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

Sea el campo vectorial \[ \vec{F}=(2y-z)\,\boldsymbol{i}+(3x+1)\,\boldsymbol{j}+(xy)\,\boldsymbol{k}. \] Para hallar el rotacional \(\nabla \times \vec{F}\), recordamos que si \(\vec{F}=(F_x,F_y,F_z)\), entonces \[ \nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right)\boldsymbol{i} - \left( \frac{\partial F_z}{\partial x}-\frac{\partial F_x}{\partial z}\right)\boldsymbol{j} + \left( \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\boldsymbol{k}. \] Identificamos las componentes: \[ F_x=2y-z,\quad F_y=3x+1,\quad F_z=xy. \] 1. Componente \( \boldsymbol{i} \): \[ \frac{\partial F_z}{\partial y} = \frac{\partial (xy)}{\partial y} = x,\quad \frac{\partial F_y}{\partial z} = \frac{\partial (3x+1)}{\partial z} = 0. \] Por lo tanto, \[ \text{Componente } \boldsymbol{i} = x - 0 = x. \] 2. Componente \( \boldsymbol{j} \): Primero, \[ \frac{\partial F_z}{\partial x} = \frac{\partial (xy)}{\partial x} = y,\quad \frac{\partial F_x}{\partial z} = \frac{\partial (2y-z)}{\partial z} = -1. \] Entonces, la componente \( \boldsymbol{j} \) es \[ -\left( y - (-1) \right) = -(y + 1). \] 3. Componente \( \boldsymbol{k} \): \[ \frac{\partial F_y}{\partial x} = \frac{\partial (3x+1)}{\partial x} = 3,\quad \frac{\partial F_x}{\partial y} = \frac{\partial (2y-z)}{\partial y} = 2. \] Por lo tanto, \[ \text{Componente } \boldsymbol{k} = 3 - 2 = 1. \] Uniendo las componentes, obtenemos \[ \nabla \times \vec{F} = x\,\boldsymbol{i} - (y+1)\,\boldsymbol{j} + 1\,\boldsymbol{k}. \] Comparando con las opciones facilitadas, la forma correcta es: \[ O(x)\,\boldsymbol{i} - (y+1)\,\boldsymbol{j} + (1)\,\boldsymbol{k}. \]