Hello World

ello, calculamos el vector [math]\vec{v} = \vec{e_z} \times \vec{u} [/math].

[math] \vec{v} = \vec{e_z} \times f(ρ)\vec{e_θ} =(\frac{2}{3}ρ- \frac{2}{3ρ})( \vec{e_z} \times \vec{e_θ}) = ( \frac{2}{3ρ}-\frac{2}{3}ρ)\vec{e_ρ} [/math]

Una vez hallado el vector [math]\vec{v} [/math], hallamos su potencial escalar, que representa la función de corriente de [math]\vec{u} [/math], es decir, hallamos la función ψ que cumpla [math]∇ψ = \vec{v}[/math]

[math] ∇ψ = \vec{v} = (\frac{2}{3ρ} - \frac{2ρ}{3}) \vec{e_ρ} [/math] [math] ∇ψ = \frac{\partial ψ}{\partial ρ} \vec{e_ρ} + \frac{1}{ρ}\frac{\partial ψ}{\partial θ} \vec{e_θ} + \frac{\partial ψ}{\partial z}\vec{e_z} 0 = (\frac{2}{3ρ} - \frac{2ρ}{3}) \vec{e_ρ}[/math] [math]\frac{\partial ψ}{\partial ρ} = \frac{2}{3ρ} - \frac{2ρ}{3}[/math] [math]ψ=\int (\frac{2}{3ρ}-\frac{2ρ}{3}) \partial ρ [/math] [math]ψ=\frac{2}{3}(ln(ρ)-\frac{ρ^2}{2}) \partial ρ [/math]