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

Passo 1

O enunciado pede para calcular (a) o rotacional e (b) a divergente do campo vetorial
\mathbf{F}(x,y,z)=x\,y\,\mathbf{j}+x\,y\,z\,\mathbf{k}.
O conteúdo envolvido é cálculo vetorial em coordenadas cartesianas, usando definições de derivadas parciais.

Passo 2

Primeiro, identificamos as componentes de \mathbf{F}:
F_x=0,\quad F_y=x\,y,\quad F_z=x\,y\,z.

Passo 3

Para o rotacional usamos a definição
\nabla\times\mathbf{F} =\Bigl(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\Bigr)\mathbf{i} -\Bigl(\frac{\partial F_z}{\partial x}-\frac{\partial F_x}{\partial z}\Bigr)\mathbf{j} +\Bigl(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\Bigr)\mathbf{k}.
Calculando as derivadas:
\frac{\partial F_z}{\partial y} =\frac{\partial (x\,y\,z)}{\partial y}=x\,z, \quad \frac{\partial F_y}{\partial z} =\frac{\partial (x\,y)}{\partial z}=0,
\frac{\partial F_z}{\partial x} =\frac{\partial (x\,y\,z)}{\partial x}=y\,z, \quad \frac{\partial F_x}{\partial z}=0,
\frac{\partial F_y}{\partial x} =\frac{\partial (x\,y)}{\partial x}=y, \quad \frac{\partial F_x}{\partial y}=0.
Substituindo,
\nabla\times\mathbf{F} =(x\,z-0)\,\mathbf{i} -(y\,z-0)\,\mathbf{j} +(y-0)\,\mathbf{k} = x\,z\,\mathbf{i}-y\,z\,\mathbf{j}+y\,\mathbf{k}.

Passo 4

Para a divergente usamos
\nabla\cdot\mathbf{F} =\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}.
Como \partial F_x/\partial x=0,
\frac{\partial F_y}{\partial y} =\frac{\partial (x\,y)}{\partial y}=x, \quad \frac{\partial F_z}{\partial z} =\frac{\partial (x\,y\,z)}{\partial z}=x\,y,
temos
\nabla\cdot\mathbf{F}=0 + x + x\,y = x + x\,y.

Resposta

(a) \nabla\times\mathbf{F}=x\,z\,\mathbf{i}-y\,z\,\mathbf{j}+y\,\mathbf{k}.
(b) \nabla\cdot\mathbf{F}=x+x\,y.

Exercícios relacionados