Question

Pergunta de Cálculo III - Engenharia:

1) Encontre o rotacional de F= (x² - y) i + 4z j + x²k

Answer 1

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

          $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\textrm{rot}\:\vec{F} = -4\vec{i} - 2x\vec{j} + 1\vec{k}\:\:\:}}\end{gathered} }$

   

Seja:

                   $\Large\displaystyle\begin{gathered} F = (x^{2} - y) i + 4z + x^{2}k\end{gathered}$

Organizando o campo vetorial, temos:

        $\Large\displaystyle\begin{gathered} \vec{F}(x, y, z) = (x^{2} - y)\vec{i} + (4z)\vec{j} + (x^{2})\vec{k}\end{gathered}$

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:

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

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

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

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

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

                   $\Large\displaystyle\begin{gathered} = (0 - 4)\vec{i} - (2x - 0)\vec{j} + (0 - (-1))\vec{k}\end{gathered}$

                   $\Large\displaystyle\begin{gathered} = -4\vec{i} - 2x\vec{j} + 1\vec{k}\end{gathered}$      

✅ Portanto, a resposta é:

         $\Large\displaystyle\begin{gathered} \textrm{rot}\:\vec{F} = -4\vec{i} - 2x\vec{j} + 1\vec{k}\end{gathered}$

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