Question

高数问题，计算对弧长的曲线积分。

求高手
急用。

Answer 1

x² + y² = 9，左半圆为x = - √(9 - y²)
令x = 3cosθ，y = 3sinθ，π/2 ≤ θ ≤ 3π/2
dx/dθ = - 3sinθ，dy/dθ = 3cosθ
ds = √[x'(θ)² + y'(θ)²] dθ = √(9sin²θ + 9cos²θ) dθ = 3dθ
∫_L y² ds
= ∫(π/2-->3π/2) 9sin²θ · 3dθ
= 27/2 · ∫(π/2-->3π/2) (1 - cos2θ) dθ
= 27/2 · (θ - 1/2 · sin2θ) |(π/2-->3π/2)
= 27/2 · [(3π/2) - (π/2)]
= 27π/2

Answer 2

x^2+y^2=9
x=-√(9-y^2)
dx/dy=-y/√(9-y^2)
ds=√(dx)^2+(dy)^2=√[(dx/dy)^2+1] dy
=√[y^2/(9-y^2)+1]dy
=[3/√(9-y^2)] dy
∫y^2ds=∫[-3,3] y^2*3dy/√(9-y^2)
=3∫[-3,3]y^2dy/√(9-y^2)
=3*[(-1/2)y√(9-y^2)+(1/2)arcsin(y/3)]|[-3,3]
=3*[(1/2)*(π/2)-(-1/2)(π/2)]
=3π/2

∫y^2dy/√(9-y^2)=-∫√(9-y^2)dy+∫dy/√(9-y^2)
=-y√(9-y^2)-∫y^2dy/√(9-y^2)+∫dy/√(9-y^2)
2∫y^2dy/√(9-y^2)=-y√(9-y^2)+∫dy/√(9-y^2)
∫y^2dy/√(9-y^2)=(-1/2)y√(9-y^2)+(1/2)∫d(y/3)/√(1-(y^2/9))
=(-1/2)y√(9-y^2)+(1/2)arcsin(y/3)

Answer 3

用极坐标
ds = r dt， t是极角
y = r sin t
然后t从Pi/2积分到3Pi/2
就行了，得到
r^3 Pi/2=27Pi/2