Question

seja (x,y,z) = xzi+xyzj-y²k,determine a rotacional de F

Answer 1

$rot( F) = \left[\begin{array}{ccc}\vec i&\vec j&\vec k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\P&Q&R\end{array}\right] = ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} ) \vec i + ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} )\vec j + ( \frac{\partial Q}{\partial x} - \frac{\partial P }{\partial y} ) \vec k$

temos
$\boxed{\boxed{(x;y;z) = (xz) \vec i + (xyz) \vec j+ (-y^2) \vec k}}$

temos
$P = xz$

pela formula temos que derivar o P em relaçao a z ..e em relaçao a y
$\frac{\partial P}{\partial z} = x\\\\ \frac{\partial P }{\partial y} =0$

para Q
$Q = xyz\\\\ \frac{\partial Q}{\partial x}=yz\\\\ \frac{\partial Q}{\partial z}=xy$


para R
$R = -y^2\\\\ \frac{\partial R}{\partial x} = 0\\\\ \frac{\partial R}{\partial y}= -2y$

substituindo 

$rot (F)= (-2y - xy ) \vec i + ( x - 0 )\vec j + ( yz- 0 ) \vec k\\\\\ \boxed{\boxed{\ rot (F) = -y(2+x)\vec i + x\vec j+ yz \vec k}}$

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:

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

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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