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Problems of Green's Formula in Higher Mathematics
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)?