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 Incoming links: Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 3 / a

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 3 / d / i / Solution 2026-10-25

Choose the outward orientation on the upper half of the ellipsoid, giving counterclockwise boundary orientation on

C

as viewed from

z > 0

. Parametrize the boundary circle by

c (t) = (cos t, sin t, 0), 0 \leq t \leq 2 π .

Then

u (c (t)) = (- sin t, cos t, 0) = c^{'} (t)

, so the line integral is

\oint_{C} u \cdot d r = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = 2 π .

For a direct surface integral, use the parametrized surface

s (φ, t) = (sin φ cos t, sin φ sin t, α cos φ), 0 \leq φ \leq \frac{π}{2}, 0 \leq t \leq 2 π .

Its outward vector surface element is supplied by the cross product

s_{φ} \times s_{t} = (α sin^{2} φ cos t, α sin^{2} φ sin t, sin φ cos φ) .

Using the curl from a,

\iint_{S_{α}} (\nabla \times u) \cdot n d S = \int_{0}^{2 π} \int_{0}^{π /2} 2 sin φ cos φ d φ d t = 2 π .

The two integrals agree for every

α > 0

, verifying the Stokes theorem. Only the vertical component of the vector surface element enters, which explains the absence of

α

in the answer.

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