Dado el campo vectorial \( \mathbf{F}(x, y, z)=\left(y^{2}+3 z\right) \mathbf{i}+(2 x+z) \mathbf{j}+\left(x^{2}-y\right) \mathbf{k} \) ¿Cuál es el rotacional \( \nabla \times F \) ?

Para calcular el rotacional de un campo vectorial \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), utilizamos la fórmula: \[ \nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \] Dado que \( P = y^{2} + 3z \), \( Q = 2x + z \) y \( R = x^{2} - y \), procedemos a calcular cada uno de los componentes: 1. **Componente \( \mathbf{i} \)**: \[ \frac{\partial R}{\partial y} = -1,\quad \frac{\partial Q}{\partial z} = 1 \quad \Rightarrow \quad \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = -1 - 1 = -2 \] 2. **Componente \( \mathbf{j} \)**: \[ \frac{\partial R}{\partial x} = 2x,\quad \frac{\partial P}{\partial z} = 3 \quad \Rightarrow \quad -\left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) = -(2x - 3) = 3 - 2x \] 3. **Componente \( \mathbf{k} \)**: \[ \frac{\partial Q}{\partial x} = 2,\quad \frac{\partial P}{\partial y} = 2y \quad \Rightarrow \quad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 2y \] Finalmente, combinando todo, obtenemos el rotacional: \[ \nabla \times \mathbf{F} = -2 \mathbf{i} - (3 - 2x) \mathbf{j} + (2 - 2y) \mathbf{k} \] Por lo tanto, \[ \nabla \times \mathbf{F} = -2 \mathbf{i} - (3 - 2x) \mathbf{j} + (2 - 2y) \mathbf{k} \]