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

Passo 1

O enunciado pede (a) o rotacional e (b) a divergente do campo vetorial
\mathbf{F}(x,y,z)=x e^{y}\,\mathbf{i}\;-\;z e^{-y}\,\mathbf{j}\;+\;y\ln z\,\mathbf{k}.
O conteúdo envolve operações com derivadas parciais e o uso dos operadores diferenciais de vetor:
– Para o rotacional, vamos usar \nabla\times \mathbf{F}.
– Para a divergente, usamos \nabla\cdot \mathbf{F}.

Passo 2

Para obter o rotacional de \mathbf{F}=(P,Q,R), lembramos que
\nabla\times \mathbf{F} =\Bigl(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\Bigr)\mathbf{i} +\Bigl(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\Bigr)\mathbf{j} +\Bigl(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Bigr)\mathbf{k}.
Aqui P=x e^{y}, Q=-z e^{-y} e R=y\ln z.

Passo 3

Calculamos as derivadas parciais necessárias:
\frac{\partial R}{\partial y} =\frac{\partial}{\partial y}\bigl(y\ln z\bigr) =\ln z, \qquad \frac{\partial Q}{\partial z} =\frac{\partial}{\partial z}\bigl(-z e^{-y}\bigr) =-e^{-y},
\frac{\partial P}{\partial z} =\frac{\partial}{\partial z}\bigl(x e^{y}\bigr) =0, \qquad \frac{\partial R}{\partial x} =\frac{\partial}{\partial x}\bigl(y\ln z\bigr) =0,
\frac{\partial Q}{\partial x} =\frac{\partial}{\partial x}\bigl(-z e^{-y}\bigr) =0, \qquad \frac{\partial P}{\partial y} =\frac{\partial}{\partial y}\bigl(x e^{y}\bigr) =x e^{y}.
Substituindo no determinante do rotacional, obtemos
\nabla\times \mathbf{F} =(\ln z -(-e^{-y}))\,\mathbf{i} +(0-0)\,\mathbf{j} +(0 - x e^{y})\,\mathbf{k} =(\ln z + e^{-y})\,\mathbf{i} - x e^{y}\,\mathbf{k}.

Passo 4

Para a divergente de \mathbf{F}, usamos
\nabla\cdot \mathbf{F} =\frac{\partial P}{\partial x} +\frac{\partial Q}{\partial y} +\frac{\partial R}{\partial z}.
Já sabemos P=x e^{y}, Q=-z e^{-y} e R=y\ln z.

Passo 5

Calculamos as parciais:
\frac{\partial P}{\partial x} =\frac{\partial}{\partial x}(x e^{y}) =e^{y}, \qquad \frac{\partial Q}{\partial y} =\frac{\partial}{\partial y}(-z e^{-y}) =-z(-e^{-y}) =z e^{-y},
\frac{\partial R}{\partial z} =\frac{\partial}{\partial z}(y\ln z) =y\frac{1}{z} =\frac{y}{z}.
Somando, a divergente é
\nabla\cdot \mathbf{F} =e^{y}+z e^{-y}+\frac{y}{z}.

Resposta

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

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