Question

∫∫∫(x+y+z)dxdydz 积分区域Ω是由四个平面：x=0、y=0、z=0和x+y+z=1围成的。

帮忙解一下，答案好像是1/8
所围的立体整个表面外侧

Answer 1

Ω就是0<x<1，0<y<1-x，0<z<1-x-y
∫∫∫(x+y+z)dxdydz
= ∫(0,1)dx∫(0,1-x)dy∫(0,1-x-y)(x+y+z)dz
= ∫(0,1)dx∫(0,1-x)dy[x(1-x-y) + y(1-x-y) + (1-x-y)²/2]
= ∫(0,1)dx [(1-x)(1-x²)/2 - x(1-x)²/2 - (1-x)³/6]
= [(x^4)/24 - x²/4+ x/3]|(0,1)
= 1/8

Answer 2

=∫（0,1）dx∫（0,1）dy∫（0,1-x-y）x+y+zdz
=∫（0,1）dx∫（0,1）-xy-1/2x^2-1/2y^2+1/2dy
=∫（0,1）-1/2x-1/2x^2+1/3dx
=-1/12