In each of the following parts, find a numerical value for div \mathbf{D} at the point specified:
\mathbf{D}=2 \rho z^{2} \sin ^{2} \varphi \mathbf{a}_{\rho}+\rho z^{2} \sin 2 \varphi \mathbf{a}_{\varphi}+2 \rho^{2} z \sin ^{2} \varphi \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} at P_{B}\left(\rho=2, \varphi=110^{\circ}, z=-1\right);

Passo 1

O enunciado pede calcular a divergência \nabla\!\cdot\mathbf{D} no ponto P_{B}(\rho=2,\;\varphi=110^\circ,\;z=-1). O conteúdo é divergência em coordenadas cilíndricas.

Passo 2

Em coordenadas cilíndricas a divergência de \mathbf{D}=D_{\rho}\,\mathbf{a}_{\rho}+D_{\varphi}\,\mathbf{a}_{\varphi}+D_{z}\,\mathbf{a}_{z} vem dada por
\nabla\!\cdot\mathbf{D} =\frac{1}{\rho}\frac{\partial(\rho\,D_{\rho})}{\partial\rho} +\frac{1}{\rho}\frac{\partial D_{\varphi}}{\partial\varphi} +\frac{\partial D_{z}}{\partial z}.

Passo 3

Calcular o primeiro termo:
\rho\,D_{\rho} =\rho\,(2\,\rho\,z^{2}\sin^{2}\varphi) =2\,\rho^{2}z^{2}\sin^{2}\varphi,
\frac{\partial(\rho\,D_{\rho})}{\partial\rho} =4\,\rho\,z^{2}\sin^{2}\varphi, \quad \frac{1}{\rho}\frac{\partial(\rho\,D_{\rho})}{\partial\rho} =4\,z^{2}\sin^{2}\varphi.

Passo 4

Calcular o segundo termo:
D_{\varphi} =\rho\,z^{2}\sin2\varphi =2\,\rho\,z^{2}\sin\varphi\cos\varphi,
\frac{\partial D_{\varphi}}{\partial\varphi} =2\,\rho\,z^{2}(\cos^{2}\varphi-\sin^{2}\varphi), \quad \frac{1}{\rho}\frac{\partial D_{\varphi}}{\partial\varphi} =2\,z^{2}(\cos^{2}\varphi-\sin^{2}\varphi).

Passo 5

Calcular o terceiro termo e somar tudo:
D_{z}=2\,\rho^{2}z\sin^{2}\varphi, \quad \frac{\partial D_{z}}{\partial z} =2\,\rho^{2}\sin^{2}\varphi.
Logo
\nabla\!\cdot\mathbf{D} =4\,z^{2}\sin^{2}\varphi +2\,z^{2}(\cos^{2}\varphi-\sin^{2}\varphi) +2\,\rho^{2}\sin^{2}\varphi =2\,z^{2}+2\,\rho^{2}\sin^{2}\varphi.

Passo 6

Agora avaliamos em \rho=2,\;z=-1,\;\varphi=110^\circ:
z^{2}=1,\quad \rho^{2}=4,\quad \sin^{2}(110^\circ)=\sin^{2}(70^\circ)\approx0{,}8830,
\nabla\!\cdot\mathbf{D} =2\cdot1+2\cdot4\cdot0{,}8830 \approx2+7{,}064=9{,}064.

Resposta

\nabla\!\cdot\mathbf{D}\big|_{P_{B}}\approx9{,}06\quad\bigl(\mathrm{C}/\mathrm{m}^{3}\bigr).

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