Question

O rotacional do campo vetorial F(x,y,z)=xzi + xyz j - y²k, é o vetor:

Answer 1

Boa noite!!

Para calcular o rotacional de F, temos a definição:

$rot(F) = \nabla\times F(x,y,z)\\\\\Rightarrow rot(F) = \nabla\times (xz, xyz, -y^2)$

Temos também que $\nabla$ é definido por:

$\nabla = (\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$

Assim:

$rot(F) = (\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z})\times (xz, xyz, -y^2)$

Pode-se calcular o produto vetorial através do determinante dessa matriz:

$\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\xz&xyz&-y^2\end{array}\right]$

Ou, bastará calcular:

$rot(F) = (\dfrac{\partial (-y^2)}{\partial y}-\dfrac{\partial xyz}{\partial z})i +(\dfrac{\partial xz)}{\partial z}-\dfrac{\partial (-y^2)}{\partial x})j + (\dfrac{\partial xyz}{\partial x}-\dfrac{\partial xz}{\partial y})k$

Feito os cálculos teremos o seguinte resultado:

$rot(F) = i(-xy -2y) +xj +yzk\\\\\Rightarrow rot(F) = (-xy -2y,~x,~yz)$

Bons estudos!

Answer 2

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

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

   

Seja a função:

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

Organizando o campo vetorial, temos:

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

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:

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

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

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

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

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

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

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

✅ Portanto, a resposta é:

         $\Large\displaystyle\begin{gathered} \textrm{rot}\:\vec{F} = -y(2 + x)\vec{i} + x\vec{j} + yz\vec{k}\end{gathered}$

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