Question Number 101220 by Ar Brandon last updated on 01/Jul/20

∫∫_D (√(x^2 +y^2 ))dxdy   D= { (((x,y)∈R, x^2 +y^2 ≥2y, x^2 +y^2 ≤1)),((x≥0 , y≥0)) :}

Answered by Ar Brandon last updated on 03/Jul/20

     I=∫∫_D (√(x^2 +y^2 ))dxdy , D= { (((x,y) ∈R, x^2 +y^2 ≥2y, x^2 +y^2 ≤1)),((x≥0 , y≥0)) :} {: (),() }  En coordonne^� es polaires;   { ((x=rcosθ)),((y=rsinθ)) :}  ⇒   { ((r^2 ≥2rsinθ)),((r^2 ≤1)) :}  ⇒2sinθ≤r≤1  x^2 +y^2 ≥2y ⇒ r^2 ≥2rsinθ ⇒(1/2)≥sinθ ⇒ (π/6)≥θ≥0  ⇒I=∫_0 ^(π/6) ∫_(2sinθ) ^1 r^2 drdθ=∫_0 ^(π/6) [(r^3 /3)]_(2sinθ) ^1 dθ=∫_0 ^(π/6) [(1/3)−((8sin^3 θ)/3)]dθ  ⇒3I=∫_0 ^(π/6) (1−8sin^3 θ)dθ=[θ]_0 ^(π/6) −8∫_0 ^(π/6) (1−cos^2 θ)sinθdθ             =(π/6)+8[cosθ−((cos^3 θ)/3)]_0 ^(π/6) =(π/6)+8[((√3)/2)−((√3)/8)−1+(1/3)]             =(π/6)+8(((3(√3))/8)−(2/3))=(π/6)+3(√3)−((16)/3)  ⇒I=(π/(18))+(√3)−((16)/9)