Dado o campo vetorial! F(x,y,z)=2xyoverrightarrow (i)+yz^2overrightarrow (j)+lnzoverrightarrow (k)' Calcule seu rotacional e assinale a alternativa correta Dados: Rotacional através do determinante rot F = r=tvert } i&j&k (partial )/(partial x)&(partial )/(partial y)&(partial )/(partial z) M&N&P vert ou por rot F=(partial P)/(partial y)overrightarrow (i)+(partial M)/(partial z)overrightarrow (j)+(partial N)/(partial x)overrightarrow (k)-(partial N)/(partial z)overrightarrow (i)-(partial P)/(partial x)overrightarrow (j)-(partial M)/(partial y)overrightarrow (k) square

Para calcular o rotacional do campo vetorial \( F(x,y,z) = 2xy\overrightarrow{i} + yz^2\overrightarrow{j} + \ln(z)\overrightarrow{k} \), podemos usar a fórmula do rotacional através do determinante:<br /><br />\[ \text{rot} \, F = \nabla \times F = \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy & yz^2 & \ln(z) \end{vmatrix} \]<br /><br />Calculando o determinante, obtemos:<br /><br />\[ \text{rot} \, F = \left( \frac{\partial(\ln(z))}{\partial y} - \frac{\partial(yz^2)}{\partial z} \right) \overrightarrow{i} - \left( \frac{\partial(2xy)}{\partial z} - \frac{\partial(\ln(z))}{\partial x} \right) \overrightarrow{j} + \left( \frac{\partial(2xy)}{\partial y} - \frac{\partial(2xy)}{\partial x} \right) \overrightarrow{k} \]<br /><br />Simplificando as derivadas, temos:<br /><br />\[ \text{rot} \, F = \left( 0 - 2yz \right) \overrightarrow{i} - \left( 0 - 0 \right) \overrightarrow{j} + \left( 2x - 2x \right) \overrightarrow{k} \]<br /><br />Portanto, o rotacional do campo vetorial \( F(x,y,z) \) é:<br /><br />\[ \text{rot} \, F = -2yz \overrightarrow{i} + 0 \overrightarrow{j} + 0 \overrightarrow{k} \]<br /><br />A alternativa correta é:<br /><br />\[ \text{rot} \, F = -2yz \overrightarrow{i} \]