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高数题求大佬解答
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解:由题可知,Ω可以表示为:
0≤x≤1/2,
0≤y≤1-2x,
0≤z≤1-2x-y,
所以所求三重积分可计算如下:
ʃʃʃΩ xdxdydz
=ʃ₀¹⸍²xdxʃ₀¹ᐨ²ˣdyʃ₀¹ᐨ²ˣᐨʸdz
=ʃ₀¹⸍²xdxʃ₀¹ᐨ²ˣ(1-2x-y)dy
=ʃ₀¹⸍²x[(1-2x)y-y²/2]₀¹ᐨ²ˣdx
=ʃ₀¹⸍²x[(1-2x)²/2]dx
=ʃ₀¹⸍²(x/2-2x²+2x³)dx
=(x²/4-2x³/3+x⁴/2)|₀¹⸍²
=1/16-1/12+1/32
=1/96 .
0≤x≤1/2,
0≤y≤1-2x,
0≤z≤1-2x-y,
所以所求三重积分可计算如下:
ʃʃʃΩ xdxdydz
=ʃ₀¹⸍²xdxʃ₀¹ᐨ²ˣdyʃ₀¹ᐨ²ˣᐨʸdz
=ʃ₀¹⸍²xdxʃ₀¹ᐨ²ˣ(1-2x-y)dy
=ʃ₀¹⸍²x[(1-2x)y-y²/2]₀¹ᐨ²ˣdx
=ʃ₀¹⸍²x[(1-2x)²/2]dx
=ʃ₀¹⸍²(x/2-2x²+2x³)dx
=(x²/4-2x³/3+x⁴/2)|₀¹⸍²
=1/16-1/12+1/32
=1/96 .
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∫∫∫<Ω>xdxdydz = ∫<0, 1>dz∫<0, 1-z>dy∫<0, (1-y-z)/2>xdx
= (1/2)∫<0, 1>dz∫<0, 1-z>dy[x^2]<0, (1-y-z)/2>
= (1/8)∫<0, 1>dz∫<0, 1-z>(1-y-z)^2dy
= (-1/24)∫<0, 1>dz[(1-y-z)^3]<0, 1-z>
= (1/24)∫<0, 1>(1-z)^4dz
= (-1/120)[(1-z)^5]<0, 1> = 1/120
= (1/2)∫<0, 1>dz∫<0, 1-z>dy[x^2]<0, (1-y-z)/2>
= (1/8)∫<0, 1>dz∫<0, 1-z>(1-y-z)^2dy
= (-1/24)∫<0, 1>dz[(1-y-z)^3]<0, 1-z>
= (1/24)∫<0, 1>(1-z)^4dz
= (-1/120)[(1-z)^5]<0, 1> = 1/120
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