 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-3/11c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For a bounded volume

Ω

with piecewise smooth boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem states

\oint_{\partial Ω} F \cdot n d S = \int_{Ω} \nabla \cdot F d V,

where

n

is the outward unit normal.

The boundary here consists of a paraboloid cap over the unit disk, a cylindrical wall of height one, and the bottom disk. Their joined surface encloses

Ω = {(x, y, z) : x^{2} + y^{2} \leq 1, 0 \leq z \leq 3 - 2 (x^{2} + y^{2})} .

The cap meets the wall at

r = 1, z = 1

, and the wall meets the base at

r = 1, z = 0

. The sketches show each boundary piece and their closed union.

Differentiate the field components to find the divergence:

\nabla \cdot F = (y + 6 x^{5}) + (- y + 8 y^{7}) + 1 = 1 + 6 x^{5} + 8 y^{7} .

The volume is invariant under reflecting

x

y

, so odd-integrand cancellation by reflection removes both odd-power contributions. The outward flux equals the volume:

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

The corners between the three pieces are permitted by the piecewise smooth boundary hypothesis.

 New to topics? Read the docs here!