Nos problemas 17-20, determine o rotacional e o divergente do campo vetorial indicado.
\boldsymbol{F}(x, y, z)=x y e^{x} \boldsymbol{i}-x^{3} y z e^{z} \boldsymbol{j}+x y^{2} e^{y} \boldsymbol{k}

Passo 1

O enunciado pede para calcular o divergente e o rotacional (curl) do campo vetorial
\boldsymbol{F}(x,y,z)=x\,y\,e^{x}\,\boldsymbol{i}\;-\;x^{3}y\,z\,e^{z}\,\boldsymbol{j}\;+\;x\,y^{2}\,e^{y}\,\boldsymbol{k}.
O conteúdo necessário é a definição de divergente,
\nabla\cdot\boldsymbol{F} =\frac{\partial F_{1}}{\partial x} +\frac{\partial F_{2}}{\partial y} +\frac{\partial F_{3}}{\partial z},
e de rotacional,
\nabla\times\boldsymbol{F} =\bigl(\tfrac{\partial F_{3}}{\partial y}-\tfrac{\partial F_{2}}{\partial z}\bigr)\boldsymbol{i} -\bigl(\tfrac{\partial F_{3}}{\partial x}-\tfrac{\partial F_{1}}{\partial z}\bigr)\boldsymbol{j} +\bigl(\tfrac{\partial F_{2}}{\partial x}-\tfrac{\partial F_{1}}{\partial y}\bigr)\boldsymbol{k}.

Passo 2

Vamos ao divergente. Primeiro identificamos
F_{1}=x\,y\,e^{x},\quad F_{2}=-x^{3}y\,z\,e^{z},\quad F_{3}=x\,y^{2}\,e^{y}.
Calculamos cada derivada:
\frac{\partial F_{1}}{\partial x} =\frac{\partial}{\partial x}(x\,y\,e^{x}) =y\,e^{x}+x\,y\,e^{x}, \quad \frac{\partial F_{2}}{\partial y} =\frac{\partial}{\partial y}(-x^{3}y\,z\,e^{z}) =-x^{3}z\,e^{z}, \quad \frac{\partial F_{3}}{\partial z} =\frac{\partial}{\partial z}(x\,y^{2}\,e^{y}) =0.
Somando, obtemos
\nabla\cdot\boldsymbol{F} =y\,e^{x}+x\,y\,e^{x}-x^{3}z\,e^{z}.

Passo 3

Agora o rotacional. Usamos a fórmula componente a componente:
  1. Componente \boldsymbol{i}:
\frac{\partial F_{3}}{\partial y} =\frac{\partial}{\partial y}(x\,y^{2}\,e^{y}) =x\bigl(2y\,e^{y}+y^{2}e^{y}\bigr) =x\,y\,(2+y)\,e^{y}, \quad \frac{\partial F_{2}}{\partial z} =\frac{\partial}{\partial z}(-x^{3}y\,z\,e^{z}) =-x^{3}y\bigl(e^{z}+z\,e^{z}\bigr) =-x^{3}y\,(1+z)\,e^{z}.
Logo
(\nabla\times\boldsymbol{F})_{i} =x\,y\,(2+y)\,e^{y} +\;x^{3}y\,(1+z)\,e^{z}.
  1. Componente \boldsymbol{j}:
\frac{\partial F_{3}}{\partial x} =\frac{\partial}{\partial x}(x\,y^{2}\,e^{y}) =y^{2}\,e^{y}, \quad \frac{\partial F_{1}}{\partial z} =\frac{\partial}{\partial z}(x\,y\,e^{x}) =0.
Então
(\nabla\times\boldsymbol{F})_{j} =-\bigl(y^{2}\,e^{y}-0\bigr) =-\,y^{2}\,e^{y}.
  1. Componente \boldsymbol{k}:
\frac{\partial F_{2}}{\partial x} =\frac{\partial}{\partial x}(-x^{3}y\,z\,e^{z}) =-3x^{2}y\,z\,e^{z}, \quad \frac{\partial F_{1}}{\partial y} =\frac{\partial}{\partial y}(x\,y\,e^{x}) =x\,e^{x}.
Logo
(\nabla\times\boldsymbol{F})_{k} =-3x^{2}y\,z\,e^{z}-x\,e^{x}.

Resposta

Divergente:
\nabla\cdot\boldsymbol{F} =y\,e^{x}+x\,y\,e^{x}-x^{3}z\,e^{z}.
Rotacional:
\nabla\times\boldsymbol{F} =\bigl(x\,y\,(2+y)\,e^{y}+x^{3}y\,(1+z)\,e^{z}\bigr)\boldsymbol{i} -\bigl(y^{2}\,e^{y}\bigr)\boldsymbol{j} +\bigl(-3x^{2}y\,z\,e^{z}-x\,e^{x}\bigr)\boldsymbol{k}.

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