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

Passo 1

O enunciado pede (a) o rotacional e (b) a divergente do campo vetorial
\mathbf{F}(x,y,z)=x e^{yz}\,\mathbf{i} \;+\;y e^{xz}\,\mathbf{j}\;+\;z e^{xy}\,\mathbf{k}.
Isso faz parte de cálculo vetorial: vamos usar a definição do operador nabla
\nabla=(\partial/\partial x,\;\partial/\partial y,\;\partial/\partial z)
para calcular
\nabla\times\mathbf{F} (rotacional) e
\nabla\cdot\mathbf{F} (divergente).

Passo 2

Primeiro, identificamos as componentes escalares de \mathbf{F}:
P(x,y,z)=x e^{yz},\quad Q(x,y,z)=y e^{xz},\quad R(x,y,z)=z e^{xy}.
As fórmulas são
\displaystyle\nabla\times\mathbf{F} = \bigl(\tfrac{\partial R}{\partial y}-\tfrac{\partial Q}{\partial z},\; \tfrac{\partial P}{\partial z}-\tfrac{\partial R}{\partial x},\; \tfrac{\partial Q}{\partial x}-\tfrac{\partial P}{\partial y}\bigr)
e
\displaystyle\nabla\cdot\mathbf{F} = \tfrac{\partial P}{\partial x}+\tfrac{\partial Q}{\partial y}+\tfrac{\partial R}{\partial z}.

Passo 3

Calculamos as derivadas parciais necessárias, usando regra do produto:
\begin{aligned} \tfrac{\partial P}{\partial x}&=e^{yz},& \tfrac{\partial P}{\partial y}&=x z\,e^{yz},& \tfrac{\partial P}{\partial z}&=x y\,e^{yz},\\\ \tfrac{\partial Q}{\partial x}&=y z\,e^{xz},& \tfrac{\partial Q}{\partial y}&=e^{xz},& \tfrac{\partial Q}{\partial z}&=x y\,e^{xz},\\\ \tfrac{\partial R}{\partial x}&=y z\,e^{xy},& \tfrac{\partial R}{\partial y}&=x z\,e^{xy},& \tfrac{\partial R}{\partial z}&=e^{xy}. \end{aligned}

Passo 4

Substituindo na definição de rotacional:
\nabla\times\mathbf{F}= \Bigl(\underbrace{\tfrac{\partial R}{\partial y}-\tfrac{\partial Q}{\partial z}}_{x z\,e^{xy}-x y\,e^{xz}},\; \underbrace{\tfrac{\partial P}{\partial z}-\tfrac{\partial R}{\partial x}}_{x y\,e^{yz}-y z\,e^{xy}},\; \underbrace{\tfrac{\partial Q}{\partial x}-\tfrac{\partial P}{\partial y}}_{y z\,e^{xz}-x z\,e^{yz}}\Bigr).

Passo 5

Substituindo na definição de divergente:
\nabla\cdot\mathbf{F} = \underbrace{e^{yz}}_{\partial P/\partial x} +\underbrace{e^{xz}}_{\partial Q/\partial y} +\underbrace{e^{xy}}_{\partial R/\partial z} = e^{yz}+e^{xz}+e^{xy}.

Resposta

(a)
\nabla\times\mathbf{F}= \bigl(x z\,e^{xy}-x y\,e^{xz},\; x y\,e^{yz}-y z\,e^{xy},\; y z\,e^{xz}-x z\,e^{yz}\bigr).
(b)
\nabla\cdot\mathbf{F}=e^{yz}+e^{xz}+e^{xy}.

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