 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/9b/b/solution

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

I = \oint_{\partial S} A \cdot d r .

For the upward orientation, the outer circle

r = 2, z = 1

is counterclockwise from above and the inner circle

r = 1, z = 4

is clockwise. On a counterclockwise circle of radius

R

z

\oint A \cdot d r = - R^{2} \int_{0}^{2 π} (3 sin^{2} θ + z cos^{2} θ) d θ = - π R^{2} (3 + z) .

The outer circle contributes

- 16 π

. The inner circle would contribute

- 7 π

counterclockwise and hence contributes

+ 7 π

with its induced clockwise orientation. Thus

I = - 16 π + 7 π = - 9 π,

agreeing with the direct calculation.

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