计算曲线积分：
∫(L) (2xy^3 - y^2cosx) dx + (1 - 2ysinx + 3x^2y^2) dy
其中L是在抛物线2x = πy^2上由点(0,0)到(π/2,1)的一段弧.
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补线：
L1：x = π/2、逆时针方向、dx = 0、由y = 0变化到y = 1
L2：y = 0、逆时针方向、dy = 0、由x = 0变化到x = π/2
由于L是顺时针方向,现在设L⁻是L的逆时针方向
∮(L⁻+L1+L2) (2xy^3 - y^2cosx) dx + (1 - 2ysinx + 3x^2y^2) dy
= ∫∫D [∂/∂x (1 - 2ysinx + 3x^2y^2) - ∂/∂y (2xy^3 - y^2cosx)] dxdy、用Green公式
= ∫∫D [(- 2ycosx + 6xy^2) - (6xy^2 - 2ycosx)] dxdy
= ∫∫D (- 2ycosx + 6xy^2 - 6xy^2 + 2ycosx) dxdy
= 0
而∫(L1) (2xy^3 - y^2cosx) dx + (1 - 2ysinx + 3x^2y^2) dy
= ∫(0→1) [0 + 1 - 2y + 3(π/2)^2y^2] dy
= ∫(0→1) [1 - 2y + (3/4)π^2 * y^2] dy
= y - y^2 + (3/4)π^2 * (1/3)y^3：(0→1)
= 1 - 1 + (3/4)π^2 * 1/3
= (1/4)π^2
而∫(L2) (2xy^3 - y^2cosx) dx + (1 - 2ysinx + 3x^2y^2) dy
= ∫(L2) 0 dx
= 0
于是∫(L⁻) + ∫(L1) + ∫(L2) = ∮(L⁻+L1+L2)
∫(L⁻) + (1/4)π^2 + 0 = 0
∫(L⁻) = - (1/4)π^2
∫(L) = (1/4)π^2
即原式∫(L) (2xy^3 - y^2cosx) dx + (1 - 2ysinx + 3x^2y^2) dy = (1/4)π^2