∫∫(D) (y^2) dσ,其中D是圆 x^2+y^2=1 和 x^2+y^2=4(π^2)之间的环形区域

极坐标
x^2 + y^2 = 1 ==> r = 1
x^2 + y^2 = 4π^2 ==> r = 2π
∫∫Dy^2 dxdy
= ∫(0 , 2π)∫(1 , 2π) (r * sinθ)^2 rdrdθ
= ∫(0 , 2π) (1 - cos2θ)/2 dθ * ∫(1 , 2π) r^3 dr
= [(1/2)θ - (1/4)sin2θ] |(0 , 2π) * r^4/4 |(1 , 2π)
= π * (4π^4- 1/4)
= 4π^5 - π/4