 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-3/3b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

For an arbitrary constant vector

c

\nabla \cdot (ϕ c) = c \cdot \nabla ϕ .

The divergence theorem therefore implies

c \cdot \int_{V} \nabla ϕ d V = \int_{V} \nabla \cdot (ϕ c) d V = \int_{S} ϕ c \cdot d S = c \cdot \int_{S} ϕ d S .

Since this holds for every

c

\int_{V} \nabla ϕ d V = \int_{S} ϕ d S .

For

ϕ = x + y

\nabla ϕ = (1, 1, 0)

. The ball of radius

a

4 π a^{3} /3

, so

\int_{V} \nabla ϕ d V = \frac{4 π a ^{3}}{3} (1, 1, 0) .

On the sphere, use

x = a sin θ cos φ

y = a sin θ sin φ

and the stated outward oriented surface element. The first component of the surface integral is

a^{3} \int_{0}^{π} sin^{3} θ d θ \int_{0}^{2 π} (cos^{2} φ + sin φ cos φ) d φ = a^{3} \frac{4}{3} π,

and symmetry gives the same second component and a zero third component. The two sides agree.

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