Determine(a)a rotacional e(b)a divergente do campo vetorial.
\mathbf{F}(x, y, z)=\operatorname{sen} x \mathbf{i}+\cos x \mathbf{j}+z^{2} \mathbf{k}

Passo 1

O enunciado pede (a) o rotacional e (b) o divergente do campo vetorial
\mathbf{F}(x,y,z)=\operatorname{sen} x\,\mathbf{i}+\cos x\,\mathbf{j}+z^{2}\,\mathbf{k}.
O conteúdo aqui é cálculo vetorial: usaremos as definições
\nabla\times\mathbf{F} =\Bigl(\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z}\Bigr)\mathbf{i} +\Bigl(\frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x}\Bigr)\mathbf{j} +\Bigl(\frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y}\Bigr)\mathbf{k},
\nabla\cdot\mathbf{F} =\frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y}+\frac{\partial F_{z}}{\partial z}.

Passo 2

Identificamos as componentes de \mathbf{F}:
F_{x}=\operatorname{sen} x,\quad F_{y}=\cos x,\quad F_{z}=z^{2}.

Passo 3

Calculamos o rotacional \nabla\times\mathbf{F}:
Para a componente em \mathbf{i}:
\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z} =\frac{\partial (z^{2})}{\partial y}-\frac{\partial (\cos x)}{\partial z} =0-0=0.
Para a componente em \mathbf{j}:
\frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x} =\frac{\partial (\operatorname{sen} x)}{\partial z}-\frac{\partial (z^{2})}{\partial x} =0-0=0.
Para a componente em \mathbf{k}:
\frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y} =\frac{\partial (\cos x)}{\partial x}-\frac{\partial (\operatorname{sen} x)}{\partial y} =-\operatorname{sen} x-0=-\operatorname{sen} x.
Logo
\nabla\times\mathbf{F}=0\,\mathbf{i}+0\,\mathbf{j}+(-\operatorname{sen} x)\,\mathbf{k} =-\operatorname{sen} x\,\mathbf{k}.

Passo 4

Calculamos o divergente \nabla\cdot\mathbf{F}:
\frac{\partial F_{x}}{\partial x} +\frac{\partial F_{y}}{\partial y} +\frac{\partial F_{z}}{\partial z} =\frac{\partial (\operatorname{sen} x)}{\partial x} +\frac{\partial (\cos x)}{\partial y} +\frac{\partial (z^{2})}{\partial z} =\cos x+0+2z.

Resposta

(a) \nabla\times\mathbf{F}=-\operatorname{sen} x\,\mathbf{k}.
(b) \nabla\cdot\mathbf{F}=\cos x+2z.

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