Verified answer

The area inside the inner loop is given by A = ∫ (1 - 2 cos θ)^2 dθ, with limits from θ = 0 to θ = π/3. This area is approximately 0.544.

The area inside the outer loop but outside the inner loop, is given by ∫ [(1 - 2 sin θ)^2, from θ = π/3 to θ = π, then subtracting the previously found area. That area is approximately 8.881 - 0.544 = 8.337.

R = cos? is genuinely interior of r = a million+cos? discipline (r = a million+cos?) = 0.5 of ? [0 to 2?] r² d? = 0.5 of ? [0 to 2?] (a million + cos?)² d? = 0.5 ? [0 to 2?] (a million + 2cos? + cos²?) d? = 0.5 of ? [0 to 2?] (a million + 2cos? + a million/2 + 0.5 of cos(2?)) d? = a million/2 ? [0 to 2?] (3/2 + 2cos? + a million/2 cos(2?)) d? = a million/2 (3/2 ? + 2sin? + a million/4 sin(2?)) 0.5 of [(3/2 (2?) + 0 + 0) ? (0 + 0 + 0)] = three?/2 container (r = cos?) = a million/2 ? [0 to ?] r² d? = 0.5 of ? [0 to ?] cos²? d? = a million/2 ? [0 to ?] (0.5 + a million/2 cos(2?)) d? = 0.5 of (0.5 of ? + a million/4 [0 to ?] = 0.5 [(a million/2 (?) + 0 + 0) ? ( 0 + 0)] = ?/4 A = section (r = a million+cos?) ? discipline (r = cos?) A = 3?/2 ? ?/4 A = 5?/4

Solving 1-2cos θ = 0 will give the limits of integration for the inside loop . . . pi/3 and 5pi/3.

Polar areas are not my strong point.