Determine (a) o rotacional e (b) o divergente do campo vetorial.
\mathbf{F}(x, y, z)=\left\langle e^{x} \operatorname{sen} y, e^{y} \operatorname{sen} \mathrm{z}, e^{z} \operatorname{sen} \mathrm{x}\right\rangle

Passo 1

O enunciado pede (a) o rotacional e (b) o divergente do campo vetorial \mathbf{F}(x,y,z)=\bigl\langle e^{x}\sin y,\;e^{y}\sin z,\;e^{z}\sin x\bigr\rangle.
O conteúdo é cálculo vetorial: usamos as definições \nabla\times\mathbf{F} e \nabla\cdot\mathbf{F}.

Passo 2

A definição do rotacional em componentes é
\nabla\times\mathbf{F} =\Bigl\langle \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\rangle.
E a definição do divergente é
\nabla\cdot\mathbf{F} =\frac{\partial F_{1}}{\partial x} +\frac{\partial F_{2}}{\partial y} +\frac{\partial F_{3}}{\partial z}.

Passo 3

Para a primeira componente do rotacional:
\frac{\partial F_{3}}{\partial y} =\frac{\partial}{\partial y}\bigl(e^{z}\sin x\bigr)=0, \qquad \frac{\partial F_{2}}{\partial z} =\frac{\partial}{\partial z}\bigl(e^{y}\sin z\bigr) =e^{y}\cos z,
logo
\frac{\partial F_{3}}{\partial y}-\frac{\partial F_{2}}{\partial z} =0 - e^{y}\cos z =-\,e^{y}\cos z.

Passo 4

Para a segunda componente do rotacional:
\frac{\partial F_{1}}{\partial z} =\frac{\partial}{\partial z}\bigl(e^{x}\sin y\bigr)=0, \qquad \frac{\partial F_{3}}{\partial x} =\frac{\partial}{\partial x}\bigl(e^{z}\sin x\bigr) =e^{z}\cos x,
logo
\frac{\partial F_{1}}{\partial z}-\frac{\partial F_{3}}{\partial x} =0 - e^{z}\cos x =-\,e^{z}\cos x.

Passo 5

Para a terceira componente do rotacional:
\frac{\partial F_{2}}{\partial x} =\frac{\partial}{\partial x}\bigl(e^{y}\sin z\bigr)=0, \qquad \frac{\partial F_{1}}{\partial y} =\frac{\partial}{\partial y}\bigl(e^{x}\sin y\bigr) =e^{x}\cos y,
logo
\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} =0 - e^{x}\cos y =-\,e^{x}\cos y.

Passo 6

Para o divergente:
\frac{\partial F_{1}}{\partial x} =\frac{\partial}{\partial x}\bigl(e^{x}\sin y\bigr) =e^{x}\sin y, \quad \frac{\partial F_{2}}{\partial y} =e^{y}\sin z, \quad \frac{\partial F_{3}}{\partial z} =e^{z}\sin x,
portanto
\nabla\cdot\mathbf{F} =e^{x}\sin y + e^{y}\sin z + e^{z}\sin x.

Resposta

(a) \displaystyle \nabla\times\mathbf{F} =\bigl\langle -e^{y}\cos z,\;-e^{z}\cos x,\;-e^{x}\cos y\bigr\rangle. (b) \displaystyle \nabla\cdot\mathbf{F} =e^{x}\sin y + e^{y}\sin z + e^{z}\sin x.

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