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

Passo 1

O enunciado pede o rotacional e a divergente do campo vetorial
\mathbf{F}(x,y,z)=e^{xyz}\,\mathbf{i}+\sin(x-y)\,\mathbf{j}-\dfrac{xy}{z}\,\mathbf{k}.
Vamos usar as definições
\nabla \times \mathbf{F} =\Bigl(\frac{\partial F_k}{\partial y}-\frac{\partial F_j}{\partial z},\, \frac{\partial F_i}{\partial z}-\frac{\partial F_k}{\partial x},\, \frac{\partial F_j}{\partial x}-\frac{\partial F_i}{\partial y}\Bigr)
e
\nabla \cdot \mathbf{F} =\frac{\partial F_i}{\partial x}+\frac{\partial F_j}{\partial y}+\frac{\partial F_k}{\partial z}.

Passo 2

Identificamos as componentes de \mathbf{F}:
F_i = e^{xyz},\quad F_j = \sin(x-y),\quad F_k = -\frac{xy}{z}.

Passo 3

Calculamos \frac{\partial F_k}{\partial y} e \frac{\partial F_j}{\partial z}.
\frac{\partial}{\partial y}\Bigl(-\frac{xy}{z}\Bigr) = -\frac{x}{z}, \qquad \frac{\partial}{\partial z}\bigl(\sin(x-y)\bigr)=0.
Logo a primeira componente do rotacional é
(\nabla \times \mathbf{F})_1 = -\frac{x}{z}.

Passo 4

Calculamos \frac{\partial F_i}{\partial z} e \frac{\partial F_k}{\partial x}.
\frac{\partial}{\partial z}\bigl(e^{xyz}\bigr)=xy\,e^{xyz}, \qquad \frac{\partial}{\partial x}\Bigl(-\frac{xy}{z}\Bigr)=-\frac{y}{z}.
Assim a segunda componente é
(\nabla \times \mathbf{F})_2 =xy\,e^{xyz} -\bigl(-\tfrac{y}{z}\bigr) =xy\,e^{xyz}+\frac{y}{z}.

Passo 5

Calculamos \frac{\partial F_j}{\partial x} e \frac{\partial F_i}{\partial y}.
\frac{\partial}{\partial x}\bigl(\sin(x-y)\bigr)=\cos(x-y), \qquad \frac{\partial}{\partial y}\bigl(e^{xyz}\bigr)=xz\,e^{xyz}.
Portanto a terceira componente é
(\nabla \times \mathbf{F})_3 =\cos(x-y)-xz\,e^{xyz}.

Passo 6

Reunindo as componentes, o rotacional fica
\nabla \times \mathbf{F} =-\frac{x}{z}\,\mathbf{i} +\Bigl(xy\,e^{xyz}+\frac{y}{z}\Bigr)\,\mathbf{j} +\bigl(\cos(x-y)-xz\,e^{xyz}\bigr)\,\mathbf{k}.

Passo 7

Para a divergente, calculamos
\frac{\partial F_i}{\partial x}, \frac{\partial F_j}{\partial y} e \frac{\partial F_k}{\partial z}:
\frac{\partial}{\partial x}\bigl(e^{xyz}\bigr)=yz\,e^{xyz}, \quad \frac{\partial}{\partial y}\bigl(\sin(x-y)\bigr)=-\cos(x-y), \quad \frac{\partial}{\partial z}\Bigl(-\frac{xy}{z}\Bigr)=\frac{xy}{z^2}.

Passo 8

Somando essas derivadas, a divergente é
\nabla \cdot \mathbf{F} =yz\,e^{xyz}-\cos(x-y)+\frac{xy}{z^2}.

Resposta

(a) \nabla \times \mathbf{F} =-\tfrac{x}{z}\,\mathbf{i} +\bigl(xy\,e^{xyz}+\tfrac{y}{z}\bigr)\,\mathbf{j} +\bigl(\cos(x-y)-xz\,e^{xyz}\bigr)\,\mathbf{k}.
(b) \nabla \cdot \mathbf{F} =yz\,e^{xyz}-\cos(x-y)+\tfrac{xy}{z^2}.

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