所求面积是心形线 ρ=a(1+cosθ)的面积,减去圆 ρ=acosθ 的面积,
由对称性只考虑x轴上方部分,再2倍即可,
S = 2{∫<0,π>(1/2)[a(1+cosθ)]^2dθ-∫<0,π/2>(1/2)(acosθ)^2dθ}
= a^2[∫<0,π>(1+cosθ)^2dθ-∫<0,π/2>(cosθ)^2dθ]
= a^2[∫<0,π>[1+2cosθ+(cosθ)^2]dθ-∫<0,π/2>(cosθ)^2dθ]
= a^2[∫<0,π>[3/2+2cosθ+(1/2)cos2θ]dθ-(1/2)∫<0,π/2>(1+cos2θ)dθ]
= a^2{3θ/2+2sinθ+(1/4)sin2θ]<0,π>-(1/2)[θ+(1/2)sin2θ]<0,π/2>}
= a^2{3π/2-(1/2)(π/2+1/2)] = (5π-1)a^2/4
追问首先,很感谢你,为我花了那么多时间,但是,没有符合的答案啊,不过还是谢谢你,以后你直接用稿纸写出来拍照可能会更快一些
追答最后一步有错,重算如下:
S = 2{∫(1/2)[a(1+cosθ)]^2dθ-∫(1/2)(acosθ)^2dθ}
= a^2[∫(1+cosθ)^2dθ-∫(cosθ)^2dθ]
= a^2[∫[1+2cosθ+(cosθ)^2]dθ-∫(cosθ)^2dθ]
= a^2[∫[3/2+2cosθ+(1/2)cos2θ]dθ-(1/2)∫(1+cos2θ)dθ]
= a^2{3θ/2+2sinθ+(1/4)sin2θ]-(1/2)[θ+(1/2)sin2θ]}
= a^2(3π/2-π/4) = 5πa^2/4, 选 C。