计算弧长积分:∫(x^2+y^2-2x+1)^nds,其中L为圆周x^2+y^2-2x=0

Question

计算弧长积分:∫(x^2+y^2-2x+1)^nds,其中L为圆周x^2+y^2-2x=0

Answer 1

∵x^2+y^2-2x=0 ==>y^2=2x-x^2
==>y=±√(2x-x^2)
==>y'=±(1-x)/√(2x-x^2)
∴ds=√[1+(y')^2]dx=dx/√(2x-x^2)
故 ∫(x^2+y^2-2x+1)^nds=2∫[1^n/√(2x-x^2)]dx
=2∫dx/√(2x-x^2)]
=2∫dx/√[1-(x-1)^2)]
=2∫dt (令x-1=sint,并化简)
=2(π/2+π/2)
=2π.