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r=cosθ
r=1−cosθ
Let's set the two functions equal to each other and solve for θ to find the points of intersection:
cosθ=1−cosθ
2cosθ=1
cosθ=12
θ=arccos(12)=π3and−π3
The enclosed areas are between 0andπ3 and between 0and−π3, and the two are symmetrical. We can figure out the area enclosed between 0andπ3 and multiply it by two:
∫π30(cos2θ−(1−cosθ)2)(d(θ))
∫π30(cos2θ−(1+cos2θ−2cosθ))(d(θ))
∫π30(cos2θ−1−cos2θ+2cosθ)(d(θ))
=∫π30(2cosθ−1)(d(θ))=2∫π30cosθ(d(θ))−∫π30d(θ)
=(2sinθ−θ)π30=2(√32)−π3−2sin0−0=√3−π3−0−0=√3−π3
Now we multiply it by 2 to get the two enclosed areas together:
=2√3−2π3
