Matéria: Matemática
Autor: amiltonsilvacorrea
Perguntado 3 anos atrás

calcular o rotacional x^2yz+xy^2z+xyz^2

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{V}= x(z^{2} - y^{2})\,\vec{i} - y(z^{2} - x^{2})\,\vec{j} + z(y^{2} - x^{2})\,\vec{k}\:\:\:}}\end{gathered}$}$

Seja o dado:

$\Large\displaystyle\text{$\begin{gathered} x^{2}yz + xy^{2}z + xyz^{2}\end{gathered}$}$

Organizando o campo vetorial, temos:

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

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

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

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

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

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

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

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

✅ Portanto, a resposta é:

$\Large\displaystyle\text{$\begin{gathered} \textrm{rot}\: \vec{V}= x(z^{2} - y^{2})\,\vec{i} - y(z^{2} - x^{2})\,\vec{j} + z(y^{2} - x^{2})\,\vec{k}\end{gathered}$}$