Determine(a)a rotacional e(b)a divergente do campo vetorial.
\boldsymbol{F}(x, y, z)-(x-2 z) \mathbf{i}+(x+y+z) \mathbf{j}+(x-2 y) \mathbf{k}

Passo 1

O enunciado dá o campo vetorial
\boldsymbol{F}(x,y,z) = (x-2z)\,\mathbf{i} + (x+y+z)\,\mathbf{j} + (x-2y)\,\mathbf{k}
e pede (a) o rotacional e (b) a divergente. O conteúdo é cálculo vetorial, usando as definições de \nabla\times\boldsymbol{F} e \nabla\cdot\boldsymbol{F}.

Passo 2

Para o rotacional usamos a fórmula
\nabla\times\boldsymbol{F} = \bigl(\tfrac{\partial R}{\partial y}-\tfrac{\partial Q}{\partial z}\bigr)\,\mathbf{i} - \bigl(\tfrac{\partial R}{\partial x}-\tfrac{\partial P}{\partial z}\bigr)\,\mathbf{j} + \bigl(\tfrac{\partial Q}{\partial x}-\tfrac{\partial P}{\partial y}\bigr)\,\mathbf{k},
em que P=x-2z, Q=x+y+z e R=x-2y.

Passo 3

Calculando as derivadas parciais:
\tfrac{\partial R}{\partial y} = -2,\quad \tfrac{\partial Q}{\partial z} = 1 \;\Longrightarrow\; \mathbf{i}\text{-componente} = -2 - 1 = -3,
\tfrac{\partial R}{\partial x} = 1,\quad \tfrac{\partial P}{\partial z} = -2 \;\Longrightarrow\; \mathbf{j}\text{-componente} = -\bigl(1 - (-2)\bigr) = -3,
\tfrac{\partial Q}{\partial x} = 1,\quad \tfrac{\partial P}{\partial y} = 0 \;\Longrightarrow\; \mathbf{k}\text{-componente} = 1 - 0 = 1.
Logo
\nabla\times\boldsymbol{F} = -3\,\mathbf{i} - 3\,\mathbf{j} + 1\,\mathbf{k}.

Passo 4

Para a divergente usamos
\nabla\cdot\boldsymbol{F} = \tfrac{\partial P}{\partial x} + \tfrac{\partial Q}{\partial y} + \tfrac{\partial R}{\partial z}.

Passo 5

Calculando as derivadas:
\tfrac{\partial P}{\partial x} = 1,\quad \tfrac{\partial Q}{\partial y} = 1,\quad \tfrac{\partial R}{\partial z} = 0 \;\Longrightarrow\; \nabla\cdot\boldsymbol{F} = 1 + 1 + 0 = 2.

Resposta

(a) \nabla\times\boldsymbol{F} = -3\,\mathbf{i} - 3\,\mathbf{j} + \mathbf{k}.
(b) \nabla\cdot\boldsymbol{F} = 2.

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