设是由平面x+y+z=1及三坐标平面围成的区域,则∫∫∫(x+y+z)dv=
问题描述：
设是由平面x+y+z=1及三坐标平面围成的区域,则∫∫∫(x+y+z)dv=
答
x->(0,1）
y->(0,1-x)
z->(0,1-x-y)
=>
∫∫∫(x+y+z)dv=∫(x:0,1)∫(y:0,1-x)∫(z:0,1-x-y)(x+y+z)dzdydx
=∫(x:0,1)∫(y:0,1-x)[(x+y)(1-x-y)+(1-x-y)^2/2]dydx
=1/2*∫(x:0,1)∫(y:0,1-x)[1-(x+y)^2]dydx
=1/2*∫(x:0,1)[1-x -1/3*(x+1-x)^3+1/3*x^3]dx
=1/6*∫(x:0,1)[2-3x+x^3]dx
=1/8