题目

计算[0,π/2]∫dΘ[0-cosΘ]∫rdr,

提问时间：2021-03-23

答案

∫[0,π/2]dΘ∫[0-cosΘ]rdr
=∫[0,π/2]1/2r² |[0-cosΘ] dΘ
=1/2∫[0,π/2]cos²Θ] dΘ
=1/4∫[0,π/2](1+cos2Θ) dΘ
=1/4(Θ+1/2sin2Θ) |[0,π/2]
=π/8