TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Scalar surface integrals on a surface of revolution

Statement

Let $a < b$ , and let $r : [a, b] \to [0, \infty)$ be $C^{1}$ on a neighbourhood of $[a, b]$ , positive on $(a, b)$ , and allowed to vanish only at the endpoints. Put $φ (s, t) = (s, r (s) \cos t, r (s) \sin t), (s, t) \in [a, b] \times [0, 2 π] .$ For every continuous real-valued function $q$ on $φ ([a, b] \times [0, 2 π])$ , the scalar surface integral is $\int_{S} q d S = \int_{0}^{2 π} \int_{a}^{b} q (φ (s, t)) r (s) \sqrt{1 + r^{'} (s)^{2}} d s d t,$ where $S$ is the patch with its displayed parametrization.

Facts & Assumptions

Given: The nondegenerate interval, radius function, parametrization, and continuous scalar field $q$ .

[L1]

The sine and cosine derivative formulas and derivative algebra compute the parameter tangents, and $\sin^{2} t + \cos^{2} t = 1$ (The derivatives of sine and cosine are cosine and minus sine, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Parity and the Pythagorean identity for sine and cosine).

[L2]

A parametrization with nonzero cross product in the parameter interior and no interior parameter point sharing its image with another point of the region is a regular patch; its scalar integral uses the cross-product norm as density, and Jordan-Fubini identifies the rectangle integral with the stated iterated integral (Regular parametrized surface patches on compact Jordan parameter regions, The surface area density is the norm of the cross product of the parameter tangents, Surface area and scalar surface integrals on a regular patch, Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable).

Proof

technique · direct

1.1

By [L1], $φ_{s} = (1, r^{'} \cos t, r^{'} \sin t)$ and $φ_{t} = (0, - r \sin t, r \cos t)$ , and direct expansion gives $∥ φ_{s} \times φ_{t} ∥_{2} = r (s) \sqrt{1 + r^{'} (s)^{2}}$ .

givenL1algebra

2.1

In the rectangle interior, $r (s) > 0$ , so the cross product is nonzero. The first coordinate determines $s$ , and the angle determines the point on the positive-radius circle for $0 < t < 2 π$ ; only the angular seam and possible endpoint-axis collapses lie on the boundary. Thus [L2] makes $φ$ a regular patch.

givenstep 1.1L2

3.1

Substitute the density from step 1.1 into the scalar surface-integral definition and use the Jordan-Fubini clause in [L2] to obtain the stated iterated form.

step 1.1step 2.1L2

4.1

Endpoint zeros and the $t = 0, 2 π$ seam occur only on the content-zero parameter boundary, so they do not add terms or change the integral.

step 2.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

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Sources

M. E. Taylor, Introduction to Analysis in Several Variables, Exercise 17 (standard reference, not scraped)
University of Toronto MAT237 notes, Section 5.3 (standard reference, not scraped)