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

Passo 1

O enunciado pede para calcular (a) o rotacional
\nabla\times\mathbf{F}
e (b) a divergente
\nabla\cdot\mathbf{F}
do campo
\mathbf{F}(x,y,z)=x e^{y z}\,\mathbf{i}+y e^{x z}\,\mathbf{j}+z e^{y}\,\mathbf{k}.
Para isso vamos usar as definições:
\nabla\times\mathbf{F} =\bigl(\tfrac{\partial F_{3}}{\partial y}-\tfrac{\partial F_{2}}{\partial z}\bigr)\mathbf{i} +\bigl(\tfrac{\partial F_{1}}{\partial z}-\tfrac{\partial F_{3}}{\partial x}\bigr)\mathbf{j} +\bigl(\tfrac{\partial F_{2}}{\partial x}-\tfrac{\partial F_{1}}{\partial y}\bigr)\mathbf{k},
\nabla\cdot\mathbf{F} =\tfrac{\partial F_{1}}{\partial x}+\tfrac{\partial F_{2}}{\partial y}+\tfrac{\partial F_{3}}{\partial z}.

Passo 2

Calcular as derivadas parciais dos componentes
F_{1}=x e^{yz}, F_{2}=y e^{xz}, F_{3}=z e^{y}:
\tfrac{\partial F_{1}}{\partial x}=e^{yz}, \quad \tfrac{\partial F_{1}}{\partial y}=x\,z\,e^{yz}, \quad \tfrac{\partial F_{1}}{\partial z}=x\,y\,e^{yz};
\tfrac{\partial F_{2}}{\partial x}=y\,z\,e^{xz}, \quad \tfrac{\partial F_{2}}{\partial y}=e^{xz}, \quad \tfrac{\partial F_{2}}{\partial z}=x\,y\,e^{xz};
\tfrac{\partial F_{3}}{\partial x}=0, \quad \tfrac{\partial F_{3}}{\partial y}=z\,e^{y}, \quad \tfrac{\partial F_{3}}{\partial z}=e^{y}.

Passo 3

Montar o rotacional usando as fórmulas:
(\nabla\times\mathbf{F})_{x} =\tfrac{\partial F_{3}}{\partial y}-\tfrac{\partial F_{2}}{\partial z} =z\,e^{y}-x\,y\,e^{xz},
(\nabla\times\mathbf{F})_{y} =\tfrac{\partial F_{1}}{\partial z}-\tfrac{\partial F_{3}}{\partial x} =x\,y\,e^{yz}-0,
(\nabla\times\mathbf{F})_{z} =\tfrac{\partial F_{2}}{\partial x}-\tfrac{\partial F_{1}}{\partial y} =y\,z\,e^{xz}-x\,z\,e^{yz}.
Logo,
\nabla\times\mathbf{F} =\bigl(z\,e^{y}-x\,y\,e^{xz}\bigr)\mathbf{i} +\bigl(x\,y\,e^{yz}\bigr)\mathbf{j} +\bigl(y\,z\,e^{xz}-x\,z\,e^{yz}\bigr)\mathbf{k}.

Passo 4

Montar a divergente:
\nabla\cdot\mathbf{F} =\tfrac{\partial F_{1}}{\partial x} +\tfrac{\partial F_{2}}{\partial y} +\tfrac{\partial F_{3}}{\partial z} =e^{yz}+e^{xz}+e^{y}.

Resposta

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

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