Calcule o rotacional e o divergente para os seguintes campos vetoriais:
\vec{F}(x, y, z)=\left(y^{2}+z^{2}\right) \widehat{i}+x e^{y} \operatorname{Cos}(z) \widehat{j}-y e^{x} \operatorname{Sen}(z) \widehat{k}

Passo 1

O enunciado pede para calcular o divergente e o rotacional do campo vetorial
\vec{F}(x,y,z) = \bigl(y^{2}+z^{2}\bigr)\,\widehat{\imath} \;+\; x\,e^{y}\cos(z)\,\widehat{\jmath} \;-\; y\,e^{x}\sin(z)\,\widehat{k}.
O conteúdo aqui é cálculo vetorial:
  1. divergente: \nabla\cdot\vec{F}=\partial_xF_{1}+\partial_yF_{2}+\partial_zF_{3},
  2. rotacional: \nabla\times\vec{F}=\bigl(\partial_yF_{3}-\partial_zF_{2}\bigr)\,\widehat{\imath} + \bigl(\partial_zF_{1}-\partial_xF_{3}\bigr)\,\widehat{\jmath} + \bigl(\partial_xF_{2}-\partial_yF_{1}\bigr)\,\widehat{k}.

Passo 2

Calculando o divergente:
\nabla\cdot\vec{F} = \frac{\partial}{\partial x}\bigl(y^{2}+z^{2}\bigr) + \frac{\partial}{\partial y}\bigl(x\,e^{y}\cos z\bigr) + \frac{\partial}{\partial z}\bigl(-\,y\,e^{x}\sin z\bigr).
Agora cada termo:
\frac{\partial}{\partial x}(y^{2}+z^{2}) = 0, \quad \frac{\partial}{\partial y}(x\,e^{y}\cos z) = x\,e^{y}\cos z, \quad \frac{\partial}{\partial z}(-\,y\,e^{x}\sin z) = -\,y\,e^{x}\cos z.
Logo
\nabla\cdot\vec{F} = 0 + x\,e^{y}\cos z \;-\; y\,e^{x}\cos z = x\,e^{y}\cos z \;-\; y\,e^{x}\cos z.

Passo 3

Calculando o rotacional: use
\nabla\times\vec{F} = \bigl(\partial_{y}F_{3}-\partial_{z}F_{2}\bigr)\,\widehat{\imath} + \bigl(\partial_{z}F_{1}-\partial_{x}F_{3}\bigr)\,\widehat{\jmath} + \bigl(\partial_{x}F_{2}-\partial_{y}F_{1}\bigr)\,\widehat{k}.
Primeiro os parciais:
\partial_{y}F_{3} = \frac{\partial}{\partial y}\bigl(-y\,e^{x}\sin z\bigr) = -\,e^{x}\sin z, \quad \partial_{z}F_{2} = \frac{\partial}{\partial z}\bigl(x\,e^{y}\cos z\bigr) = -\,x\,e^{y}\sin z,
\partial_{z}F_{1} = \frac{\partial}{\partial z}\bigl(y^{2}+z^{2}\bigr) = 2\,z, \quad \partial_{x}F_{3} = \frac{\partial}{\partial x}\bigl(-y\,e^{x}\sin z\bigr) = -\,y\,e^{x}\sin z,
\partial_{x}F_{2} = \frac{\partial}{\partial x}\bigl(x\,e^{y}\cos z\bigr) = e^{y}\cos z, \quad \partial_{y}F_{1} = \frac{\partial}{\partial y}\bigl(y^{2}+z^{2}\bigr) = 2\,y.
Agora cada componente do rotacional:
(\nabla\times\vec{F})_{x} = (-\,e^{x}\sin z)\;-\;( -\,x\,e^{y}\sin z) = -\,e^{x}\sin z \;+\; x\,e^{y}\sin z,
(\nabla\times\vec{F})_{y} = 2\,z \;-\;( -\,y\,e^{x}\sin z) = 2\,z \;+\; y\,e^{x}\sin z,
(\nabla\times\vec{F})_{z} = e^{y}\cos z \;-\; 2\,y.
Logo
\nabla\times\vec{F} = \bigl(-\,e^{x}\sin z + x\,e^{y}\sin z\bigr)\widehat{\imath} + \bigl(2\,z + y\,e^{x}\sin z\bigr)\widehat{\jmath} + \bigl(e^{y}\cos z - 2\,y\bigr)\widehat{k}.

Resposta

Divergente:
\nabla\cdot\vec{F} = x\,e^{y}\cos z \;-\; y\,e^{x}\cos z.
Rotacional:
\nabla\times\vec{F} = \bigl(-\,e^{x}\sin z + x\,e^{y}\sin z\bigr)\widehat{\imath} + \bigl(2\,z + y\,e^{x}\sin z\bigr)\widehat{\jmath} + \bigl(e^{y}\cos z - 2\,y\bigr)\widehat{k}.

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