求教高数平面曲线的弧长

求该曲线相应于指定两点间弧段的弧长...
2026年09月25日 22:54
有2个网友回答
网友(1):

y'=√(3-x²)
s=∫(-√3,√3)√(1+y'²)dx
=∫(-√3,√3)√(1+3-x²)dx
=∫(-√3,√3)√(4-x²)dx
=2∫(0,√3)√(4-x²)dx
设x=2sinv,x=0~√3,v=0~π/3
dx=2cosvdv
s=4∫(0,π/3)cosv.2cosvdv
=4∫(0,π/3)2cos²vdv
=4∫(0,π/3)(1+cos2v)dv
=4[v+sin2v/2](0,π/3)
=4[π/3+sin2π/3/2]
=4[π/3+√3/2]
=4π/3+2√3

网友(2):