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

Para calcular el rotacional \(\nabla \times \mathbf{F}\) de \[ \mathbf{F}=(2y-z)\,\mathbf{i}+(3x+1)\,\mathbf{j}+(xy)\,\mathbf{k}, \] utilizamos la definición del rotacional en función de determinantes: \[ \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2y-z & 3x+1 & xy \end{vmatrix}. \] Identificamos: - \(P=2y-z\) - \(Q=3x+1\) - \(R=xy\) **Paso 1. Componente \(\mathbf{i}\):** \[ (\nabla \times \mathbf{F})_x = \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}. \] Calculamos: - \(\frac{\partial R}{\partial y} = \frac{\partial (xy)}{\partial y} = x\). - \(\frac{\partial Q}{\partial z} = \frac{\partial (3x+1)}{\partial z} = 0\). Entonces: \[ (\nabla \times \mathbf{F})_x = x - 0 = x. \] **Paso 2. Componente \(\mathbf{j}\):** \[ (\nabla \times \mathbf{F})_y = -\left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right). \] Calculamos: - \(\frac{\partial R}{\partial x} = \frac{\partial (xy)}{\partial x} = y\). - \(\frac{\partial P}{\partial z} = \frac{\partial (2y-z)}{\partial z} = -1\). Por lo tanto: \[ \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} = y - (-1) = y+1, \] y \[ (\nabla \times \mathbf{F})_y = -\,(y+1). \] **Paso 3. Componente \(\mathbf{k}\):** \[ (\nabla \times \mathbf{F})_z = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}. \] Calculamos: - \(\frac{\partial Q}{\partial x} = \frac{\partial (3x+1)}{\partial x} = 3\). - \(\frac{\partial P}{\partial y} = \frac{\partial (2y-z)}{\partial y} = 2\). Entonces: \[ (\nabla \times \mathbf{F})_z = 3-2 = 1. \] **Resultado final:** El rotacional del campo es \[ \nabla \times \mathbf{F} = x\,\mathbf{i} - (y+1)\,\mathbf{j} + 1\,\mathbf{k}. \] Comparando con las opciones propuestas, la respuesta correcta es: \[ x\,\mathbf{i} - (y+1)\,\mathbf{j} + (1)\,\mathbf{k}. \]