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)