Solution: green's formula: [C]∮Pdx+Qdy=[C]∫∫ (? Q/? x-? P/? y)dxdy,P=x? -2y; Q=3x+ye^y.
Among them? Q/? x = 3; ? P/? y =-2; Replaced:
[c]∮pdx+qdy=[c]∫∫(3+2)dxdy=[d]5∫∫dxdy
Will the linear equation x=2-2y get (2-2y)? +y? =4-8y+5y? = 1, which means there is 5y? -8y+3=(5y-3)(y- 1)=0
Therefore, the coordinate of the intersection of a straight line and a circle is A(4/5, 3/5); B(0, 1)。
The integral ∫∫dxdy is the area of the graph FOECAB = the area of the sector FOB+the area of the triangle BOC.
=π/4+( 1/2)×2× 1=π/4+ 1
∴[d]∮(x^2-2y)dx+(3x+ye^y)dy=[c]∫∫(3+2)dxdy=[d]5∫∫dxdy=5(π/4+ 1)?