Question

Find the area lying outside r=6cos(theta) and inside r=3+3cos(theta)

Answer 1

The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b... 

the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. 
2cosθ = 1 => θ = ±π/3 

A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ 
= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ 
= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. 
.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] 
= 3θ/2 +sin(2θ) - sin(θ) 

Area = A(π/3) - A(-π/3) 
= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) 
= π.