在极坐标中r = sinθ从0到2π的弧长是π吗?

2024-12-16 05:44:32
推荐回答(1个)
回答1:

s=∫ {[r'(θ)]^2+[r(θ)]^2}^(1/2)dθ
r=sinθ,r'=cosθ
s=∫(0,2π) {[r'(θ)]^2+[r(θ)]^2}^(1/2)dθ=∫(0,2π) dθ=2π