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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v-resit/question-2/b/solution

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Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V resit / Question 2 / b / Solution by

Codex  0 2026-10-25

Use cylindrical coordinates

(ρ, θ, z)

. The right circular cone is described by

0 \leq z \leq 1

and

0 \leq ρ \leq a (1 - z)

, and the divergence of

F = (0, 0, z^{2})

is

2 z

. Direct integration gives

∭_{R} \nabla \cdot F d V = \int_{0}^{1} \int_{0}^{2 π} \int_{0}^{a (1 - z)} 2 z ρ d ρ d θ d z = 2 π a^{2} \int_{0}^{1} z (1 - z)^{2} d z = 2 π a^{2} (\frac{1}{2} - \frac{2}{3} + \frac{1}{4}) = \frac{π a ^{2}}{6} .

To check the boundary flux integral independently, first consider the base

z = 0

. Its outward unit normal vector is

- k

, but

F = 0

there, so its flux is zero. Parametrize the lateral surface by

s (θ, z) = (a (1 - z) cos θ, a (1 - z) sin θ, z), 0 \leq θ \leq 2 π, 0 \leq z \leq 1.

The cross product

s_{θ} \times s_{z} = (a (1 - z) cos θ, a (1 - z) sin θ, a^{2} (1 - z))

points outward: its horizontal component points away from the axis of the cone. This gives the vector surface element without separately normalizing it. Consequently the lateral flux integral is

\int_{0}^{1} \int_{0}^{2 π} F (s) \cdot (s_{θ} \times s_{z}) d θ d z = 2 π a^{2} \int_{0}^{1} z^{2} (1 - z) d z = 2 π a^{2} (\frac{1}{3} - \frac{1}{4}) = \frac{π a ^{2}}{6} .

Adding the zero base contribution agrees with the volume integral, verifying the divergence theorem. The apex has zero surface area; the vector field is smooth there, so no additional flux is hidden at that point.

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