 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/10b/ii/solution

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

D

in the plane

z = 0

. Its outward normal is

- e_{z}

, but

F \cdot (- e_{z}) = - z^{4} = 0

there. Also

\nabla \cdot F = 2 x - 2 y + 4 z^{3} .

The

x

and

y

terms integrate to zero by symmetry. The divergence theorem therefore gives

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

 New to topics? Read the docs here!