Dado el campo vectorial \( \vec{F}=(2 y-z) i+(3 x+1) j+(x y) \boldsymbol{k} \) ¿cuál es el rotacional \( \nabla \times F ? \)

Sea el campo vectorial \[ \vec{F} = (2y - z)\,\mathbf{i} + (3x + 1)\,\mathbf{j} + (xy)\,\mathbf{k}. \] Recordemos que el rotacional de \(\vec{F}\) se define como \[ \nabla \times \vec{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k}, \] donde \(F_1 = 2y-z\), \(F_2 = 3x+1\) y \(F_3 = xy\). Procedemos a calcular las derivadas parciales: 1. Calculemos las derivadas de \(F_1\): \[ \frac{\partial F_1}{\partial z} = \frac{\partial (2y - z)}{\partial z} = -1, \] \[ \frac{\partial F_1}{\partial y} = \frac{\partial (2y - z)}{\partial y} = 2. \] 2. Calculemos las derivadas de \(F_2\): \[ \frac{\partial F_2}{\partial z} = \frac{\partial (3x + 1)}{\partial z} = 0, \] \[ \frac{\partial F_2}{\partial x} = \frac{\partial (3x + 1)}{\partial x} = 3. \] 3. Calculemos las derivadas de \(F_3\): \[ \frac{\partial F_3}{\partial y} = \frac{\partial (xy)}{\partial y} = x, \] \[ \frac{\partial F_3}{\partial x} = \frac{\partial (xy)}{\partial x} = y. \] Ahora, sustituyendo en la fórmula del rotacional: - Componente en \(\mathbf{i}\): \[ \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = x - 0 = x. \] - Componente en \(\mathbf{j}\): \[ \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = (-1) - y = -1 - y. \] - Componente en \(\mathbf{k}\): \[ \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 3 - 2 = 1. \] Por lo tanto, el rotacional es \[ \nabla \times \vec{F} = x\,\mathbf{i} + (-1-y)\,\mathbf{j} + 1\,\mathbf{k}. \]