Se F=y i^+x j^, o rotor ×F e: a) 2 k^ b) 2 k^ c) k^ d) 0

Para calcular o rotor (ou curl) de um campo vetorial \( \mathbf{F} = y \mathbf{i} + x \mathbf{j} \), utilizamos a seguinte fórmula: \[ \nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k} \] Onde: - \( F_x = y \) - \( F_y = x \) - \( F_z = 0 \) Agora, vamos calcular cada componente do rotor: 1. Componente \( \mathbf{i} \): \[ \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0 - 0 = 0 \] 2. Componente \( \mathbf{j} \): \[ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0 - 0 = 0 \] 3. Componente \( \mathbf{k} \): \[ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 1 - 1 = 0 \] Assim, temos que: \[ \nabla \times \mathbf{F} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} = 0 \] Portanto, a resposta correta é: d) 0.