Question

求由p=3cosθ和p=1+cosθ所围成的图形的面积

Answer 1

简单分析一下，答案如图所示

Answer 2

求由ρ=3cosθ和ρ=1+cosθ所围成的图形的面积
由ρ=3cosθ得x²+y²=3x；即(x-3/2)²+y²=9/4是一个圆心在(3/2,0),
半径R=3/2的园.在极点(原点)处的切线是y轴.
阴影上半部分的面积S/2=【D】∫∫ρdρdθ
=【0,π/3】∫dθ【0,1+cosθ】∫ρdρ+【π/3,π/2】∫dθ【0,3cosθ】∫ρdρ
=【0,π/3】(1/2)∫(1+cosθ)²dθ+【π/3,π/2】(9/2)∫cos²θdθ
=【0,π/3】(1/2)∫(1+2cosθ+cos²θ)dθ+【π/3,π/2】(9/4)∫(1+cos2θ)dθ
=(1/2)[θ+2sinθ+(1/2)θ+(1/4)sin2θ]【0,π/3】
+(9/4)[θ+(1/2)sin2θ]【π/3,π/2】
=(1/2)[(π/3)+√3+(π/6)+(√3/8)]+(9/4)[(π/2)-(π/3)-(√3/4)]
=(1/2)[(π/2)+9/8)(√3)]+(9/4)[(π/6)-√3/4)]
=(5/8)π-(9/16)(√3-1)
即S=(5/4)π-(9/8)(√3-1)