Dado el campo vectorial \( \mathbf{F}(x, y, z)=\left(y^{2}+3 z\right) \mathbf{i}+(2 x+z) \mathbf{j}+\left(x^{2}-y\right) \mathbf{k} \) ¿cuál es el rotacional \( \nabla \times \mathbf{F} \) ?

Sean \[ F_1(x,y,z)=y^2+3z,\quad F_2(x,y,z)=2x+z,\quad F_3(x,y,z)=x^2-y. \] El rotacional se calcula como \[ \nabla \times \mathbf{F}=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\mathbf{i}+\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)\mathbf{j}+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\mathbf{k}. \] 1. Calculemos la componente en \(\mathbf{i}\): \[ \frac{\partial F_3}{\partial y} = \frac{\partial (x^2-y)}{\partial y} = -1, \] \[ \frac{\partial F_2}{\partial z} = \frac{\partial (2x+z)}{\partial z} = 1. \] Por lo tanto, \[ \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z} = -1-1 = -2. \] 2. Calculemos la componente en \(\mathbf{j}\): \[ \frac{\partial F_1}{\partial z} = \frac{\partial (y^2+3z)}{\partial z} = 3, \] \[ \frac{\partial F_3}{\partial x} = \frac{\partial (x^2-y)}{\partial x} = 2x. \] Así, \[ \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x} = 3-2x. \] 3. Calculemos la componente en \(\mathbf{k}\): \[ \frac{\partial F_2}{\partial x} = \frac{\partial (2x+z)}{\partial x} = 2, \] \[ \frac{\partial F_1}{\partial y} = \frac{\partial (y^2+3z)}{\partial y} = 2y. \] Entonces, \[ \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y} = 2-2y. \] Por lo tanto, el rotacional es \[ \nabla \times \mathbf{F} = -2\,\mathbf{i}+(3-2x)\,\mathbf{j}+(2-2y)\,\mathbf{k}. \]