Nos problemas 17-20, determine o rotacional e o divergente do campo vetorial indicado.
\boldsymbol{F}(x, y, z)=(x-y)^{3} \boldsymbol{i}+e^{-y z} \boldsymbol{j}+x y e^{2 y} \boldsymbol{k}

Passo 1

O campo vetorial é
\boldsymbol{F}(x,y,z) = (x - y)^{3}\,\mathbf{i} + e^{-y z}\,\mathbf{j} + x y e^{2y}\,\mathbf{k}.
Precisamos de duas operações de cálculo vetorial: o divergente e o rotacional. Usamos as definições
\nabla\cdot\boldsymbol{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}
e
\nabla\times\boldsymbol{F} = \begin{pmatrix} \displaystyle \frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}\\[6pt] \displaystyle \frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}\\[6pt] \displaystyle \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y} \end{pmatrix}.]

Passo 2

Calcular o divergente componente a componente:
\frac{\partial}{\partial x}(x-y)^{3} = 3(x-y)^{2}, \qquad \frac{\partial}{\partial y}e^{-y z} = -z\,e^{-y z}, \qquad \frac{\partial}{\partial z}\bigl(xy e^{2y}\bigr) = 0.
Logo
\nabla\cdot\boldsymbol{F} = 3(x-y)^{2} \;-\; z\,e^{-y z}.

Passo 3

Calcular o rotacional em cada componente:
Para a componente x:
\frac{\partial}{\partial y}\bigl(xy e^{2y}\bigr) = x\bigl(e^{2y} + 2y e^{2y}\bigr) = x(1+2y)e^{2y},
\frac{\partial}{\partial z}e^{-y z} = -y\,e^{-y z},
(\nabla\times\boldsymbol{F})_{x} = x(1+2y)e^{2y} - \bigl(-y\,e^{-y z}\bigr) = x(1+2y)e^{2y} + y\,e^{-y z}.
Para a componente y:
\frac{\partial}{\partial z}(x-y)^{3} = 0, \qquad \frac{\partial}{\partial x}\bigl(xy e^{2y}\bigr) = y\,e^{2y},
(\nabla\times\boldsymbol{F})_{y} = 0 - y\,e^{2y} = -y\,e^{2y}.
Para a componente z:
\frac{\partial}{\partial x}e^{-y z} = 0, \qquad \frac{\partial}{\partial y}(x-y)^{3} = -3(x-y)^{2},
(\nabla\times\boldsymbol{F})_{z} = 0 - \bigl(-3(x-y)^{2}\bigr) = 3(x-y)^{2}.
Assim
\nabla\times\boldsymbol{F} = \bigl(x(1+2y)e^{2y} + y\,e^{-y z}\bigr)\,\mathbf{i} \;-\; y\,e^{2y}\,\mathbf{j} \;+\; 3(x-y)^{2}\,\mathbf{k}.

Resposta

\nabla\cdot\boldsymbol{F} = 3(x-y)^{2} - z\,e^{-y z},
\nabla\times\boldsymbol{F} = \bigl(x(1+2y)e^{2y} + y\,e^{-y z}\bigr)\,\mathbf{i} - y\,e^{2y}\,\mathbf{j} + 3(x-y)^{2}\,\mathbf{k}.

Exercícios relacionados