求助高数题!!
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dx=-3acos²θsinθdθ
dy=3asin²θcosθdθ
dl=√[(dx)²十(dy)²]
=√[9a²cos^4(θ)sin²θ十9a²sin^4(θ)cos²θ]dθ
=3asinθcosθ√[cos²θ十sin²θ]dθ
=(3/2)asin2θdθ
l=∫dl
=(3a/2)∫(0,π/2)sin2θdθ
=(3a/4)[-cos2θ](0,π/2)
=-(3a/4)[-1-1]
=3a/2
dy=3asin²θcosθdθ
dl=√[(dx)²十(dy)²]
=√[9a²cos^4(θ)sin²θ十9a²sin^4(θ)cos²θ]dθ
=3asinθcosθ√[cos²θ十sin²θ]dθ
=(3/2)asin2θdθ
l=∫dl
=(3a/2)∫(0,π/2)sin2θdθ
=(3a/4)[-cos2θ](0,π/2)
=-(3a/4)[-1-1]
=3a/2
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