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闭路积分假设是逆时针方向:
A(0,0); B(1,0); C(0,1)
AB: y = 0
BC: x+y = 1, ds = √2 dx
CA: x = 0
∫AB + ∫BC + ∫CA
= ∫[0,1] xdx + ∫[1,0](x+y)√2 dx + ∫[1,0]ydy
= ∫[0,1] xdx + ∫[1,0]√2 dx + ∫[1,0]ydy
= -√2
A(0,0); B(1,0); C(0,1)
AB: y = 0
BC: x+y = 1, ds = √2 dx
CA: x = 0
∫AB + ∫BC + ∫CA
= ∫[0,1] xdx + ∫[1,0](x+y)√2 dx + ∫[1,0]ydy
= ∫[0,1] xdx + ∫[1,0]√2 dx + ∫[1,0]ydy
= -√2
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