Question

使用极坐标求二重积分

Answer 1

对积分区域D作极坐标变换
0<=x<=a：0<=rcosθ<=a，0<=r<=asecθ
0<=y<=a：0<=rsinθ<=a，0<=r<=acscθ
原式=∫(0,π/4)dθ∫(0,asecθ)rdr/(a^2+r^2)^(3/2)+∫(π/4,π/2)dθ∫(0,acscθ)rdr/(a^2+r^2)^(3/2)
=(1/2)*∫(0,π/4)dθ∫(0,asecθ)d(a^2+r^2)/(a^2+r^2)^(3/2)+(1/2)*∫(π/4,π/2)dθ∫(0,acscθ)d(a^2+r^2)/(a^2+r^2)^(3/2)
=-∫(0,π/4)dθ*1/√(a^2+r^2)|(0,asecθ)-∫(π/4,π/2)dθ*1/√(a^2+r^2)|(0,acscθ)
=(1/a)*∫(0,π/4)[1-1/√(1+sec^2θ)]dθ+(1/a)*∫(π/4,π/2)[1-1/√(1+csc^2θ)]dθ
=(1/a)*[∫(0,π/4)dθ-∫(0,π/4)cosθdθ/√(cos^2θ+1)+∫(π/4,π/2)dθ-∫(π/4,π/2)sinθdθ/√(sin^2θ+1)]
=(1/a)*[π/2-∫(0,π/4)d(sinθ)/√(2-sin^2θ)+∫(π/4,π/2)d(cosθ)/√(2-cos^2θ)]
=(1/a)*[π/2-arcsin(sinθ/√2)|(0,π/4)+arcsin(cosθ/√2)|(π/4,π/2)]
=(1/a)*(π/2-π/6-π/6)
=π/6a