大一高数定积分面积应用题
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先求交点:
3cosθ=1+cosθ,cosθ=1/2,θ=±π/3
ρ=3cosθ,是直径3的圆,ρ=1+cosθ是心形线,关于极轴对称。
重合部分,±π/3之间,外轮廓为心形线,面积:
S1=2∫(0,π/3)0.5ρ²dθ
=∫(0,π/3)(1+cosθ)²dθ
=∫(0,π/3)(1+2cosθ+cos²)dθ
=∫(0,π/3)(1+2cosθ+(1+cos2θ)/2)dθ
=(3/2.θ+2sinθ+sin2θ/4)|(0,π/3)
=(3/2)(π/3)+√3+√3/8
=π/2+9√3/8
余下是两个弓形部分,边界是圆:
S2=2∫(π/3,π/2)0.5ρ²dθ
=∫(π/3,π/2)9cos²θdθ
=(9/2)∫(π/3,π/2)(1+cos2θ)dθ
=(9/2)(θ+sin2θ/2)|(π/3,π/2)
=(9/2)(π/2-π/3+0.5(sinπ-sin2π/3))
=(9/2)(π/6+0.5(-√3/2)
=3π/4-√3/4
S1+S2即得。
3cosθ=1+cosθ,cosθ=1/2,θ=±π/3
ρ=3cosθ,是直径3的圆,ρ=1+cosθ是心形线,关于极轴对称。
重合部分,±π/3之间,外轮廓为心形线,面积:
S1=2∫(0,π/3)0.5ρ²dθ
=∫(0,π/3)(1+cosθ)²dθ
=∫(0,π/3)(1+2cosθ+cos²)dθ
=∫(0,π/3)(1+2cosθ+(1+cos2θ)/2)dθ
=(3/2.θ+2sinθ+sin2θ/4)|(0,π/3)
=(3/2)(π/3)+√3+√3/8
=π/2+9√3/8
余下是两个弓形部分,边界是圆:
S2=2∫(π/3,π/2)0.5ρ²dθ
=∫(π/3,π/2)9cos²θdθ
=(9/2)∫(π/3,π/2)(1+cos2θ)dθ
=(9/2)(θ+sin2θ/2)|(π/3,π/2)
=(9/2)(π/2-π/3+0.5(sinπ-sin2π/3))
=(9/2)(π/6+0.5(-√3/2)
=3π/4-√3/4
S1+S2即得。
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