Calcule \nabla \times \mathbf{A} para os seguintes campos:
\mathbf{A}=r^{2} \cdot \sin (\theta) \mathbf{a}_{r}+\frac{r}{\cos (\varphi)} \mathbf{a}_{\theta}.

Passo 1

O enunciado pede o rotacional (curl) de um campo vetorial dado em coordenadas esféricas. O conteúdo é cálculo vetorial em coordenadas curvilíneas: usaremos as expressões do rotacional no sistema (r,\theta,\varphi), em que \mathbf{a}_r,\mathbf{a}_\theta,\mathbf{a}_\varphi são os versores locais, e o campo é
\mathbf{A}=A_r\,\mathbf{a}_r + A_\theta\,\mathbf{a}_\theta + A_\varphi\,\mathbf{a}_\varphi
com
A_r = r^2\sin\theta,\quad A_\theta=\frac{r}{\cos\varphi},\quad A_\varphi=0.

Passo 2

As componentes do rotacional em esféricas são
(\nabla\times\mathbf{A})_r = \frac{1}{r\sin\theta}\Bigl[\frac{\partial}{\partial\theta}\bigl(A_\varphi\sin\theta\bigr)-\frac{\partial A_\theta}{\partial\varphi}\Bigr],
(\nabla\times\mathbf{A})_\theta = \frac{1}{r}\Bigl[\frac{1}{\sin\theta}\frac{\partial A_r}{\partial\varphi} - \frac{\partial}{\partial r}\bigl(r\,A_\varphi\bigr)\Bigr],
(\nabla\times\mathbf{A})_\varphi = \frac{1}{r}\Bigl[\frac{\partial}{\partial r}\bigl(r\,A_\theta\bigr) - \frac{\partial A_r}{\partial\theta}\Bigr].

Passo 3

Calculemos a componente radial:
A_\varphi =0 implica \frac{\partial}{\partial\theta}(A_\varphi\sin\theta)=0.
\frac{\partial A_\theta}{\partial\varphi} =\frac{\partial}{\partial\varphi}\Bigl(\frac{r}{\cos\varphi}\Bigr) =r\;\frac{\sin\varphi}{\cos^2\varphi}.
Logo
(\nabla\times\mathbf{A})_r =-\frac{1}{r\sin\theta}\;r\,\frac{\sin\varphi}{\cos^2\varphi} =-\frac{\sin\varphi}{\sin\theta\,\cos^2\varphi}.

Passo 4

Para a componente \theta:
\frac{\partial A_r}{\partial\varphi} =\frac{\partial}{\partial\varphi}\bigl(r^2\sin\theta\bigr)=0, \qquad \frac{\partial}{\partial r}(r\,A_\varphi)=0.
Assim
(\nabla\times\mathbf{A})_\theta =\frac{1}{r}\Bigl[0-0\Bigr]=0.

Passo 5

Finalmente, a componente \varphi:
r\,A_\theta = \frac{r^2}{\cos\varphi} \quad\Longrightarrow\quad \frac{\partial}{\partial r}(r\,A_\theta) =\frac{2r}{\cos\varphi},
\frac{\partial A_r}{\partial\theta} =\frac{\partial}{\partial\theta}(r^2\sin\theta) =r^2\cos\theta.
Portanto
(\nabla\times\mathbf{A})_\varphi =\frac{1}{r}\Bigl[\frac{2r}{\cos\varphi}-r^2\cos\theta\Bigr] =\frac{2}{\cos\varphi}-r\cos\theta.

Resposta

\nabla\times\mathbf{A} =-\frac{\sin\varphi}{\sin\theta\,\cos^2\varphi}\,\mathbf{a}_r \;+\;0\;\mathbf{a}_\theta \;+\;\Bigl(\frac{2}{\cos\varphi}-r\cos\theta\Bigr)\,\mathbf{a}_\varphi.

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