2 Questão

Determine \nabla \times \vec{F} e \nabla \cdot \vec{F}, se
\vec{F}(x, y, z)=e^{x} \sin y \vec{i}+e^{y} \sin z \vec{j}+e^{z} \sin x \vec{k}

Passo 1

O enunciado pede para calcular a divergência e o rotacional do campo vetorial
\vec{F}(x,y,z)=e^{x}\sin y\,\vec{i}+e^{y}\sin z\,\vec{j}+e^{z}\sin x\,\vec{k}.
Vamos usar as definições de divergência e rotacional, que envolvem derivadas parciais.

Passo 2

Para achar a divergência usamos
\nabla \cdot \vec{F} =\frac{\partial F_x}{\partial x} +\frac{\partial F_y}{\partial y} +\frac{\partial F_z}{\partial z}.

Passo 3

Calculamos as derivadas parciais:
\displaystyle \frac{\partial}{\partial x}\bigl(e^{x}\sin y\bigr)=e^{x}\sin y.
\displaystyle \frac{\partial}{\partial y}\bigl(e^{y}\sin z\bigr)=e^{y}\sin z.
\displaystyle \frac{\partial}{\partial z}\bigl(e^{z}\sin x\bigr)=e^{z}\sin x.
Somando, obtemos a divergência.

Passo 4

Para o rotacional usamos
\nabla \times \vec{F} =\Bigl(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\Bigr)\vec{i} +\Bigl(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\Bigr)\vec{j} +\Bigl(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\Bigr)\vec{k}.

Passo 5

Componente \vec{i}: \displaystyle \frac{\partial F_z}{\partial y} =\frac{\partial}{\partial y}(e^{z}\sin x)=0, \displaystyle \frac{\partial F_y}{\partial z} =\frac{\partial}{\partial z}(e^{y}\sin z)=e^{y}\cos z,
logo (\nabla\times\vec{F})_x=0-e^{y}\cos z=-e^{y}\cos z.

Passo 6

Componente \vec{j}: \displaystyle \frac{\partial F_x}{\partial z} =\frac{\partial}{\partial z}(e^{x}\sin y)=0, \displaystyle \frac{\partial F_z}{\partial x} =\frac{\partial}{\partial x}(e^{z}\sin x)=e^{z}\cos x,
logo (\nabla\times\vec{F})_y=0-e^{z}\cos x=-e^{z}\cos x.

Passo 7

Componente \vec{k}: \displaystyle \frac{\partial F_y}{\partial x} =\frac{\partial}{\partial x}(e^{y}\sin z)=0, \displaystyle \frac{\partial F_x}{\partial y} =\frac{\partial}{\partial y}(e^{x}\sin y)=e^{x}\cos y,
logo (\nabla\times\vec{F})_z=0-e^{x}\cos y=-e^{x}\cos y.

Resposta

\nabla \times \vec{F} = -e^{y}\cos z\,\vec{i} - e^{z}\cos x\,\vec{j} - e^{x}\cos y\,\vec{k}
\nabla \cdot \vec{F} = e^{x}\sin y + e^{y}\sin z + e^{z}\sin x

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