confused on a calculus question

Question

Determine the area of the region between the inner and outer loop of the limacon r = 0.5 + cos θ.

Type your answer in the space below and give 3 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.482)

Answer 1

Polar Area = ∫_a^b(1/2)r²dθ

The top half of the outer loop domain is [0,2π/3] and the top half of the inner loop domain is [π,4π/3].

So the area between the loops will be twice the difference of the two areas.

A = 2((area of outer top) - (area of inner top)) = 2A₁ - 2A₂

r² = cos²(θ) + cos(θ) + 0.25

2A₁ = ∫r²dθ = ∫cos²(θ) + cos(θ) + 0.25 dθ evaluated from 0 to 2π/3

= (sin(2x) + 4sin(x) + 3x)/4 ]₀^2π/3 = (-√3/2 +4(√3/2 ) + 3(2π/3))/4 = (4π+3^3/2)/8

2A₂ = (sin(2x) + 4sin(x) + 3x)/4 evaluated from π to 4π/3

= (2π-3^3/2)/8

2A₁ - 2A₂ = (2π+2(3^3/2))/8 ≈ 2.084

Answer 2

r = 0.5 + cosθ; For outer loop r ≥ 0, 0.5 + cosθ ≥ 0; cosθ ≥ - 0.5; 0 ≤ θ ≤ 2π/3 or 4π/3 ≤ θ ≤ 2π.

Inner loop for r ≤ 0 for 2π/3 ≤ θ ≤ 4π/3; So, aria what we looking for is A = 2·1/2∫₀^2π/3(0.5 + cosθ)²dθ -

2·1/2∫_2π/3^π(0.5 + cosθ)²dθ = ∫₀^2π/3(0.25 + cosθ + cos²θ)dθ - ∫_2π/3^π(0.25 + cosθ + cos²θ)dθ =

∫₀^2π/3(0.75 + cosθ + 1/2cos2θ)dθ - ∫_2π/3^π(0.75 + cosθ + 1/2cos2θ)dθ = (0.75θ + sinθ + 1/4sin2θ)₀^2π/3 -

(0.75θ + sinθ + 1/4sin2θ)_2π/3^π = π/2 + sin2π/3 + 1/4sin4π/3 - 3/4π - sinπ -1/4sin2π + π/2 + sin2π/3 + 1/4sin4π/3 = π/4 +√3/2 - √3/8 +√3/2 - √3/8 = π/4 + 3√3/4 = (π + 3√3)/4.

We use here symmetry about x-axis