Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 10B / ii

Table of contents
- Solution ii

Solution

Close the surface with the unit disk

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} .

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