Calcule o rotacional de cada um dos seguintes campos vetoriais:
\mathbf{B}=\rho z^{2} \mathbf{a}_{\rho}+\rho \operatorname{sen}^{2}(\varphi) \mathbf{a}_{\varphi}+2 \rho z \operatorname{sen}^{2}(\varphi) \mathbf{a}_{\mathbf{z}}

Passo 1

O enunciado pede o rotacional (curl) do campo vetorial em coordenadas cilíndricas
\mathbf{B}=\rho\,z^{2}\,\mathbf{a}_{\rho}+\rho\,\operatorname{sen}^{2}(\varphi)\,\mathbf{a}_{\varphi}+2\,\rho\,z\,\operatorname{sen}^{2}(\varphi)\,\mathbf{a}_{z}.
O conteúdo aqui é o cálculo de \nabla\times\mathbf{B} em coordenadas cilíndricas, usando a forma conhecida das componentes do rotacional nesse sistema.

Passo 2

Em coordenadas cilíndricas (\rho,\varphi,z), para
\mathbf{B}=A_{\rho}\,\mathbf{a}_{\rho}+A_{\varphi}\,\mathbf{a}_{\varphi}+A_{z}\,\mathbf{a}_{z},
as componentes do rotacional são
(\nabla\times\mathbf{B})_{\rho} =\frac{1}{\rho}\Bigl(\frac{\partial A_{z}}{\partial\varphi}-\frac{\partial(\rho\,A_{\varphi})}{\partial z}\Bigr),
(\nabla\times\mathbf{B})_{\varphi} =\frac{\partial A_{\rho}}{\partial z}-\frac{\partial A_{z}}{\partial \rho},
(\nabla\times\mathbf{B})_{z} =\frac{1}{\rho}\Bigl(\frac{\partial(\rho\,A_{\varphi})}{\partial \rho}-\frac{\partial A_{\rho}}{\partial\varphi}\Bigr).

Passo 3

Identificamos
A_{\rho}=\rho\,z^{2},\quad A_{\varphi}=\rho\,\operatorname{sen}^{2}(\varphi),\quad A_{z}=2\,\rho\,z\,\operatorname{sen}^{2}(\varphi).
Para a componente radial:
\frac{\partial A_{z}}{\partial\varphi} =\frac{\partial}{\partial\varphi}\bigl(2\,\rho\,z\,\operatorname{sen}^{2}\varphi\bigr) =2\,\rho\,z\cdot2\,\operatorname{sen}\varphi\cos\varphi =4\,\rho\,z\,\operatorname{sen}\varphi\cos\varphi,
\frac{\partial(\rho\,A_{\varphi})}{\partial z} =\frac{\partial}{\partial z}\bigl(\rho^{2}\,\operatorname{sen}^{2}\varphi\bigr) =0.
Logo,
(\nabla\times\mathbf{B})_{\rho} =\frac{1}{\rho}\bigl(4\,\rho\,z\,\operatorname{sen}\varphi\cos\varphi-0\bigr) =4\,z\,\operatorname{sen}\varphi\cos\varphi.

Passo 4

Para a componente azimutal:
\frac{\partial A_{\rho}}{\partial z} =\frac{\partial}{\partial z}(\rho\,z^{2}) =2\,\rho\,z,
\frac{\partial A_{z}}{\partial \rho} =\frac{\partial}{\partial \rho}\bigl(2\,\rho\,z\,\operatorname{sen}^{2}\varphi\bigr) =2\,z\,\operatorname{sen}^{2}\varphi.
Logo,
(\nabla\times\mathbf{B})_{\varphi} =2\,\rho\,z-2\,z\,\operatorname{sen}^{2}\varphi =2\,z\bigl(\rho-\operatorname{sen}^{2}\varphi\bigr).

Passo 5

Para a componente axial:
\frac{\partial(\rho\,A_{\varphi})}{\partial \rho} =\frac{\partial}{\partial \rho}\bigl(\rho^{2}\,\operatorname{sen}^{2}\varphi\bigr) =2\,\rho\,\operatorname{sen}^{2}\varphi,
\frac{\partial A_{\rho}}{\partial\varphi} =\frac{\partial}{\partial\varphi}(\rho\,z^{2}) =0.
Logo,
(\nabla\times\mathbf{B})_{z} =\frac{1}{\rho}\bigl(2\,\rho\,\operatorname{sen}^{2}\varphi-0\bigr) =2\,\operatorname{sen}^{2}\varphi.

Passo 6

Reunindo as três componentes no sistema cilíndrico, obtemos
\nabla\times\mathbf{B} =4\,z\,\operatorname{sen}\varphi\cos\varphi\,\mathbf{a}_{\rho} +2\,z\bigl(\rho-\operatorname{sen}^{2}\varphi\bigr)\,\mathbf{a}_{\varphi} +2\,\operatorname{sen}^{2}\varphi\,\mathbf{a}_{z}.

Resposta

\nabla\times\mathbf{B} =4\,z\,\operatorname{sen}\varphi\cos\varphi\;\mathbf{a}_{\rho} +2\,z\bigl(\rho-\operatorname{sen}^{2}\varphi\bigr)\;\mathbf{a}_{\varphi} +2\,\operatorname{sen}^{2}\varphi\;\mathbf{a}_{z}.

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