Question

∫∫∫x^2+y^2+z^2dxdydz,Ω由x^2+y^2+z^2<=1

∫∫∫x^2+y^2+z^2dxdydz,Ω由x^2+y^2+z^2<=1

Answer 1

作球坐标变换：x=rsinucosv,y=rsinusinv,z=rcosu,
则dxdydz=r^2sinudrdudv,0<=u<=2π，0<=v<=π，
原式=∫<0,π>dv∫<0,2π>du∫<0,1>r^4*sinudr
=(π/5)∫<0,2π>sinudu
=0.
仅供参考。