Divergencia Rotacional 37B961

1. **Planteamiento del problema:** Calcular la divergencia y el rotacional del campo vectorial $$\mathbf{F}(x,y,z) = e^{-x} \sin y \mathbf{i} + e^{-y} \sin z \mathbf{j} + e^{-z} \sin x \mathbf{k}$$. 2. **Fórmulas importantes:** - La divergencia de un campo vectorial $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$$ es $$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$$. - El rotacional de $$\mathbf{F}$$ es $$\nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}$$. 3. **Cálculo de la divergencia:** - $$P = e^{-x} \sin y$$, $$Q = e^{-y} \sin z$$, $$R = e^{-z} \sin x$$. - Derivadas parciales: $$\frac{\partial P}{\partial x} = \frac{\partial}{\partial x} (e^{-x} \sin y) = -e^{-x} \sin y$$ $$\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y} (e^{-y} \sin z) = -e^{-y} \sin z$$ $$\frac{\partial R}{\partial z} = \frac{\partial}{\partial z} (e^{-z} \sin x) = -e^{-z} \sin x$$ - Sumamos para obtener la divergencia: $$\nabla \cdot \mathbf{F} = -e^{-x} \sin y - e^{-y} \sin z - e^{-z} \sin x$$ 4. **Cálculo del rotacional:** - Derivadas parciales necesarias: $$\frac{\partial R}{\partial y} = \frac{\partial}{\partial y} (e^{-z} \sin x) = 0$$ (no depende de y) $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z} (e^{-y} \sin z) = e^{-y} \cos z$$ $$\frac{\partial R}{\partial x} = \frac{\partial}{\partial x} (e^{-z} \sin x) = e^{-z} \cos x$$ $$\frac{\partial P}{\partial z} = \frac{\partial}{\partial z} (e^{-x} \sin y) = 0$$ (no depende de z) $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (e^{-y} \sin z) = 0$$ (no depende de x) $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (e^{-x} \sin y) = e^{-x} \cos y$$ - Componentes del rotacional: $$\left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) = 0 - e^{-y} \cos z = -e^{-y} \cos z$$ $$- \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) = - (e^{-z} \cos x - 0) = -e^{-z} \cos x$$ $$\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) = 0 - e^{-x} \cos y = -e^{-x} \cos y$$ - Por lo tanto: $$\nabla \times \mathbf{F} = -e^{-y} \cos z \mathbf{i} - e^{-z} \cos x \mathbf{j} - e^{-x} \cos y \mathbf{k}$$ 5. **Respuesta final:** - Divergencia: $$\boxed{\nabla \cdot \mathbf{F} = -e^{-x} \sin y - e^{-y} \sin z - e^{-z} \sin x}$$ - Rotacional: $$\boxed{\nabla \times \mathbf{F} = -e^{-y} \cos z \mathbf{i} - e^{-z} \cos x \mathbf{j} - e^{-x} \cos y \mathbf{k}}$$