∫Lyds,其中L为心形线r＝a(1+cosθ)的下半部分.

Question

∫L y ds ,其中L为心形线r＝a(1+cosθ)的下半部分.

Answer 1

心形线下半部分,θ：π---->2π
ds=√(r²+(r')²) dθ
=√(a²(1+cosθ)²+(a²sin²θ)) dθ
=a√(1+2cosθ+cos²θ+sin²θ) dθ
=a√(2+2cosθ) dθ
=a√[4cos²(θ/2)] dθ
=2a√[cos²(θ/2)] dθ
由于θ：π---->2π,则θ/2：π/2---->π,余弦值为负
=-2acos(θ/2) dθ
∫L yds
=-∫[π---->2π] rsinθ*2acos(θ/2) dθ
=-2a∫[π---->2π] a(1+cosθ)sinθcos(θ/2) dθ
=-2a²∫[π---->2π] 2cos²(θ/2)*(1/2)sin(θ/2)cos(θ/2)cos(θ/2) dθ
=-2a²∫[π---->2π] cos⁴(θ/2)sin(θ/2)dθ
=-4a²∫[π---->2π] cos⁴(θ/2)sin(θ/2)d(θ/2)
=4a²∫[π---->2π] cos⁴(θ/2)d(cos(θ/2))
=(4a²/5)*cos⁵(θ/2) |[π---->2π]
=-4a²/5-0
=-4a²/5