Question

求r=asin^3(θ/3)弧长，0<=θ<3π

Answer 1

你好
dr=a*3sin^2(θ/3)*1/3dθ=asin^2(θ/3)dθ
弧长=∫θdr=∫(上3π下0)asin^2(θ/3)θdθ
∫asin^2(θ/3)θdθ
=a/2∫(1-cos2θ/3)θdθ
=a/2*θ^2/2-a/2*3/2∫θd(sin2θ/3)
=a/2*θ^2/2-3a/4sin2θ/3*θ+3a/4∫sin2θ/3dθ
=aθ^2/4-3a/4sin2θ/3*θ-3a/4*3/2cos2θ/3+c
=aθ^2/4-3a/4sin2θ/3*θ-9a/8cos2θ/3+c
弧长=∫(上3π下0)asin^2(θ/3)θdθ
=aθ^2/4-3a/4sin2θ/3*θ-9a/8cos2θ/3│上3π下0
=9aπ^2/4
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