[2.0 pt] Determine o fluxo do campo vetorial
\mathbf{F}=\ln \left(x^{2}+y^{2}\right) \mathbf{i}-\left(\frac{2 z}{x} \operatorname{tg}^{-1} \frac{y}{x}\right) \mathbf{j}+z \sqrt{x^{2}+y^{2}} \mathbf{k}
Sendo S as superfícies entre os cilindro de paredes espessas 1 \leq x^{2}+y^{2} \leq 2 e \quad-1 \leq z \leq 2.

Passo 1

O enunciado pede o fluxo de
\mathbf{F}=\ln\bigl(x^{2}+y^{2}\bigr)\,\mathbf{i}\;-\;\frac{2z}{x}\,\tan^{-1}\!\!\bigl(\tfrac{y}{x}\bigr)\,\mathbf{j}\;+\;z\,\sqrt{x^{2}+y^{2}}\,\mathbf{k}
através da superfície fechada S que envolve a região
1\le x^{2}+y^{2}\le2,\quad -1\le z\le2.
Como S é fechado, vamos usar o Teorema da Divergência:
\iint_{S}\mathbf{F}\cdot d\mathbf{S} =\iiint_{V}\nabla\cdot\mathbf{F}\,dV.

Passo 2

Calculamos \nabla\cdot\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}, onde
P=\ln(x^{2}+y^{2}),\quad Q=-\frac{2z}{x}\,\tan^{-1}\!\bigl(\tfrac{y}{x}\bigr),\quad R=z\,\sqrt{x^{2}+y^{2}}.
  1. \displaystyle\frac{\partial P}{\partial x} =\frac{2x}{x^{2}+y^{2}}.
  2. Usando \theta=\tan^{-1}(y/x),
    \displaystyle \frac{\partial Q}{\partial y} =-2z\;\frac{\partial}{\partial y}\!\Bigl(\frac{\theta}{x}\Bigr) =-2z\,\frac{1}{x^{2}+y^{2}}.
  3. \displaystyle\frac{\partial R}{\partial z} =\sqrt{x^{2}+y^{2}}.
    Logo
\nabla\cdot\mathbf{F} =\frac{2x}{x^{2}+y^{2}} \;-\;\frac{2z}{x^{2}+y^{2}} \;+\;\sqrt{x^{2}+y^{2}}.

Passo 3

Passamos a coordenadas cilíndricas
x=r\cos\theta,\;y=r\sin\theta,\;dV=r\,dr\,d\theta\,dz,
com 1\le r\le\sqrt2, 0\le\theta\le2\pi, -1\le z\le2.
Assim
\iiint_{V}\nabla\cdot\mathbf{F}\,dV =I_{1}+I_{2}+I_{3},
onde
I_{1}=\iiint\frac{2x}{r^{2}}\,dV,\quad I_{2}=\iiint\Bigl(-\frac{2z}{r^{2}}\Bigr)\,dV,\quad I_{3}=\iiint r\,dV.

Passo 4

O termo I_{1} é zero por simetria em \theta.
Para I_{2}:
I_{2} =\int_{0}^{2\pi}\!\int_{1}^{\sqrt2}\!\int_{-1}^{2} \Bigl(-\frac{2z}{r^{2}}\Bigr)\,r\,dz\,dr\,d\theta =-2\int_{0}^{2\pi}d\theta\int_{1}^{\sqrt2}\frac{dr}{r} \int_{-1}^{2}z\,dz.
Calculamos
\int_{-1}^{2}z\,dz=\Bigl[\tfrac{z^{2}}{2}\Bigr]_{-1}^{2} =\frac{4}{2}-\frac{1}{2}=\frac{3}{2},\quad \int_{1}^{\sqrt2}\frac{dr}{r}=\ln\sqrt2=\frac{1}{2}\ln2, \quad\int_{0}^{2\pi}d\theta=2\pi.
Logo
I_{2} =-2\cdot2\pi\cdot\frac{1}{2}\ln2\cdot\frac{3}{2} =-3\pi\ln2.

Passo 5

Para I_{3}:
I_{3} =\int_{0}^{2\pi}\!\int_{1}^{\sqrt2}\!\int_{-1}^{2}r\,(r\,dz\,dr\,d\theta) =\int_{0}^{2\pi}d\theta\int_{-1}^{2}dz\int_{1}^{\sqrt2}r^{2}dr.
Calculamos
\int_{-1}^{2}dz=3,\quad \int_{1}^{\sqrt2}r^{2}dr=\Bigl[\tfrac{r^{3}}{3}\Bigr]_{1}^{\sqrt2} =\frac{( \sqrt2)^{3}-1}{3}=\frac{2\sqrt2-1}{3}, \quad \int_{0}^{2\pi}d\theta=2\pi.
Portanto
I_{3} =2\pi\cdot3\cdot\frac{2\sqrt2-1}{3} =2\pi\,(2\sqrt2-1).

Passo 6

Somando os termos não nulos, o fluxo é
\iint_{S}\mathbf{F}\cdot d\mathbf{S} =I_{2}+I_{3} =-3\pi\ln2 +2\pi\bigl(2\sqrt2-1\bigr) =\pi\bigl(4\sqrt2-2-3\ln2\bigr).

Resposta

\boxed{\pi\bigl(4\sqrt2-2-3\ln2\bigr)}.

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