求∫(e∧xsiny-y)dx+(e∧xcosy-1)dy,其中L为点A（2,0）到点B（0,0）的

网友回答

补上线段y = 0
则令P = e^xsiny - y,dP/dy = e^xcosy - 1
Q = e^xcosy - 1,dQ/dx = e^xcosy
∫_L (e^xsiny - y) dx + (e^xcosy - 1) dy
= ∫∫_D [(e^xcosy) - (e^xcosy - 1)] dxdy
= ∫∫_D dxdy
= 1/2 • π(1)²
= π/2