Let (√2 + 1) sin θ = 1 + cos θ
(3 + 2√2) sin² θ = 1 + 2 cos θ + cos² θ
(3 + 2√2) (1 - cos² θ) = 1 + 2 cos θ + cos² θ
(2 + √2) cos² θ + cos θ - (√2 +1) = 0
cos θ = √2 / 2, θ = ¼ π , or
cos θ = 1, θ = π
So the area overlaped (重复区域的面积)
= ∫ (θ:0→¼ π) ∫[θ:0→(√2 + 1)sin θ rdr ]dθ
+ ∫ (θ:¼ π→π) ∫[θ:0→(1 + cos θ) rdr ]dθ
= ¼(3 + 2√2)(¼π - ½) + 9π/16 - 1/√2 - 1/8
= ¾ π - ½ - ¾ √2 + √2 π/8
≈ 1.35089
曲线 r = (√2 + 1) sin θ 所围成的面积
= ¼(3 + 2√2)π
曲线 r = 1 + cos θ 所围成的面积
= 2 ∫ (θ:0→π) ∫[θ:0→1+cos θ] rdr dθ
= ³/₂π
总面积：
Total Area = ¼(3 + 2√2)π + ³/₂π - [¾ π - ½ - ¾ √2 + √2 π/8]
= ³/₂π + ½ + (3√2)π/8 + ¾ √2
≈ 7.9391