Resposta:  [tex]\mathrm{rot}(\mathbf{F})=(0,\,8z,\,2x).[/tex]

Explicação passo a passo:

Calcular o rotacional do campo vetorial no [tex]\mathbb{R}^3[/tex]

    [tex]\mathbf{F}(x,\,y,\,z)=P(x,\,y,\,z)\mathbf{i}+Q(x,\,y,\,z)\mathbf{j}+R(x,\,y,\,z)\mathbf{k}[/tex]

com

    [tex]\begin{cases}~P(x,\,y,\,z)=2x^3+4z^2\\ ~Q(x,\,y,\,z)=x^2-y\\ ~R(x,\,y,\,z)=z \end{cases}[/tex]

O rotacional de [tex]\mathbf{F}[/tex] é dado por

    [tex]\mathrm{rot}(\mathbf{F})=\nabla\times \mathbf{F}\\\\\\=\left(\dfrac{\partial}{\partial x},\,\dfrac{\partial}{\partial y},\,\dfrac{\partial}{\partial z}\right)\times \Big(P(x,\,y,\,z),\,Q(x,\,y,\,z),\,R(x,\,y,\,z)\Big)[/tex]

Podemos escrever o cálculo acima na forma de um determinante simbólico:

    [tex]=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\\ P&Q&R \end{vmatrix}[/tex]

    [tex]=\begin{vmatrix}\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\\\ Q&R \end{vmatrix}\mathbf{i}-\begin{vmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial z}\\\\ P&R \end{vmatrix}\mathbf{j}+\begin{vmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\\\\ P&Q \end{vmatrix}\mathbf{k}[/tex]

    [tex]=\left(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z}\right)\!\mathbf{i}-\left(\dfrac{\partial R}{\partial x}-\dfrac{\partial P}{\partial z}\right)\!\mathbf{j}+\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\!\mathbf{k}[/tex]

Substituindo as coordenadas [tex]P,\,Q[/tex] e [tex]R[/tex] do campo vetorial e calculando as derivadas parciais indicadas, temos

    [tex]=\left(\dfrac{\partial}{\partial y}(z)-\dfrac{\partial}{\partial z}(x^2-y)\right)\!\mathbf{i}-\left(\dfrac{\partial}{\partial x}(z)-\dfrac{\partial}{\partial z}(2x^3+4z^2)\right)\!\mathbf{j}+\left(\dfrac{\partial}{\partial x}(x^2-y)-\dfrac{\partial}{\partial y}(2x^3+4z^2)\right)\!\mathbf{k}[/tex]

    [tex]=(0-0)\mathbf{i}-(0-8z)\mathbf{j}+(2x-0)\mathbf{k}\\\\\\=(0)\mathbf{i}+(8z)\mathbf{j}+(2x)\mathbf{k}\\\\\\=(0,\,8z,\,2x)\quad\longleftarrow\quad\mathsf{resposta.}[/tex]

Dúvidas? Comente.

Bons estudos!