Determine (a) a rotacional e (b) a divergente do campo vetorial.
F(x, y, z)-x y j+x y z k

Passo 1

O enunciado pede o cálculo do rotacional e da divergente do campo vetorial \r
\mathbf{F}(x,y,z) = -\,x\,y\,\mathbf{j} + x\,y\,z\,\mathbf{k}. \r
Para isso, vamos usar as definições em coordenadas cartesianas em \mathbb{R}^3.

Passo 2

Definimos \mathbf{F}=(P,Q,R) com \r
P(x,y,z)=0,\quad Q(x,y,z)=-x\,y,\quad R(x,y,z)=x\,y\,z.\r
O rotacional é \r
\nabla\times\mathbf{F} =\Bigl( \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\; \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\; \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \Bigr).\r
Calculando cada derivada: \r
\frac{\partial R}{\partial y} = x\,z, \r
\frac{\partial Q}{\partial z} = 0, \r
\frac{\partial P}{\partial z} = 0, \r
\frac{\partial R}{\partial x} = y\,z, \r
\frac{\partial Q}{\partial x} = -\,y, \r
\frac{\partial P}{\partial y}=0. \r
Logo \r
\nabla\times\mathbf{F} =(x\,z - 0)\,\mathbf{i}+(0 - y\,z)\,\mathbf{j}+(-y - 0)\,\mathbf{k} = x\,z\,\mathbf{i} - y\,z\,\mathbf{j} - y\,\mathbf{k}.

Passo 3

A divergente é \r
\nabla\cdot\mathbf{F} =\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}.\r
Usando
\frac{\partial P}{\partial x}=0,
\frac{\partial Q}{\partial y}=-x,
\frac{\partial R}{\partial z}=x\,y,
temos
\nabla\cdot\mathbf{F} =0 + (-x) + x\,y = x\,(y-1).

Resposta

(a) \nabla\times\mathbf{F} = x\,z\,\mathbf{i} - y\,z\,\mathbf{j} - y\,\mathbf{k}. \r
(b) \nabla\cdot\mathbf{F} = x\,(y-1).

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