计算I=三重积分z^2dxdydz,其中积分区域为x^2+y^2+z^2<a^2,x^2+y^2+(z-a)^2<=a^2的公共部分！

Question

计算I=三重积分z^2dxdydz,其中积分区域为x^2+y^2+z^2<a^2,x^2+y^2+(z-a)^2<=a^2的公共部分！ fenge2009@qq.com

Answer 1

解：原式=∫<0,2π>dθ∫<0,√3a/2>rdr∫<a-√(a²-r²),√(a²-r²)>z²dz (做柱面坐标变换)
=(2π/3)∫<0,√3a/2>[(a²-r²)^(3/2)-(a-√(a²-r²))³]rdr
=(π/3)∫<0,√3a/2>[a³-3a²(a²-r²)^(1/2)+3a(a²-r²)-2(a²-r²)^(3/2)]d(a²-r²)
=(π/3)[a³(a²-r²)-2a²(a²-r²)^(3/2)+(3/2)a(a²-r²)²-(4/5)(a²-r²)^(5/2)]│<0,√3a/2>
=(π/3)(a^5/4-a^5/4+3a^5/32-a^5/40-a^5+2a^5-3a^5/2+4a^5/5)
=59π/480。