Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 1

Table of contents
- a 1
  - Solution a
- b 1
  - Solution b
- c 1
  - Solution c

a

Solution

Use Einstein summation convention, writing

\partial_{j} = \partial / \partial x_{j}

. The contraction of two Levi-Civita symbols gives

[\nabla \times (a \times b)]_{i} = ϵ_{ijk} \partial_{j} (ϵ_{k l m} a_{l} b_{m}) = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) \partial_{j} (a_{l} b_{m}) = \partial_{j} (a_{i} b_{j} - a_{j} b_{i}) = a_{i} \partial_{j} b_{j} - b_{i} \partial_{j} a_{j} + b_{j} \partial_{j} a_{i} - a_{j} \partial_{j} b_{i} .

The last equality is the product rule. Recognizing the divergences and directional derivatives proves the curl of a cross product identity:

\nabla \times (a \times b) = a (\nabla \cdot b) - b (\nabla \cdot a) + (b \cdot \nabla) a - (a \cdot \nabla) b .

b

Solution

The divergence theorem states that, for a continuously differentiable vector field

H

on a bounded region

V

with piecewise smooth boundary

S

and outward unit normal

\hat{n}

,

\int_{V} \nabla \cdot H d V = \int_{S} H \cdot \hat{n} d S .

To apply it, first derive the divergence of a cross product using the Levi-Civita symbol and the product rule:

\nabla \cdot (F \times G) = ϵ_{ijk} \partial_{i} (F_{j} G_{k}) = ϵ_{ijk} (\partial_{i} F_{j}) G_{k} + ϵ_{ijk} F_{j} \partial_{i} G_{k} = G \cdot (\nabla \times F) - F \cdot (\nabla \times G) .

In the second term the minus sign comes from interchanging the first two indices of the Levi-Civita symbol. The divergence theorem with

H = F \times G

now yields

\int_{V} [G \cdot (\nabla \times F) - F \cdot (\nabla \times G)] d V = \int_{S} (F \times G) \cdot \hat{n} d S .

c

Solution

Close the hyperboloid segment with the two disks at

z = \pm 1

, and let

V

be its interior. The flux through either disk is zero because

A_{z} = 0

. Thus the desired surface integral is the total outward flux through the closed boundary. Since

\nabla \cdot A = 3 x^{2} + 3 y^{2},

the divergence theorem and cylindrical coordinates give

\int_{S} A \cdot \hat{n} d S = \int_{- 1}^{1} \int_{0}^{2 π} \int_{0}^{1 + z^{2}} 3 r^{2} r d r d θ d z = \frac{3 π}{2} \int_{- 1}^{1} (1 + z^{2})^{2} d z = \frac{3 π}{2} (2 + \frac{4}{3} + \frac{2}{5}) = \frac{28 π}{5} .

The extra factor

r

is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

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