[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^{2} z^{2} \mathbf{i}+x^{2} y z^{2} \mathbf{j}+x^{2} y^{2} z \mathbf{k}

Passo 1

O enunciado pede o divergente e o rotacional do campo
\mathbf{F}(x,y,z)=x\,y^{2}z^{2}\,\mathbf{i}+x^{2}y\,z^{2}\,\mathbf{j}+x^{2}y^{2}z\,\mathbf{k}.
O conteúdo aqui é cálculo vetorial: usamos as fórmulas
\mathrm{div}\,\mathbf{F}=\frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}+\frac{\partial F_{3}}{\partial z}, \quad \mathrm{rot}\,\mathbf{F} =\Bigl(\frac{\partial F_{3}}{\partial y}-\frac{\partial F_{2}}{\partial z},\, \frac{\partial F_{1}}{\partial z}-\frac{\partial F_{3}}{\partial x},\, \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}\Bigr).

Passo 2

Para facilitar o divergente, note que podemos escrever
\mathbf{F}=f\,\mathbf{G}, \quad f(x,y,z)=x\,y\,z, \quad \mathbf{G}=(y\,z,\;x\,z,\;x\,y).
Pela identidade
\mathrm{div}(f\,\mathbf{G})=f\,\mathrm{div}\,\mathbf{G}+\nabla f\cdot\mathbf{G},
calculamos primeiro
\mathrm{div}\,\mathbf{G} =\frac{\partial (y\,z)}{\partial x} +\frac{\partial (x\,z)}{\partial y} +\frac{\partial (x\,y)}{\partial z} =0+0+0=0,
e em seguida
\nabla f=(y\,z,\;x\,z,\;x\,y),
\nabla f\cdot\mathbf{G} =(y\,z)(y\,z)+(x\,z)(x\,z)+(x\,y)(x\,y) =y^{2}z^{2}+x^{2}z^{2}+x^{2}y^{2}.
Logo
\mathrm{div}\,\mathbf{F} =0+ \bigl(y^{2}z^{2}+x^{2}z^{2}+x^{2}y^{2}\bigr) =x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}.

Passo 3

Agora calculamos o rotacional componente a componente:
(\mathrm{rot}\,\mathbf{F})_{1} =\frac{\partial F_{3}}{\partial y}-\frac{\partial F_{2}}{\partial z} =\frac{\partial (x^{2}y^{2}z)}{\partial y} -\frac{\partial (x^{2}y\,z^{2})}{\partial z} =2x^{2}y\,z-2x^{2}y\,z=0,
(\mathrm{rot}\,\mathbf{F})_{2} =\frac{\partial F_{1}}{\partial z}-\frac{\partial F_{3}}{\partial x} =\frac{\partial (x\,y^{2}z^{2})}{\partial z} -\frac{\partial (x^{2}y^{2}z)}{\partial x} =2x\,y^{2}z-2x\,y^{2}z=0,
(\mathrm{rot}\,\mathbf{F})_{3} =\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} =\frac{\partial (x^{2}y\,z^{2})}{\partial x} -\frac{\partial (x\,y^{2}z^{2})}{\partial y} =2x\,y\,z^{2}-2x\,y\,z^{2}=0.

Passo 4

Como todas as três componentes deram zero, concluímos
\mathrm{rot}\,\mathbf{F}=\mathbf{0}.

Resposta

Divergente:
\mathrm{div}\,\mathbf{F} =x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}.
Rotacional:
\mathrm{rot}\,\mathbf{F}=\mathbf{0}.

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