x^2-6x+y^2=0 - уравнение окружности со смещенным центром: (x-3)^2+y^2=1
Область D ( см. рис.)
[m]S= ∫∫_{D} dxdy[/m]
Полярные координаты
x= ρ cos θ
y= ρ sin θ
x^2+y^2=(ρ cos θ )^2+(ρ sin θ)^2= ρ ^2*(cos^2 θ +sin^2 θ )= ρ ^2*1= ρ ^2
x^2-2x+y^2=0 ⇒ ρ^2-2*ρ cos θ =0 ⇒ ρ =2 cos θ
x^2-6x+y^2=0 ⇒ ρ^2-6*ρ cos θ =0 ⇒ ρ =6 cos θ
0 ≤ θ ≤ (π/4)
2 cos θ ≤ ρ ≤ 6 cos θ
[m]S= ∫∫_{D} dxdy= ∫_{0} ^{\frac{π}{4}}(∫_{2 cos θ }^{6 cos θ } ρd ρ )d θ= ∫_{0} ^{\frac{π}{4}}(\frac{ρ^2}{2})| _{2 cos θ }^{6 cos θ } )d θ= ∫_{0} ^{\frac{π}{4}}(\frac{(6cos θ )^2}{2}-\frac{(2cos θ )^2}{2} )d θ= [/m]
[m]=∫_{0} ^{\frac{π}{4}}(\frac{36cos^2 θ }{2}-\frac{4cos^2 θ }{2} )d θ=∫_{0} ^{\frac{π}{4}}16 cos^2 θ d θ= 8∫ _{0} ^{\frac{π}{4}} (1+cos2 θ) d θ = 8\cdot (θ +\frac{1}{2}sin2 θ )|∫_{0} ^{\frac{π}{4}}=...[/m]