Determine (a) o rotacional e (b) a divergencia do campo vetorial.
\mathbf{F}(x, y, z)=x y z \mathbf{i}+x^{2} y \mathbf{k}

Passo 1

O enunciado nos dá o campo vetorial
\mathbf{F}(x,y,z)=xyz\,\mathbf{i} + x^{2}y\,\mathbf{k}
e pede (a) o rotacional \nabla\times\mathbf{F} e (b) a divergência \nabla\cdot\mathbf{F}. Vamos usar derivadas parciais e as fórmulas padrão de cálculo vetorial.

Passo 2

Identificamos as componentes escalares de \mathbf{F}:
P(x,y,z)=xyz,\quad Q(x,y,z)=0,\quad R(x,y,z)=x^{2}y.

Passo 3

O rotacional é dado por
\nabla \times \mathbf{F} =\biggl(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\biggr)\mathbf{i} +\biggl(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\biggr)\mathbf{j} +\biggl(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\biggr)\mathbf{k}.
Calculando cada derivada:
\frac{\partial R}{\partial y}=\frac{\partial(x^2y)}{\partial y}=x^2,\quad \frac{\partial Q}{\partial z}=0,
\frac{\partial P}{\partial z}=\frac{\partial(xyz)}{\partial z}=xy,\quad \frac{\partial R}{\partial x}=\frac{\partial(x^2y)}{\partial x}=2xy,
\frac{\partial Q}{\partial x}=0,\quad \frac{\partial P}{\partial y}=\frac{\partial(xyz)}{\partial y}=xz.
Substituindo na fórmula, obtemos
\nabla \times \mathbf{F} =(x^2-0)\,\mathbf{i}+(xy-2xy)\,\mathbf{j}+(0-xz)\,\mathbf{k} =x^2\,\mathbf{i}-xy\,\mathbf{j}-xz\,\mathbf{k}.

Passo 4

A divergência é
\nabla\cdot\mathbf{F} =\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.
Calculando:
\frac{\partial P}{\partial x}=\frac{\partial(xyz)}{\partial x}=yz,\quad \frac{\partial Q}{\partial y}=0,\quad \frac{\partial R}{\partial z}=\frac{\partial(x^2y)}{\partial z}=0.
Logo
\nabla\cdot\mathbf{F}=yz.

Resposta

(a) \nabla \times \mathbf{F}=x^2\,\mathbf{i}-xy\,\mathbf{j}-xz\,\mathbf{k}.
(b) \nabla \cdot \mathbf{F}=yz.

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