Determine (a) a rotacional e (b) a divergente do campo vetorial.
\mathbf{F}(x, y, z)=(x+3 y-5 z) \mathbf{i}+(z-3 y) \mathbf{j}+(5 x+6 y-z) \mathbf{k}

Passo 1

O enunciado dá o campo vetorial em \mathbb{R}^3
\mathbf{F}(x,y,z) = (x + 3y - 5z)\,\mathbf{i} \;+\;(z - 3y)\,\mathbf{j}\;+\;(5x + 6y - z)\,\mathbf{k}
e pede (a) o rotacional \nabla \times \mathbf{F} e (b) a divergente \nabla\cdot \mathbf{F}. Vamos usar as definições usuais de cálculo vetorial, que envolvem derivadas parciais.

Passo 2

Para calcular o rotacional aplicamos
\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\\ \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z}\\\ x+3y-5z & z-3y & 5x+6y-z \end{vmatrix}.
Isso dá as componentes
(\nabla\times\mathbf{F})_x = \frac{\partial}{\partial y}(5x+6y-z)\;-\;\frac{\partial}{\partial z}(z-3y) = 6 - 1 = 5,
(\nabla\times\mathbf{F})_y = \frac{\partial}{\partial z}(x+3y-5z)\;-\;\frac{\partial}{\partial x}(5x+6y-z) = -5 - 5 = -10,
(\nabla\times\mathbf{F})_z = \frac{\partial}{\partial x}(z-3y)\;-\;\frac{\partial}{\partial y}(x+3y-5z) = 0 - 3 = -3.
Logo
\nabla \times \mathbf{F} = 5\,\mathbf{i} - 10\,\mathbf{j} - 3\,\mathbf{k}.

Passo 3

Para a divergente usamos
\nabla\cdot \mathbf{F} = \frac{\partial}{\partial x}(x+3y-5z) + \frac{\partial}{\partial y}(z-3y) + \frac{\partial}{\partial z}(5x+6y-z).
Calculando cada termo:
\partial_x(x+3y-5z)=1,
\partial_y(z-3y)=-3,
\partial_z(5x+6y-z)=-1.
Somando, obtemos
\nabla\cdot \mathbf{F} = 1 - 3 - 1 = -3.

Resposta

(a) \nabla \times \mathbf{F} = 5\,\mathbf{i} - 10\,\mathbf{j} - 3\,\mathbf{k}.
(b) \nabla\cdot \mathbf{F} = -3.

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