∫( e^x sin y- y )dx + (e^x cos y - 1)dy,是(2,0)的半圆周y=√2x-x^2麻烦您了
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∫( e^x sin y- y )dx + (e^x cos y - 1)dy,是(2,0)的半圆周y=√2x-x^2麻烦您了
∫( e^x sin y- y )dx + (e^x cos y - 1)dy,是(2,0)的半圆周y=√2x-x^2
麻烦您了
∫( e^x sin y- y )dx + (e^x cos y - 1)dy,是(2,0)的半圆周y=√2x-x^2麻烦您了
利用格林公式:∮cP(x,y)dx+Q(x,y)dy=∫∫D(dQ/dx-dP/dy)dxdy
首先需要构造封闭曲线.
∫(x沿半圆周y=√2x-x^2从2积到0)( e^x sin y- y )dx + (e^x cos y - 1)dy+
∫(x沿x轴从0积到2)( e^x sin y- y )dx + (e^x cos y - 1)dy=
∫(x沿半圆周y=√2x-x^2从2积到0)( e^x sin y- y )dx + (e^x cos y - 1)dy+
∫( e^x sin 0- 0)dx + (e^x cos 0- 1)d0=
∫(x沿半圆周y=√2x-x^2从2积到0)( e^x sin y- y )dx + (e^x cos y - 1)dy=
=∫∫D(pQ/px-pP/py)dxdy (D为该圆的上半圆)
=∫∫D(e^x cosy-e^x cosy+1)dxdy=∫∫D dxdy=S(D)=π1^2/2=π/2
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