Calcule o divergente e o rotacional do campo
\vec{F} para
\vec{F}=x^{2} \vec{i}-2 \vec{j}+y z \vec{k}

Passo 1

O enunciado dá o campo vetorial
\vec{F}(x,y,z)=x^{2}\,\vec{i}-2\,\vec{j}+y\,z\,\vec{k}
e pede o divergente e o rotacional. Para isso 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}\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 2

Calcular o divergente usando F_{x}=x^{2}, F_{y}=-2 e F_{z}=y\,z:
\frac{\partial F_{x}}{\partial x}=\frac{\partial x^{2}}{\partial x}=2x,\quad \frac{\partial F_{y}}{\partial y}=\frac{\partial(-2)}{\partial y}=0,\quad \frac{\partial F_{z}}{\partial z}=\frac{\partial (y\,z)}{\partial z}=y.
Logo
\mathrm{div}\,\vec{F}=2x+0+y=2x+y.

Passo 3

Calcular o rotacional componente a componente:
Com F_{x}=x^{2}, F_{y}=-2, F_{z}=y\,z, temos
(\nabla\times\vec{F})_{x} =\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z} =\frac{\partial(y\,z)}{\partial y}-\frac{\partial(-2)}{\partial z} =z-0=z,
(\nabla\times\vec{F})_{y} =\frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x} =\frac{\partial(x^{2})}{\partial z}-\frac{\partial(y\,z)}{\partial x} =0-0=0,
(\nabla\times\vec{F})_{z} =\frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y} =\frac{\partial(-2)}{\partial x}-\frac{\partial(x^{2})}{\partial y} =0-0=0.
Portanto
\nabla\times\vec{F}=z\,\vec{i}+0\,\vec{j}+0\,\vec{k}=z\,\vec{i}.

Resposta

\mathrm{div}\,\vec{F}=2x+y, \quad \nabla\times\vec{F}=z\,\vec{i}.

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