[16.5] Determine o rotacional e o divergente do campo vetorial dado (sendo: r=|\mathbf{r}|= \left.\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\right): [Dica:7] use que \operatorname{div}(f \mathbf{F})=f \operatorname{div}(\mathbf{F})+\nabla f \bullet \mathbf{F}
\mathbf{F}(x, y, z)=x y z \mathbf{i}-x^{2} y \mathbf{k}

Passo 1

O enunciado pede o divergente e o rotacional do campo
\mathbf{F}(x,y,z)=x\,y\,z\,\mathbf{i}+0\,\mathbf{j}-x^{2}y\,\mathbf{k}.
O conteúdo é cálculo vetorial:
– Para o divergente usamos
\operatorname{div}\mathbf{F}=\frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}+\frac{\partial F_{3}}{\partial z}.
– Para o rotacional usamos
\nabla\times\mathbf{F}=\Bigl(\frac{\partial F_{3}}{\partial y}-\frac{\partial F_{2}}{\partial z}\Bigr)\mathbf{i}+\Bigl(\frac{\partial F_{1}}{\partial z}-\frac{\partial F_{3}}{\partial x}\Bigr)\mathbf{j}+\Bigl(\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}\Bigr)\mathbf{k}.

Passo 2

Calculemos o divergente:
\frac{\partial F_{1}}{\partial x} =\frac{\partial(xyz)}{\partial x}=y\,z,\quad \frac{\partial F_{2}}{\partial y} =\frac{\partial(0)}{\partial y}=0,\quad \frac{\partial F_{3}}{\partial z} =\frac{\partial(-x^{2}y)}{\partial z}=0.
Logo
\operatorname{div}\mathbf{F}=y\,z+0+0=y\,z.

Passo 3

Agora o rotacional:
Componete em \mathbf{i}:
\frac{\partial F_{3}}{\partial y}-\frac{\partial F_{2}}{\partial z} =\frac{\partial(-x^{2}y)}{\partial y}-0=-x^{2}.
Componete em \mathbf{j}:
\frac{\partial F_{1}}{\partial z}-\frac{\partial F_{3}}{\partial x} =\frac{\partial(xyz)}{\partial z}-\frac{\partial(-x^{2}y)}{\partial x} = x\,y-(-2x\,y)=3x\,y.
Componete em \mathbf{k}:
\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} =0-\frac{\partial(xyz)}{\partial y}=-x\,z.
Portanto
\nabla\times\mathbf{F}=-x^{2}\,\mathbf{i}+3x\,y\,\mathbf{j}-x\,z\,\mathbf{k}.

Resposta

\boxed{\operatorname{div}\mathbf{F}=y\,z,\quad \nabla\times\mathbf{F}=-x^{2}\,\mathbf{i}+3x\,y\,\mathbf{j}-x\,z\,\mathbf{k}.}

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