rdrdθ =∫_0 ^(√2) (r^2 /(3+r))dr .∫_0 ^(π/2) (cosθ +sinθ)dθ but ∫_0 ^(√2) (r^2 /(r+3)) dr =∫_0 ^(√2) ((r^2 −9 +9)/(r+3))dr =∫_0 ^(√2) (r−3)dr +9 ∫_1 ^(√2) (dr/(r+3)) =[(r^2 /2)−3r]_0 ^(√2) +9[ln∣r+3∣]_1 ^(√2) =1−3(√2) +9{ ln(3+(√2))−2ln(2)} =1−3ln(2)+9ln(3+(√2))−18ln(2) =1+9ln(3+(√2))−21ln(2) ∫_0 ^(π/2) (cosθ +sinθ)dθ =[sinθ −cosθ]_0 ^(π/2) =1−(−1) =2 ⇒ ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy =2 +18ln(3+(√2))−42ln(2).$
$l e t u s e t h e d i f f e o m o r p h i s m x = r c o s θ a n d y = r s i n θ w e h a v e x^{2} + y^{2} ⩽ 2 \Rightarrow$ $r^{2} ⩽ 2 \Rightarrow 0 ⩽ r ⩽ \sqrt{2} a l s o x ⩾ 0 a n d y ⩾ 0 \Rightarrow 0 ⩽ θ ⩽ \frac{π}{2} \Rightarrow$ $\int \int_{D} \frac{x + y}{3 + \sqrt{x^{2} + y^{2}}} d x d y = \int \int_{0 ⩽ r ⩽ \sqrt{2} a n d 0 ⩽ θ ⩽ \frac{π}{2}} \frac{r (c o s θ + s i n θ)}{3 + r} r d r d θ$ $= \int_{0}^{\sqrt{2}} \frac{r^{2}}{3 + r} d r . \int_{0}^{\frac{π}{2}} (c o s θ + s i n θ) d θ b u t$ $\int_{0}^{\sqrt{2}} \frac{r^{2}}{r + 3} d r = \int_{0}^{\sqrt{2}} \frac{r^{2} - 9 + 9}{r + 3} d r = \int_{0}^{\sqrt{2}} (r - 3) d r + 9 \int_{1}^{\sqrt{2}} \frac{d r}{r + 3}$ $= {[\frac{r^{2}}{2} - 3 r]}_{0}^{\sqrt{2}} + 9 {[l n ∣ r + 3 ∣]}_{1}^{\sqrt{2}} = 1 - 3 \sqrt{2} + 9 {l n (3 + \sqrt{2}) - 2 l n (2)}$ $= 1 - 3 l n (2) + 9 l n (3 + \sqrt{2}) - 18 l n (2) = 1 + 9 l n (3 + \sqrt{2}) - 21 l n (2)$ $\int_{0}^{\frac{π}{2}} (c o s θ + s i n θ) d θ = {[s i n θ - c o s θ]}_{0}^{\frac{π}{2}} = 1 - (- 1) = 2 \Rightarrow$ $\int \int_{D} \frac{x + y}{3 + \sqrt{x^{2} + y^{2}}} d x d y = 2 + 18 l n (3 + \sqrt{2}) - 42 l n (2) .$
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