Matéria: Matemática
Autor: anelisegarcia
Perguntado 5 anos atrás

O rotacional do campo vetorial F(x, y, z) =(x² + z, y - xy, z + xyz) é dado por:

Respostas

respondido por: solkarped

✅ Após resolver os cálculos, concluímos que o rotacional do referido campo vetorial é:

$\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\textrm{rot}\:\vec{F} = xz\,\vec{i} - (yz - 1)\,\vec{j} - y\,\vec{k}\:\:\:}}\end{gathered}$}$

Seja a função:

$\Large\displaystyle\text{$\begin{gathered} F(x, y, z) = (x^{2} + z,\,y - xy,\,z + xyz)\end{gathered}$}$

Organizando o campo vetorial, temos:

$\Large\displaystyle\text{$\begin{gathered} \vec{F}(x, y, z) = (x^{2} + z)\vec{i} + (y - xy)\vec{j} + (z + xyz)\vec{k}\end{gathered}$}$

Sendo F um campo vetorial em R³, podemos dizer que o rotacional de F - denotado por "rot F" - é o produto vetorial entre o operador diferencial e F, isto é:

$\Large\displaystyle\text{$\begin{gathered} \textrm{rot}\:\vec{F} = \nabla\wedge\vec{F}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \bigg(\frac{\partial}{\partial x},\,\frac{\partial}{\partial y},\,\frac{\partial}{\partial z}\bigg) \wedge(X_{F}\vec{i},\,Y_{F}\vec{j},\,Z_{F}\vec{k})\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\X_{F} & Y_{F} & Z_{F}\end{vmatrix}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \begin{vmatrix}\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\Y_{F} & Z_{F}\end{vmatrix}\vec{i} - \begin{vmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\X_{F} & Z_{F}\end{vmatrix}\vec{j} + \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y}\\X_{F} & Y_{F}\end{vmatrix}\vec{k}\end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered} = \left(\frac{\partial Z_{F}}{\partial y} - \frac{\partial Y_{F}}{\partial z}\right)\vec{i} - \left(\frac{\partial Z_{F}}{\partial x} - \frac{\partial X_{F}}{\partial z}\right)\vec{j} + \left(\frac{\partial Y_{F}}{\partial x} - \frac{\partial X_{F}}{\partial y}\right)\vec{k}\end{gathered}$}$