 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/3b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

F = (F_{1}, F_{2}, F_{3})

\nabla \cdot F + = \partial_{i} F_{i} + = \frac{\partial F _{1}}{\partial x} + + \frac{\partial F _{2}}{\partial y} + + \frac{\partial F _{3}}{\partial z}

(\nabla \times F)_{i} + = ϵ_{ijk} \partial_{j} F_{k} .

Equivalently,

\nabla \times F = + (\partial_{y} F_{3} - \partial_{z} F_{2}) e_{x} + + (\partial_{z} F_{1} - \partial_{x} F_{3}) e_{y} + + (\partial_{x} F_{2} - \partial_{y} F_{1}) e_{z} .

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