在极坐标下D={(r,θ)|0≤r≤-2asinθ，-π/4≤θ≤0}
解法一：∫∫(√(x²+y²))/(√(4a²-x²-y²)) dσ
=∫(-π/4,0)dθ∫(0,-2asinθ)[r²/√(4a²-r²)]dr=∫(-π/4,0)dθ∫(0,-2asinθ){[4a²-(4a²-r²)]/√(4a²-r²)}dr=∫(-π/4,0)dθ{∫(0,-2asinθ)[4a²/√(4a²-r²)]dr-∫(0,-2asinθ)√(4a²-r²)dr}.........①记大括号中的值为I，则I=∫(0,-2asinθ)[4a²/√(4a²-r²)]dr-∫(0,-2asinθ)√(4a²-r²)dr=4a²*arcsin(r/2a)|(0,-2asinθ)-∫(0,-2asinθ)√(4a²-r²)dr=-4a²θ-∫(0,-2asinθ)√(4a²-r²)dr....................................................②
对于积分∫(0,-2asinθ)√(4a²-r²)dr的解法有两种：方法一：套用不定积分公式∫√(a²-x²)dx=(a²/2)arcsin(x/a)+(1/2)x√(a²-x²)+C（其证明过程有两种，一种是换元法，另一种是分部积分法，证明从略）所以，∫(0,-2asinθ)√(4a²-r²)dr=[(4a²)/2]arcsin(r/2a)+(1/2)r√(4a²-r²)=-2a²θ-a²sin2θ代入②，得I=-4a²θ-[-2a²θ-a²sin2θ]=-2a²θ+a²sin2θ，代入①，得∫∫(√(x²+y²))/(√(4a²-x²-y²)) dσ=∫(-π/4,0)(-2a²θ+a²sin2θ)dθ=(π²/16-1/2)a²方法二：利用圆的几何性质，令s=√(4a²-r²)，则r²+s²=4a²，则该定积分可以看成是以(0,0)为圆心，以2a为半径的圆内r∈(0，2asin(-θ)的面积，如图阴影部分的面积即为所求。
该阴影部分可分为三角形和扇形，S三角形=(1/2)*2asin(-θ)*2acosθ)=-a²sin2θS扇形=（1/2)*(2a)²*(π/2+θ)=πa²+2a²θ所以，∫(0,-2asinθ)√(4a²-r²)dr=S阴影=S三角形+S扇形=πa²+2a²θ-a²sin2θ，代入①得∫∫(√(x²+y²))/(√(4a²-x²-y²)) dσ=∫(-π/4,0)(πa²+2a²θ-a²sin2θ)dθ=(π²/16-1/2)a²
解法二：∫∫(√(x²+y²))/(√(4a²-x²-y²)) dσ
=∫(-π/4,0)dθ∫(0,-2asinθ)[r²/√(4a²-r²)]dr=∫(-π/4,0)dθ∫(0,-θ)[(4a²sin²u)/√(4a²cos²u)]*2acosudu（令r=2asinu）=4a²∫(-π/4,0)dθ∫(0,-θ)sin²udu=4a²∫(-π/4,0)dθ∫(0,-θ)[(1-cos2u)/2]du=4a²∫(-π/4,0)[(-1/2)θ+(1/4)sin2θ)]dθ=(π²/16-1/2)a²