Dado o campo vetorial \vec{F}(x, y, z)=e^{x y} \vec{i}-\cos y \vec{j}+\sin ^{2} z \vec{k}, calcule \vec{\nabla} \cdot \vec{F} e \vec{\nabla} \times \vec{F}.

Passo 1

O enunciado pede para calcular o divergente \nabla\cdot \vec{F} e o rotacional \nabla\times \vec{F} do campo \vec{F}(x,y,z)=e^{xy}\,\vec{i}-\cos y\,\vec{j}+\sin^2 z\,\vec{k}.
Vamos usar as definições
\nabla\cdot \vec{F} =\frac{\partial F_x}{\partial x} +\frac{\partial F_y}{\partial y} +\frac{\partial F_z}{\partial z},
e
\nabla\times \vec{F} =\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\[6pt] \partial_x & \partial_y & \partial_z \\[3pt] F_x & F_y & F_z \end{vmatrix}.]

Passo 2

Para o divergente calculamos
\nabla\cdot \vec{F} =\frac{\partial}{\partial x}(e^{xy}) +\frac{\partial}{\partial y}(-\cos y) +\frac{\partial}{\partial z}(\sin^2 z).
Como \frac{\partial}{\partial x}(e^{xy})=y\,e^{xy}, \frac{\partial}{\partial y}(-\cos y)=\sin y e \frac{\partial}{\partial z}(\sin^2 z)=2\sin z\cos z,
segue
\nabla\cdot \vec{F} =y\,e^{xy}+\sin y+2\sin z\cos z.

Passo 3

Para o rotacional usamos
\nabla\times \vec{F} =\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\[6pt] \partial_x & \partial_y & \partial_z \\[3pt] e^{xy} & -\cos y & \sin^2 z \end{vmatrix}.
Expandindo:
componente \vec{i}: \partial_y(\sin^2 z)-\partial_z(-\cos y)=0-0=0.
componente \vec{j}: \partial_z(e^{xy})-\partial_x(\sin^2 z)=0-0=0.
componente \vec{k}: \partial_x(-\cos y)-\partial_y(e^{xy})=0-(x\,e^{xy})=-x\,e^{xy}.]

Passo 4

Logo
\nabla\times \vec{F} =0\,\vec{i}+0\,\vec{j}-x\,e^{xy}\,\vec{k}.

Resposta

\nabla\cdot \vec{F}=y\,e^{xy}+\sin y+2\sin z\cos z. \nabla\times \vec{F}=-x\,e^{xy}\,\vec{k}.

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