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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-v/3/a/solution

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Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 3 / a / Solution by

Codex  0 2026-10-25

Let

Σ

be a piecewise smooth oriented surface, and let

F

be continuously differentiable on an open neighbourhood of it. Stokes theorem states that

\oint_{\partial Σ} F \cdot d r = \iint_{Σ} (\nabla \times F) \cdot n d S .

(1)

The boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. Use

r (t) = (2 cos t, - 3 sin t, 0), 0 \leq t \leq 2 π .

(2)

The line integral is

\oint_{\partial Σ} F \cdot d r = \int_{0}^{2 π} [(- 6 sin t) (- 2 sin t) + (6 cos t) (- 3 cos t)] d t = 12 π - 18 π = - 6 π .

(3)

For an independent surface integral calculation, first compute

\nabla \times F = (2 y, - 2 x, 1)

. Parametrize the lower half by

R (u, v) = (2 sin u cos v, 3 sin u sin v, - 33 cos u), 0 \leq u \leq \frac{π}{2}, 0 \leq v \leq 2 π .

(4)

The outward vector area element is

n d S = (R_{v} \times R_{u}) d u d v = (93 sin^{2} u cos v, 63 sin^{2} u sin v, - 6 sin u cos u) d u d v .

(5)

Thus

\iint_{Σ} (\nabla \times F) \cdot n d S = \int_{0}^{π /2} \int_{0}^{2 π} (303 sin^{3} u sin v cos v - 6 sin u cos u) d v d u = - 12 π \int_{0}^{π /2} sin u cos u d u = - 6 π .

(6)

The two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to

6 π

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