ex2计算∫(x+y+z)S,2是由x=0y=0,z=0, ∑ x+y+z=1所围成的四面体的整个边界曲面 Solution 21:x=0,D12:0≤z≤1-y,0≤y≤1;dS=z; ∑2:y=0,D2:0≤z≤1-x,0≤x≤1;dS=4x y 23:z=0,D:0≤y≤1-x,0≤x≤1;eh; 24:z=1-x-y,Dy:0≤y≤1-x,0≤x≤1;S=3y ∫(x+y+z)ds=』+∫+∫+』=3+∫ ∑ ∑ ∑ ∑ 4 3』(x+p)bd+3c=1xp y K心1 . 2. ( ) , 0, 0, 0, 所围成的四面体的整个边界曲面 计算 是由 + + =  + +  = = =  x y z ex x y z dS x y z Solution. y z o x : 0, 1 x = D : 0  z  1− y,0  y  1; yz : 0, 2 y = D : 0  z  1− x,0  x  1; zx : 0, 3 z = D : 0  y  1− x,0  x  1; xy : 1 , 4  z = − x − y D : 0  y  1− x,0  x  1; xy            + + = + + + 1 2 3 4 (x y z)dS     = + 3 4 3 =  + +  Dxy Dxy 3 (x y)dxdy 3dxdy . 2 3 = 1+ dS = dydz; dS = dzdx; dS = dxdy; dS = 3dxdy;