(x^2+y^2)^3 = 4x^2y^2
[m]y= ρ sin θ [/m]
[m]x^2+y^2=( ρ cos θ)^2+(ρ sin θ)^2= ρ ^2(cos^2 θ +sin^2 θ )= ρ^2 [/m]
Тогда уравнение кривой принимает вид
[m] (ρ ^2)^3=4\cdot ( ρ cos θ)^2\cdot ( ρ sin θ)^2[/m]
[m] ρ ^2=4\cdot cos^2 θ\cdot sin^2 θ[/m]
Так как [m]sin2 θ =2sin θ cos θ [/m]
[m] ρ ^2= sin^2 2θ[/m]
или
[m] ρ= sin 2θ[/m]
в 1-ой четверти ⇒ 0 ≤ θ ≤ (π/2)
0 ≤ ρ ≤ sin 2θ
[m]S= ∫ ∫_{D} dxdy ⇒ 4 ∫_{0} ^{\frac{π}{2}}( ∫_{0} ^{sin 2θ } ρ d ρ )d θ =]4 ∫_{0} ^{\frac{π}{2}}(\frac{ ρ^2 }{2})|_{0} ^{sin2 θ } )d θ=4 ∫_{0} ^{\frac{π}{2}}\frac{sin^22 θ}{2}d θ= 2 ∫_{0} ^{\frac{π}{2}}sin^22 θd θ= ∫_{0} ^{\frac{π}{2}}(1-cos4 θ)d θ=( θ- \frac{1}{4}sin4 θ)|_{0} ^{\frac{π}{2}}=\frac{π}{2} [/m]