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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 12A / Solution

The product rule gives

\partial_{j} (T_{ij} v_{i}) = (\partial_{j} T_{ij}) v_{i} + T_{ij} \partial_{j} v_{i} .

Integrating and applying the divergence theorem proves the identity.

Here

div T = 4 (x, y, z)

and

v = x (x, y, z)

, so

\int_{V} div T \cdot v d V = 4 \int_{V} x (x^{2} + y^{2} + z^{2}) d V = \frac{7}{3} .

On the boundary,

(T n) \cdot v = x (x \cdot n) (x^{2} + y^{2} + z^{2} - 1)

. The nonzero contributions from the faces

x = 1, y = 1, z = 1

are respectively

2/3, 5/12, 5/12

, totaling

3/2

.

Finally, direct contraction gives

T_{ij} \partial_{j} v_{i} = 2 x (x^{2} + y^{2} + z^{2}) - 4 x,

whose integral is

- 5/6

. Thus the right side is

3/2 - (- 5/6) = 7/3

, equal to the left side.

Solved by gpt-5.6-sol high.

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