Calculo o divergente e o rotacional do campo \vec{F}(x, y, z)=\left(e^{x} y z, \cos (x y z), x y z\right).

Passo 1

O enunciado pede para calcular o divergente e o rotacional (curl) do campo \vec{F}(x,y,z) = \bigl(e^x y z,\;\cos(xyz),\;xyz\bigr).
O conteúdo é cálculo vetorial: para o divergente usamos \nabla\cdot\vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z},
e para o rotacional usamos \nabla\times\vec{F} = \bigl(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z},\;\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x},\;\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\bigr).

Passo 2

Calculamos cada termo do divergente:
\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}\bigl(e^x y z\bigr) = e^x y z,
\frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y}\bigl(\cos(xyz)\bigr) = -\sin(xyz)\,\frac{\partial (xyz)}{\partial y} = -x z\,\sin(xyz),
\frac{\partial F_3}{\partial z} = \frac{\partial}{\partial z}\bigl(xyz\bigr) = x y.

Passo 3

Somando esses termos obtemos o divergente:
\nabla\cdot\vec{F} = e^x y z \;-\; x z\,\sin(xyz)\;+\;x y.

Passo 4

Passamos ao rotacional:
\nabla\times\vec{F} = \Bigl( \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z},\; \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x},\; \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y} \Bigr).

Passo 5

Calculamos as derivadas parciais necessárias:
\frac{\partial F_3}{\partial y} = \frac{\partial}{\partial y}(xyz)=x z, \quad \frac{\partial F_2}{\partial z} = \frac{\partial}{\partial z}(\cos(xyz)) = -\sin(xyz)\,x y,
\frac{\partial F_1}{\partial z} = \frac{\partial}{\partial z}(e^x y z) = e^x y, \quad \frac{\partial F_3}{\partial x} = \frac{\partial}{\partial x}(xyz) = y z,
\frac{\partial F_2}{\partial x} = \frac{\partial}{\partial x}(\cos(xyz)) = -\sin(xyz)\,y z, \quad \frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y}(e^x y z) = e^x z.

Passo 6

Substituindo em cada componente do rotacional:
(\nabla\times\vec{F})_1 = x z -\bigl(-x y\,\sin(xyz)\bigr) = x z + x y\,\sin(xyz),
(\nabla\times\vec{F})_2 = e^x y - y z,
(\nabla\times\vec{F})_3 = -y z\,\sin(xyz) - e^x z.
Portanto,
\nabla\times\vec{F} = \bigl(x z + x y\,\sin(xyz),\;e^x y - y z,\;-y z\,\sin(xyz) - e^x z\bigr).

Resposta

O divergente é
\nabla\cdot\vec{F} = e^x y z \;-\; x z\,\sin(xyz)\;+\;x y.
O rotacional é
\nabla\times\vec{F} = \bigl(x z + x y\,\sin(xyz),\;e^x y - y z,\;-y z\,\sin(xyz) - e^x z\bigr).

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