Exercício 1. Determine o rotacional e a divergência dos campos vetoriais abaixo.
\vec{F}(x, y, z)=\left(e^{x} \sin y, e^{y} \sin z, e^{z} \sin x\right).

Passo 1

O enunciado pede para calcular a divergência e o rotacional do campo vetorial
\displaystyle \vec{F}(x,y,z)=\bigl(e^{x}\sin y,\;e^{y}\sin z,\;e^{z}\sin x\bigr).
Vamos usar as definições
\mathrm{div}\,\vec{F}=\frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y}+\frac{\partial F_{z}}{\partial z},
\nabla\times\vec{F}=\Bigl(\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z},\; \frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x},\; \frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y}\Bigr).

Passo 2

Calculamos a divergência termo a termo:
\frac{\partial}{\partial x}\bigl(e^{x}\sin y\bigr)=e^{x}\sin y,\quad \frac{\partial}{\partial y}\bigl(e^{y}\sin z\bigr)=e^{y}\sin z,\quad \frac{\partial}{\partial z}\bigl(e^{z}\sin x\bigr)=e^{z}\sin x.
Somando,
\mathrm{div}\,\vec{F} =e^{x}\sin y+e^{y}\sin z+e^{z}\sin x.

Passo 3

Para o rotacional, calculamos as seis derivadas parciais envolvidas:
\frac{\partial F_{z}}{\partial y} =\frac{\partial}{\partial y}\bigl(e^{z}\sin x\bigr)=0,\quad \frac{\partial F_{y}}{\partial z} =\frac{\partial}{\partial z}\bigl(e^{y}\sin z\bigr)=e^{y}\cos z,
\frac{\partial F_{x}}{\partial z} =\frac{\partial}{\partial z}\bigl(e^{x}\sin y\bigr)=0,\quad \frac{\partial F_{z}}{\partial x} =\frac{\partial}{\partial x}\bigl(e^{z}\sin x\bigr)=e^{z}\cos x,
\frac{\partial F_{y}}{\partial x} =\frac{\partial}{\partial x}\bigl(e^{y}\sin z\bigr)=0,\quad \frac{\partial F_{x}}{\partial y} =\frac{\partial}{\partial y}\bigl(e^{x}\sin y\bigr)=e^{x}\cos y.
Substituindo na fórmula:
\nabla\times\vec{F} =\bigl(0 - e^{y}\cos z,\;0 - e^{z}\cos x,\;0 - e^{x}\cos y\bigr) =(-e^{y}\cos z,\,-e^{z}\cos x,\,-e^{x}\cos y).

Resposta

A divergência é
\mathrm{div}\,\vec{F} =e^{x}\sin y+e^{y}\sin z+e^{z}\sin x,
e o rotacional é
\nabla\times\vec{F} =\bigl(-e^{y}\cos z,\,-e^{z}\cos x,\,-e^{x}\cos y\bigr).

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