 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-3/9a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The top and bottom fluxes cancel because

F_{z} = x^{2} + y^{2}

is independent of

z

. Parametrise the side by

(cos θ / a, sin θ / b, z)

; its outward vector area is

(\frac{c o s θ}{b}, \frac{s i n θ}{a}, 0) d θ d z .

The term involving

y^{2} z

integrates to zero, and the remaining integral is

\int_{0}^{3} \int_{0}^{2 π} \frac{3 z c o s ^{2} θ}{ab} d θ d z = \frac{27 π}{2 ab},

confirming part (a).

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