[m]y= ρ sin θ [/m]
[m]x^2+y^2=( ρ cos θ)^2+ ( ρ sin θ)^2= ρ ^2 [/m]
[m]dxdy= ρ d ρ d θ [/m]
D:
[m]0 ≤ θ ≤ 2π
a ≤ ρ ≤a^2 [/m]
[m]J= ∫ _{0}^{2π}( ∫ _{a }^{a^2}ln ρ ^2\cdot ρd ρ)d θ = ∫ _{0}^{2π}( ∫ _{a }^{a^2}2ln ρ \cdot ρd ρ)d θ =[/m]
[m]=2\cdot ∫ _{0}^{2π}( ∫ _{a }^{a^2}2ln ρ \cdot ρd ρ)d θ =[/m]
Считаем внутренний интеграл[i] по частям[/i]
[m] ∫ _{a }^{a^2}ln ρ \cdot ρ d ρ[/m]
u=ln ρ
du=(1/ ρ)d ρ
dv= ρ d ρ
v= ρ ^2/2
[m] ∫ _{a }^{a^2}ln ρ \cdot ρ d ρ=(\frac{ρ ^2ln ρ }{2})| _{a }^{a^2}- ∫ _{a }^{a^2}\frac{ρ ^2 }{2}\cdot \frac{1}{ ρ } d ρ=\frac{(a^2) ^2ln a }{2}-\frac{a ^2ln a }{2}- ∫ _{a }^{a^2}\frac{ρ }{2}d ρ = [/m]
[m]=\frac{(a^2) ^2ln a^2 }{2}-\frac{a ^2ln a }{2}-( \frac{ρ^2 }{4})| _{a }^{a^2} =\frac{(2a^4-a^2)ln a }{2}-\frac{a^4-a^2}{4} [/m]
Тогда
[m]J=2\cdot ∫ _{0}^{2π}( ∫ _{a }^{a^2}2ln ρ \cdot ρd ρ)d θ = 2\cdot( \frac{(2a^4-a^2)ln a }{2}-\frac{a^4-a^2}{4} )∫ _{0}^{2π}d θ =4π \cdot( \frac{(2a^4-a^2)ln a }{2}-\frac{a^4-a^2}{4} )[/m]