Determine(a)a rotacional e(b)a divergente do campo vetorial.
\mathbf{F}(x, y, z)=\frac{x}{z} \mathbf{i}+\frac{y}{z} \mathbf{j}-\frac{1}{z} \mathbf{k}

Passo 1

O enunciado pede para calcular o rotacional e a divergente do campo vetorial
\mathbf{F}(x,y,z)=\dfrac{x}{z}\,\mathbf{i}+\dfrac{y}{z}\,\mathbf{j}-\dfrac{1}{z}\,\mathbf{k}.
O conteúdo necessário é cálculo vetorial em \mathbb{R}^3, usando derivadas parciais para obter
\nabla\times\mathbf{F} e \nabla\cdot\mathbf{F}.

Passo 2

O rotacional de um campo \mathbf{F}=(F_x,F_y,F_z) é dado por
\nabla\times\mathbf{F} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\\ \dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\\\ F_x&F_y&F_z \end{vmatrix}.
Aqui F_x=\dfrac{x}{z}, F_y=\dfrac{y}{z}, F_z=-\dfrac1z.

Passo 3

Vamos calcular as componentes de \nabla\times\mathbf{F}:
(\nabla\times\mathbf{F})_x =\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z} =0-\bigl(-\frac{y}{z^2}\bigr)=\frac{y}{z^2},
(\nabla\times\mathbf{F})_y =\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x} =-\frac{x}{z^2}-0=-\frac{x}{z^2},
(\nabla\times\mathbf{F})_z =\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} =0-0=0.
Portanto
\nabla\times\mathbf{F} =\frac{y}{z^2}\,\mathbf{i}-\frac{x}{z^2}\,\mathbf{j}+0\,\mathbf{k} =\frac{1}{z^2}\bigl(y\,\mathbf{i}-x\,\mathbf{j}\bigr).

Passo 4

A divergente de \mathbf{F} é
\nabla\cdot\mathbf{F} =\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}.
Calculando cada termo:
\partial F_x/\partial x=1/z, \partial F_y/\partial y=1/z,
\partial F_z/\partial z=\partial(-z^{-1})/\partial z=+z^{-2}=1/z^2.
Logo
\nabla\cdot\mathbf{F} =\frac{1}{z}+\frac{1}{z}+\frac{1}{z^2} =\frac{2}{z}+\frac{1}{z^2} =\frac{2z+1}{z^2}.

Resposta

(a) \displaystyle \nabla\times\mathbf{F}=\frac{1}{z^2}\bigl(y\,\mathbf{i}-x\,\mathbf{j}\bigr).
(b) \displaystyle \nabla\cdot\mathbf{F}=\frac{2}{z}+\frac{1}{z^2}=\frac{2z+1}{z^2}.

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