求sgn(xy-1)的二重积分,D为0<=x<=2,0<=y<=2
2个回答
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注意到
sgn(x) = -1,xy<1,
= 0,xy=1,
= 1,xy>1,
把 D = D1+D2 (作图),其中,
D1 = {(x,y); xy≥1,0≤x≤2,0≤y≤2},
D2 = {(x,y); xy≤1,0≤x≤2,0≤y≤2},
这样,
∫∫(D)sgn(xy-1)dxdy
= ∫∫(D1)dxdy + ∫∫(D2)(-1)dxdy
= ∫∫(D1)dxdy - ∫∫(D2)dxdy
= ∫∫(D1)dxdy - [∫∫(D)dxdy - ∫∫(D1)dxdy]
= 2∫∫(D1)dxdy - 4
= 2∫[1/2,2]dx∫[1/x,2]dy - 4
= ……(留给你)
sgn(x) = -1,xy<1,
= 0,xy=1,
= 1,xy>1,
把 D = D1+D2 (作图),其中,
D1 = {(x,y); xy≥1,0≤x≤2,0≤y≤2},
D2 = {(x,y); xy≤1,0≤x≤2,0≤y≤2},
这样,
∫∫(D)sgn(xy-1)dxdy
= ∫∫(D1)dxdy + ∫∫(D2)(-1)dxdy
= ∫∫(D1)dxdy - ∫∫(D2)dxdy
= ∫∫(D1)dxdy - [∫∫(D)dxdy - ∫∫(D1)dxdy]
= 2∫∫(D1)dxdy - 4
= 2∫[1/2,2]dx∫[1/x,2]dy - 4
= ……(留给你)
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