Considere o campo vetorial em \mathrm{R}^{3} \mathrm{~F}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{xzi}+\mathrm{xyzj}+\mathrm{yzk}
a) Calcule o rotacional:
b) Divergente existe ? G tal que \mathrm{F}= rotacional G

Passo 1

O enunciado dá o campo vetorial em \mathbb{R}^3
F(x,y,z)=x z\,\mathbf{i}+x y z\,\mathbf{j}+y z\,\mathbf{k}
e pede:
  1. Calcular o rotacional de F.
  2. Calcular o divergente de F.
  3. Verificar se existe um vetor G tal que F=\nabla\times G.
    Para isso vamos usar as definições
\nabla\times F =\bigl(\tfrac{\partial F_z}{\partial y}-\tfrac{\partial F_y}{\partial z}\bigr)\mathbf{i} +\bigl(\tfrac{\partial F_x}{\partial z}-\tfrac{\partial F_z}{\partial x}\bigr)\mathbf{j} +\bigl(\tfrac{\partial F_y}{\partial x}-\tfrac{\partial F_x}{\partial y}\bigr)\mathbf{k}
e
\nabla\cdot F =\tfrac{\partial F_x}{\partial x}+\tfrac{\partial F_y}{\partial y}+\tfrac{\partial F_z}{\partial z}.

Passo 2

Calcular o rotacional usando as componentes F_x=xz, F_y=xyz, F_z=yz:
\nabla\times F =\bigl(\tfrac{\partial (yz)}{\partial y}-\tfrac{\partial (xyz)}{\partial z}\bigr)\mathbf{i} +\bigl(\tfrac{\partial (xz)}{\partial z}-\tfrac{\partial (yz)}{\partial x}\bigr)\mathbf{j} +\bigl(\tfrac{\partial (xyz)}{\partial x}-\tfrac{\partial (xz)}{\partial y}\bigr)\mathbf{k}.
Calculando cada termo:
\tfrac{\partial (yz)}{\partial y}=z, \tfrac{\partial (xyz)}{\partial z}=xy,
\tfrac{\partial (xz)}{\partial z}=x, \tfrac{\partial (yz)}{\partial x}=0,
\tfrac{\partial (xyz)}{\partial x}=yz, \tfrac{\partial (xz)}{\partial y}=0.
Logo,
\nabla\times F=(z-xy)\,\mathbf{i}+x\,\mathbf{j}+yz\,\mathbf{k}.

Passo 3

Calcular o divergente:
\nabla\cdot F =\tfrac{\partial (xz)}{\partial x}+\tfrac{\partial (xyz)}{\partial y}+\tfrac{\partial (yz)}{\partial z} =z+xz+y.

Passo 4

Para que exista G tal que F=\nabla\times G é necessário que \nabla\cdot F=0. Aqui
\nabla\cdot F=z+xz+y\neq 0
em geral, portanto não existe campo vetorial G com F=\nabla\times G.

Resposta

\displaystyle\nabla\times F=(z-xy)\,\mathbf{i}+x\,\mathbf{j}+yz\,\mathbf{k},
\displaystyle\nabla\cdot F=z+xz+y,
e não existe G tal que F=\nabla\times G.

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