Calcule \nabla \times \mathbf{A} para os seguintes campos:
\mathbf{A}=\rho \cdot \sin ^{2}(\varphi) \mathbf{a}_{\rho}-\rho^{2} \cdot z \cos (\varphi) \mathbf{a}_{\varphi}.

Passo 1

O enunciado pede o rotacional de
\mathbf{A} = \rho\,\sin^2(\varphi)\,\mathbf{a}_\rho \;-\;\rho^2\,z\,\cos(\varphi)\,\mathbf{a}_\varphi
em coordenadas cilíndricas (\rho,\varphi,z).
Para isso, vamos usar a fórmula do rotacional em coordenadas cilíndricas para um campo \mathbf{A}=A_\rho\,\mathbf{a}_\rho + A_\varphi\,\mathbf{a}_\varphi + A_z\,\mathbf{a}_z:
\begin{aligned} (\nabla\times\mathbf{A})_\rho &= \frac{1}{\rho}\,\frac{\partial A_z}{\partial \varphi} \;-\;\frac{\partial A_\varphi}{\partial z},\\\ (\nabla\times\mathbf{A})_\varphi &= \frac{\partial A_\rho}{\partial z} \;-\;\frac{\partial A_z}{\partial \rho},\\\ (\nabla\times\mathbf{A})_z &= \frac{1}{\rho}\left(\frac{\partial(\rho\,A_\varphi)}{\partial \rho} \;-\;\frac{\partial A_\rho}{\partial \varphi}\right). \end{aligned}

Passo 2

Identificamos as componentes:
A_\rho = \rho\,\sin^2(\varphi), \quad A_\varphi = -\,\rho^2\,z\,\cos(\varphi), \quad A_z = 0.
Começamos por (\nabla\times\mathbf{A})_\rho:
(\nabla\times\mathbf{A})_\rho = \frac{1}{\rho}\,\frac{\partial A_z}{\partial \varphi} - \frac{\partial A_\varphi}{\partial z} = 0 - \frac{\partial}{\partial z}\bigl(-\rho^2 z\cos(\varphi)\bigr).
Como \partial_z\bigl(-\rho^2 z\cos\varphi\bigr)=-\rho^2\cos\varphi, temos
(\nabla\times\mathbf{A})_\rho = -\bigl(-\rho^2\cos\varphi\bigr) = \rho^2\cos(\varphi).

Passo 3

Calculamos (\nabla\times\mathbf{A})_\varphi:
(\nabla\times\mathbf{A})_\varphi = \frac{\partial A_\rho}{\partial z} - \frac{\partial A_z}{\partial \rho} = \frac{\partial}{\partial z}\bigl(\rho\sin^2\varphi\bigr) - 0 = 0.

Passo 4

Agora (\nabla\times\mathbf{A})_z:
\frac{\partial(\rho\,A_\varphi)}{\partial \rho} = \frac{\partial}{\partial \rho}\bigl(-\rho^3 z\cos\varphi\bigr) = -3\rho^2\,z\,\cos\varphi,
\frac{\partial A_\rho}{\partial \varphi} = \frac{\partial}{\partial \varphi}\bigl(\rho\sin^2\varphi\bigr) = \rho\cdot 2\sin\varphi\cos\varphi = \rho\,\sin(2\varphi).
Logo,
(\nabla\times\mathbf{A})_z = \frac{1}{\rho}\Bigl(-3\rho^2 z\cos\varphi - \rho\sin(2\varphi)\Bigr) = -3\rho\,z\,\cos\varphi \;-\;\sin(2\varphi).

Resposta

\nabla\times\mathbf{A} = \rho^2\cos(\varphi)\,\mathbf{a}_\rho \;+\;0\;\mathbf{a}_\varphi \;+\;\bigl(-3\rho\,z\,\cos\varphi - \sin(2\varphi)\bigr)\,\mathbf{a}_z.

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