 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-3/3a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

\partial_{i} F_{i}

and curl

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

, where

ϵ_{ijk}

\partial_{j} F_{i} = 2 x_{j} x_{i} + r^{2} δ_{ij}, \nabla \cdot F = 5 r^{2}, \nabla \times F = 0 .

The derivative matrix is symmetric, so its contraction with the Levi-Civita symbol vanishes. For the second vector field,

\partial_{j} G_{i} = a_{j} x_{i} + (a \cdot x) δ_{ij}, \nabla \cdot G = 4 a \cdot x, \nabla \times G = a \times x .

For the third vector field,

H_{i} = ϵ_{ik l} a_{k} x_{l} / r

. Its divergence is zero because the Levi-Civita symbol contracts with symmetric products. For its curl, either contract two Levi-Civita symbols or use the vector triple product:

\nabla \times (\frac{a \times x}{r}) = - \frac{x}{r ^{3}} \times (a \times x) + \frac{2 a}{r} = \frac{a}{r} + \frac{( a \cdot x ) x}{r ^{3}} .

Thus

\nabla \cdot H = 0, \nabla \times H = \frac{a + ( a \cdot x ) x}{r} .

All formulas hold on

r > 0

, the domain where the radial unit vector is defined.

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