发布网友 发布时间:2024-10-23 17:55
共1个回答
热心网友 时间:2024-11-01 13:15
由r=sinθ与r=√3cosθ,得
tanθ=√3,θ=π/3.
曲线r=sinθ与r=√3cosθ所围成的公共部分的面积
=∫<0,π/3>dθ∫<0,sinθ>rdr+∫<π/3,π/2>dθ∫<0,√3cosθ>rdr
=(1/2)[∫<0,π/3>sin^θdθ+∫<π/3,π/2>3cos^θdθ]
=(1/4)[∫<0,π/3>(1-cos2θ)dθ+∫<π/3,π/2>3(1+cos2θ)dθ]
=(1/4){[θ-(1/2)sin2θ]|<0,π/3>+3[θ+(1/2)sin2θ]|<π/3,π/2>}
=(1/4){π/3-√3/4+π/2-3√3/4)
=5π/24-√3/4.