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

Para calcular el rotacional de un campo vectorial, utilizamos la fórmula \[ \nabla \times \vec{F} = \left( \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z} \right)\mathbf{i} + \left( \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x} \right)\mathbf{j} + \left( \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y} \right)\mathbf{k}. \] Dado que \[ \vec{F} = (2y - z)\mathbf{i} + (3x + 1)\mathbf{j} + (xy)\mathbf{k}, \] identificamos las componentes: \[ F_1 = 2y - z, \quad F_2 = 3x + 1, \quad F_3 = xy. \] A continuación, calculamos las derivadas parciales necesarias: 1. Para la componente en \(\mathbf{i}\): \[ \frac{\partial F_3}{\partial y} = \frac{\partial (xy)}{\partial y} = x, \] \[ \frac{\partial F_2}{\partial z} = \frac{\partial (3x+1)}{\partial z} = 0. \] Por lo tanto, \[ \left( \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z} \right) = x - 0 = x. \] 2. Para la componente en \(\mathbf{j}\): \[ \frac{\partial F_1}{\partial z} = \frac{\partial (2y-z)}{\partial z} = -1, \] \[ \frac{\partial F_3}{\partial x} = \frac{\partial (xy)}{\partial x} = y. \] Por lo tanto, \[ \left( \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x} \right) = -1 - y. \] 3. Para la componente en \(\mathbf{k}\): \[ \frac{\partial F_2}{\partial x} = \frac{\partial (3x+1)}{\partial x} = 3, \] \[ \frac{\partial F_1}{\partial y} = \frac{\partial (2y-z)}{\partial y} = 2. \] Por lo tanto, \[ \left( \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y} \right) = 3 - 2 = 1. \] Reuniendo todas las componentes, obtenemos: \[ \nabla \times \vec{F} = x\, \mathbf{i} + (-1-y)\, \mathbf{j} + 1\, \mathbf{k}. \]