R=1
r=1/2
S=π*1^2-π*(1/2)^2= [b]3π/4[/b]
ρ_(1)=sin φ - уравнение окружности
0 ≤ φ ≤ π
ρ_(2)=2sin φ - уравнение окружности
0 ≤ φ ≤ π
S= ∫ ^(π)_(0) (1/2)* [b]([/b](ρ_(2))^2-(ρ_(1))^2 [b])[/b]d φ =
=(1/2) ∫ ^(π)_(0) ((2sin2φ)^2 -sin^2 φ )d φ =
=(1/2) ∫ ^(π)_(0) (4*(1-cos4φ)/2 -(1-cos2φ)/2 )d φ =
=(1/4)( 4φ -4*(1/4)sin4 φ- φ +(1/2)sin2 φ )|^(π)_(0)=
=(1/4)*(4π - 0 - π+0)=3π/4