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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016/4/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2016 / 4 / Solution by

Codex  0 2026-10-25

The divergence theorem includes every part of the closed boundary, including the two vertical cut faces. Its volume integrand simplifies because the terms linear in

x

cancel:

\nabla \cdot F = 3 x^{2} + (3 y^{2} + 2 x) - 2 x = 3 (x^{2} + y^{2}) .

(1)

In cylindrical coordinates, the remaining integration region is

0 \leq r \leq 1

0 \leq z \leq 1 - r^{2}

π /2 \leq θ \leq 2 π

. With the Jacobian determinant

r

, the outward flux is

\int_{\partial V} F \cdot d S = \int_{π /2}^{2 π} \int_{0}^{1} \int_{0}^{1 - r^{2}} 3 r^{3} d z d r d θ = \frac{9 π}{2} (\frac{1}{4} - \frac{1}{6}) = \frac{3 π}{8} .

(2)

The angular range is three quarters of a full revolution. The volume calculation already accounts for the base and cut faces, so no additional surface terms should be appended.

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