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记 D1 中除去 D11 部分后剩余部分是 D12, 则
2∫∫<D1>√|x-y| dxdy
= 2∫∫<D11>√|x-y| dxdy + 2∫∫<D12>√|x-y| dxdy
= 2∫<0, 1>dx∫<0, x>√(x-y)dy + 2∫<0, 1>dx∫<x, 1>√(y-x)dy
= -2∫<0, 1>dx∫<0, x>√(x-y)d(x-y) + 2∫<0, 1>dx∫<x, 1>√(y-x)d(y-x)
= -2∫<0, 1>dx[(2/3)(x-y)^(3/2)]<0, x> + 2∫<0, 1>dx[(2/3)(y-x)^(3/2)]<x, 1>
= (4/3)∫<0, 1>x^(3/2)dx + (4/3)∫<0, 1>(1-x)^(3/2)dx
= (4/3)[(2/5)x^(5/2)]<0, 1> + (-4/3)[(2/5)(1-x)^(5/2)]<0, 1>
= 8/15 + 8/15 (说明前后两个积分结果相同)
= 16/15
2∫∫<D1>√|x-y| dxdy
= 2∫∫<D11>√|x-y| dxdy + 2∫∫<D12>√|x-y| dxdy
= 2∫<0, 1>dx∫<0, x>√(x-y)dy + 2∫<0, 1>dx∫<x, 1>√(y-x)dy
= -2∫<0, 1>dx∫<0, x>√(x-y)d(x-y) + 2∫<0, 1>dx∫<x, 1>√(y-x)d(y-x)
= -2∫<0, 1>dx[(2/3)(x-y)^(3/2)]<0, x> + 2∫<0, 1>dx[(2/3)(y-x)^(3/2)]<x, 1>
= (4/3)∫<0, 1>x^(3/2)dx + (4/3)∫<0, 1>(1-x)^(3/2)dx
= (4/3)[(2/5)x^(5/2)]<0, 1> + (-4/3)[(2/5)(1-x)^(5/2)]<0, 1>
= 8/15 + 8/15 (说明前后两个积分结果相同)
= 16/15
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