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Question Number 113868 by Ar Brandon last updated on 15/Sep/20

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{between}\mathrm{the}\mathrm{circle}\rho =2\mathrm{acos}\theta \mathrm{and}$$$$\mathrm{cardiode}\rho =\mathrm{a}(1+\cos \theta )$$

Commented by kaivan.ahmadi last updated on 18/Sep/20

$$2acos\theta =a+acos\theta \Rightarrow acos\theta =a\Rightarrow cos\theta =1\Rightarrow$$$$\theta =0,\theta =2\pi$$$${\int }_0^{2\pi }({\left(2acos\theta \right)}^2-{\left(a(1+cos\theta )\right)}^2)d\theta =$$$${\int }_0^{2\pi }(4a^2{cos}^2\theta -a^2{cos}^2\theta -2a^{2}cos\theta -a^2)d\theta =$$$$a^2{\int }_0^{2\pi }(3{cos}^2\theta -2cos\theta -1)d\theta =$$$$a^2(\frac{3\theta }{2}-\frac{3sin2\theta }{4}-2sin\theta -\theta {\mid }_0^{2\pi })=$$$$a^2[(3\pi -0-0-2\pi )-(0-0-0-0)]=$$$$a^2\pi$$

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