Question Number 113868 by Ar Brandon last updated on 15/Sep/20
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:\mathrm{circle}\:\rho=\mathrm{2acos}\theta\:\mathrm{and}\: \\ $$$$\mathrm{cardiode}\:\rho=\mathrm{a}\left(\mathrm{1}+\mathrm{cos}\theta\right) \\ $$
Commented by kaivan.ahmadi last updated on 18/Sep/20
![2acosθ=a+acosθ⇒acosθ=a⇒cosθ=1⇒ θ=0,θ=2π ∫_0 ^(2π) ((2acosθ)^2 −(a(1+cosθ))^2 )dθ= ∫_0 ^(2π) (4a^2 cos^2 θ−a^2 cos^2 θ−2a^2 cosθ−a^2 )dθ= a^2 ∫_0 ^(2π) (3cos^2 θ−2cosθ−1)dθ= a^2 (((3θ)/2)−((3sin2θ)/4)−2sinθ−θ∣_0 ^(2π) )= a^2 [(3π−0−0−2π)−(0−0−0−0)]= a^2 π](https://www.tinkutara.com/question/Q114349.png)
$$\mathrm{2}{acos}\theta={a}+{acos}\theta\Rightarrow{acos}\theta={a}\Rightarrow{cos}\theta=\mathrm{1}\Rightarrow \\ $$$$\theta=\mathrm{0},\theta=\mathrm{2}\pi \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\left(\mathrm{2}{acos}\theta\right)^{\mathrm{2}} −\left({a}\left(\mathrm{1}+{cos}\theta\right)\right)^{\mathrm{2}} \right){d}\theta= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−\mathrm{2}{a}^{\mathrm{2}} {cos}\theta−{a}^{\mathrm{2}} \right){d}\theta= \\ $$$${a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{3}{cos}^{\mathrm{2}} \theta−\mathrm{2}{cos}\theta−\mathrm{1}\right){d}\theta= \\ $$$${a}^{\mathrm{2}} \left(\frac{\mathrm{3}\theta}{\mathrm{2}}−\frac{\mathrm{3}{sin}\mathrm{2}\theta}{\mathrm{4}}−\mathrm{2}{sin}\theta−\theta\mid_{\mathrm{0}} ^{\mathrm{2}\pi} \right)= \\ $$$${a}^{\mathrm{2}} \left[\left(\mathrm{3}\pi−\mathrm{0}−\mathrm{0}−\mathrm{2}\pi\right)−\left(\mathrm{0}−\mathrm{0}−\mathrm{0}−\mathrm{0}\right)\right]= \\ $$$${a}^{\mathrm{2}} \pi \\ $$