计算∫L(2xy^3-y2cosx)dx+(1-2ysinx+3x2y2)dy,其中L为在抛物线2x=πy2上由(0,0)到(π/2,1)的一段弧

Question

计算∫L(2xy^3-y2cosx)dx+(1-2ysinx+3x2y2)dy,其中L为在抛物线2x=πy2上由(0,0)到(π/2,1)的一段弧

Answer 1

补线S：x = π/2、y由1变到0
补线T：y = 0、x由π/2变到0
∮(L+S+T) (2xy³ - y²cosx)dx + (1 - 2ysinx + 3x²y²)dy
= ∫∫D [ ∂/∂x (1 - 2ysinx + 3x²y²) - ∂/∂y (2xy³ - y²cosx) ] dxdy
= ∫∫D [ (- 2ycosx + 6xy²) - (6xy² - 2ycosx) ] dxdy
= 0
∫S (2xy³ - y²cosx)dx + (1 - 2ysinx + 3x²y²)dy
= ∫(1→0) (1 - 2y * 1 + 3 * π²/4 * y²) dy
= - π²/4
∫T (2xy³ - y²cosx)dx + (1 - 2ysinx + 3x²y²)dy
= ∫(π/2→0) 0 dx
= 0
==> ∫L (2xy³ - y²cosx)dx + (1 - 2ysinx + 3x²y²)dy
= 0 - (- π²/4) - 0
= π²/4