Double Integral Region

Question

∫∫_R (2xy)/(x²+y²)dA here R is the region in the first quadrant bounded by x²+y²= 4, x²+ y²= 1, y = x/√3 , and x = 0

Answer 1

∫∫_R (2xy)/(x²+y²)dA; x = rcosθ, y = rsinθ; dA = dxdy = rdrdθ; D: ∫1 ≤ r ≤ 2; π/6 ≤ θ ≤ π/2

∫∫_R (2xy)/(x²+y²)dA =∫∫_D(2rcoθrsinθrdrdθ/r² = ∫_π/6π^π/2∫₁²rsin2θdrdθ = (r²/2)₁²·(-1/2cos2θ)_π/6^π/2 = 3/2(- cosπ + cosπ/3)/2 = 3/2·3/4 = 9/8