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求∫ds,其中L為圓周x²;+y²;=4解(一):ds是弧長的微分;ds=√(dx²;+dy²;)=√[1+(dy/dx)²;]dx=[√(1+y′²;)]dx將x²;+y²;=4對x取導數得:2x+2yy′=0,故y′=-x/y.∴[L]∫ds =[-2,]2∫√(…
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