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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iv/3/b/iii/solution

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Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 3 / b / iii / Solution by

Codex  0 2026-10-25

The rotation axis is

L = {(x, 1, 0) : x \in R}

. The distance of the generating function graph from this axis is

x

, so a parametrization is

R (x, ϕ) = (x, 1 + x cos ϕ, x sin ϕ), 0 < x < 1.

The surface of revolution therefore has equation

(y - 1)^{2} + z^{2} = x

: it is a circular elliptic paraboloid opening along the positive

x

direction, with its vertex and its rim at

x = 1

excluded. Translating the axis does not change the surface area formula, so

A = 2 π \int_{0}^{1} x 1 + \frac{1}{4 x} d x = π \int_{0}^{1} 4 x + 1 d x = \frac{π}{6} (55 - 1) .

The cancellation makes the improper integral finite at the vertex. Neither the missing vertex nor the missing boundary circle changes the surface area, and no cap is included.

Figure 1.
The elliptic paraboloid obtained by rotating $y = 1 + x$ around $L$
. Details.

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