1. Cilindro espesso \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}
    D: O cilindro de paredes espessas 1 \leq x^{2}+y^{2} \leq 2, \quad-1 \leq z \leq 2.

Passo 1

Temos o campo vetorial
\mathbf{F}(x,y,z)=\ln\bigl(x^{2}+y^{2}\bigr)\,\mathbf{i}\;-\;\frac{2z}{x}\arctan\!\Bigl(\frac{y}{x}\Bigr)\,\mathbf{j}\;+\;z\sqrt{x^{2}+y^{2}}\,\mathbf{k}
e o domínio D dado pelo cilindro de paredes espessas
1 \le x^{2}+y^{2} \le 2,\quad -1 \le z \le 2.
Queremos o fluxo de \mathbf{F} pela superfície fechada \partial D. Vamos usar o Teorema da Divergência:
\iint_{\partial D}\mathbf{F}\cdot\mathbf{n}\,dS =\iiint_{D}\nabla\!\cdot\mathbf{F}\;dV.

Passo 2

Calculamos os termos de \nabla\!\cdot\mathbf{F}=\frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}+\frac{\partial F_{3}}{\partial z}.
  1. F_{1}=\ln(x^{2}+y^{2})
\frac{\partial F_{1}}{\partial x} =\frac{2x}{x^{2}+y^{2}}.
  1. F_{2}=-\frac{2z}{x}\arctan\!\bigl(\tfrac{y}{x}\bigr). Como x é constante em \partial/\partial y,
\frac{\partial F_{2}}{\partial y} =-2z\;\frac{\partial}{\partial y}\biggl[\frac{1}{x}\arctan\!\bigl(\tfrac{y}{x}\bigr)\biggr] =-2z\;\frac{1}{x}\cdot\frac{1}{1+(y/x)^{2}}\cdot\frac{1}{x} =-\frac{2z}{x^{2}+y^{2}}.
  1. F_{3}=z\sqrt{x^{2}+y^{2}}
\frac{\partial F_{3}}{\partial z} =\sqrt{x^{2}+y^{2}}.
Somando,
\nabla\!\cdot\mathbf{F} =\frac{2x}{x^{2}+y^{2}} -\frac{2z}{x^{2}+y^{2}} +\sqrt{x^{2}+y^{2}}.

Passo 3

Mudamos para coordenadas cilíndricas:
x=r\cos\theta,\;y=r\sin\theta,\;dV = r\,dr\,d\theta\,dz,\; \sqrt{x^{2}+y^{2}}=r.
No numerador temos 2x-2z=2r\cos\theta-2z, logo
\nabla\!\cdot\mathbf{F} =\frac{2r\cos\theta-2z}{r^{2}}+r =\frac{2\cos\theta}{r}-\frac{2z}{r^{2}}+r.
O fluxo é
\iiint_{D}\!\nabla\!\cdot\mathbf{F}\,dV =\int_{\theta=0}^{2\pi}\!\int_{r=1}^{\sqrt{2}}\!\int_{z=-1}^{2} \Bigl(\frac{2\cos\theta}{r}-\frac{2z}{r^{2}}+r\Bigr)\,r\,dz\,dr\,d\theta.

Passo 4

Integramos primeiro em z:
\int_{z=-1}^{2}\!\Bigl(\frac{2\cos\theta}{r}-\frac{2z}{r^{2}}+r\Bigr)r\,dz =\int_{-1}^{2}\!\Bigl(2\cos\theta-\frac{2z}{r}+r^{2}\Bigr)\,dz.
- \displaystyle\int_{-1}^{2}2\cos\theta\,dz=6\cos\theta
- \displaystyle\int_{-1}^{2}-\frac{2z}{r}\,dz =-\frac{2}{r}\Bigl[\frac{z^{2}}{2}\Bigr]_{-1}^{2} =-\frac{3}{r}
- \displaystyle\int_{-1}^{2}r^{2}\,dz=3r^{2}.
Logo a integral em z vale
6\cos\theta -\frac{3}{r}+3r^{2}.

Passo 5

Agora em \theta e r:
\int_{0}^{2\pi}\int_{1}^{\sqrt{2}} \Bigl(6\cos\theta -\frac{3}{r}+3r^{2}\Bigr)\,dr\,d\theta =\underbrace{\int_{0}^{2\pi}6\cos\theta\,d\theta}_{0} +2\pi\int_{1}^{\sqrt{2}}\Bigl(-\tfrac{3}{r}+3r^{2}\Bigr)\,dr.
Calculamos
\int_{1}^{\sqrt{2}}\Bigl(-\tfrac{3}{r}+3r^{2}\Bigr)dr =\bigl[-3\ln r + r^{3}\bigr]_{1}^{\sqrt{2}} =2\sqrt{2}-1-\tfrac{3}{2}\ln2.
Multiplicando por 2\pi, obtemos
\Phi =2\pi\Bigl(2\sqrt{2}-1-\tfrac{3}{2}\ln2\Bigr) =4\sqrt{2}\,\pi-2\pi-3\pi\ln2.

Resposta

\iint_{\partial D}\mathbf{F}\cdot\mathbf{n}\,dS =4\sqrt{2}\,\pi \;-\;2\pi\;-\;3\pi\ln2.

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