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 \[ \vec{F}(x,y,z) = (2y-z)\,\mathbf{i} + (3x+1)\,\mathbf{j} + (xy)\,\mathbf{k}. \] Definimos: \[ P(x,y,z)=2y-z,\quad Q(x,y,z)=3x+1,\quad R(x,y,z)=xy. \] El rotacional se define como: \[ \nabla \times \vec{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\mathbf{i} + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)\mathbf{j} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathbf{k}. \] 1. Componente en \(\mathbf{i}\): \[ \frac{\partial R}{\partial y} = \frac{\partial (xy)}{\partial y} = x,\quad \frac{\partial Q}{\partial z} = \frac{\partial (3x+1)}{\partial z} = 0, \] de donde: \[ (\nabla \times \vec{F})_i = x - 0 = x. \] 2. Componente en \(\mathbf{j}\): \[ \frac{\partial P}{\partial z} = \frac{\partial (2y-z)}{\partial z} = -1,\quad \frac{\partial R}{\partial x} = \frac{\partial (xy)}{\partial x} = y, \] de donde: \[ (\nabla \times \vec{F})_j = (-1) - y = -1-y. \] 3. Componente en \(\mathbf{k}\): \[ \frac{\partial Q}{\partial x} = \frac{\partial (3x+1)}{\partial x} = 3,\quad \frac{\partial P}{\partial y} = \frac{\partial (2y-z)}{\partial y} = 2, \] de donde: \[ (\nabla \times \vec{F})_k = 3 - 2 = 1. \] Por lo tanto, el rotacional es: \[ \nabla \times \vec{F} = x\,\mathbf{i} + (-1-y)\,\mathbf{j} + 1\,\mathbf{k}. \]