Question

计算∫∫xzdxdy,其中∑为平面x+y+z=1在第I卦限的部分的上侧

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Answer 1

Σ为z = 1 - x - y
∫∫Σ xz dxdy
= ∫∫D x(1 - x - y) dxdy
= ∫(0→1) x dx ∫(0→1 - x) (1 - x - y) dy
= ∫(0→1) x (y - xy - y²/2)：(0→1 - x) dx
= ∫(0→1) x [ (1 - x) - x(1 - x) - (1/2)(1 - 2x + x²) ] dx
= ∫(0→1) x [ 1 - x - x + x² - 1/2 + x - x²/2 ] dx
= ∫(0→1) x [ 1/2 - x + x²/2 ] dx
= ∫(0→1) ( x/2 - x² + x³/2 ) dx
= 1/4 - 1/3 + 1/8
= 1/24