Calculate ∮ (x 2-2y) dx+(3x+ye y) dy, where L is the boundary of the region D surrounded by a straight line y=0, x+2y=2 and an arc x 2+y 2 = 1, and the direction is counterclockwise.

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)？