Question

利用球面坐标计算三重积分∫∫∫z√x²+y²+z²dv,

Answer 1

令x=rsinψcosθ,y=rsinψsinθ,z=rcosψ
那么
∫∫∫√(x²+y²+z²)dxdydz
=∫∫∫(r*r²sinψ)drdψdθ
=∫∫∫(r³sinψ)drdψdθ
积分区域：
由x²+y²+z²≤x得：0≤r≤sinψcosθ
0≤ψ≤π,-π/2≤θ≤π/2
∫∫∫(r³sinψ)drdψdθ
=∫dθ∫dψ∫(r³sinψ)dr
=(1/4)*∫(cosθ)^4dθ*∫(sinψ)^5dψ
在0≤ψ≤π上∫(sinψ)^5dψ,相当于0≤ψ≤π/2上2∫(sinψ)^5dψ=2*(4/5)*(2/3)=16/15
在-π/2≤θ≤π/2上∫(cosθ)^4dθ,相当于0≤ψ≤π/2上2∫(cosθ)^4dθ=2*(3/4)*(1/2)*(π/2)=3π/8
故,原式=(1/4)*∫(cosθ)^4dθ*∫(sinψ)^5dψ=(1/4)*(3π/8)*(16/15)=π/10