并不难，作出图形来就容易了，如下图。
这是两个圆，而且是大圆套小圆，围成的面积就是两个圆的面积之差：
以上，请采纳。
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第1个回答 2018-12-28
联立解 r = 3acosθ 与 r = √3asinθ 得交点 r = 3a/2, θ = π/3.
S = (1/2)∫<0, π/3>(√3asinθ)^2dθ + (1/2)∫<π/3, π/2>(3acosθ)^2dθ
= (3a^2/4)∫<0, π/3>(1-cos2θ)dθ + (9a^2/4)∫<π/3, π/2>(1+cos2θ)dθ
= (3a^2/4)[θ-(1/2)sin2θ]<0, π/3> + (9a^2/4)[θ+(1/2)sin2θ]<π/3, π/2>
= (3a^2/4)(π/3-√3/4) + (9a^2/4)(π/6-√3/4) = (3a^2/4)(5π/6-√3)
S = (1/2)∫<0, π/3>(√3asinθ)^2dθ + (1/2)∫<π/3, π/2>(3acosθ)^2dθ
= (3a^2/4)∫<0, π/3>(1-cos2θ)dθ + (9a^2/4)∫<π/3, π/2>(1+cos2θ)dθ
= (3a^2/4)[θ-(1/2)sin2θ]<0, π/3> + (9a^2/4)[θ+(1/2)sin2θ]<π/3, π/2>
= (3a^2/4)(π/3-√3/4) + (9a^2/4)(π/6-√3/4) = (3a^2/4)(5π/6-√3)
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