第1个回答 2019-06-05
不太会不好意思
第2个回答 2019-06-06
解答如下图：
第3个回答 2019-06-06
x/2 + y/3 + z/4 = 1, 即 2x + (4/3)y + z = 4, z = 4-2x-(4/3)y,
∂z/∂x = -2, ∂z/∂y = -4/3,
dS = √[1+(∂z/∂x)^2+(∂z/∂y)^2]dxdy = √[1+4+(4/3)^2] = (√61/3)dxdy
I = ∫∫<∑>(2x+4y/3+z)dS = ∫∫<D> 4 (√61/3)dxdy
= (4/3)√61(1/2)*2*3 = 8√61本回答被网友采纳
∂z/∂x = -2, ∂z/∂y = -4/3,
dS = √[1+(∂z/∂x)^2+(∂z/∂y)^2]dxdy = √[1+4+(4/3)^2] = (√61/3)dxdy
I = ∫∫<∑>(2x+4y/3+z)dS = ∫∫<D> 4 (√61/3)dxdy
= (4/3)√61(1/2)*2*3 = 8√61本回答被网友采纳
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