计算曲线积分∫L(e∧xsiny－2y)dx＋(e∧xcosy－2)dy,其中L为上半圆周(x－a²)＋y²＝a²,y≥0.沿逆时针方向.

Question

计算曲线积分∫L(e∧xsiny－2y)dx＋(e∧xcosy－2)dy,其中L为上半圆周(x－a²)＋y²＝a²,y≥0.沿逆时针方向.

Answer 1

连接上半圆周(x-a)²+y²=a²(y≥0)的直径L1：y=0,0≤x≤2a.
于是,根据格林定理得
∫(e^xsiny-2y)dx+(e^xcosy-2)dy+∫(e^xsiny-2y)dx+(e^xcosy-2)dy=2∫∫dxdy
(S表示L+L1所围成的区域)
∵∫(e^xsiny-2y)dx+(e^xcosy-2)dy=0
∫∫dxdy=πa²/2
∴∫(e^xsiny-2y)dx+(e^xcosy-2)dy+0=2(πa²/2)
故∫(e^xsiny-2y)dx+(e^xcosy-2)dy=πa².