CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-24

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The surface of revolution has area $2 π \int_{a}^{b} r (s) \sqrt{1 + r^{'} (s)^{2}} d s$

Statement

Assume the hypotheses of Scalar surface integrals on a surface of revolution. The surface obtained by rotating $r$ about the axis has area $2 π \int_{a}^{b} r (s) \sqrt{1 + r^{'} (s)^{2}} d s$ .

Facts & Assumptions

Given: A radius function satisfying the surface-of-revolution hypotheses.

[L1]

With scalar field $q = 1$ , the surface integral is $\int_{0}^{2 π} \int_{a}^{b} r (s) \sqrt{1 + r^{'} (s)^{2}} d s d t$ (Scalar surface integrals on a surface of revolution).

[L2]

Jordan-Fubini separates a continuous integrand on a rectangle, and the integral of the constant $C$ over $[0, 2 π]$ is $2 π C$ (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable, If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ ).

Proof

technique · direct

1.1givenL1

Set $q = 1$ in [L1]. The inner integral is independent of $t$ .

2.1step 1.1L2

Apply [L2] to integrate that constant inner value over $0 \leq t \leq 2 π$ , obtaining the factor $2 π$ and the displayed formula.

3.1step 2.1L1∎

Possible endpoint zeros of $r$ lie on the parameter boundary already covered by [L1], so no endpoint correction is present.

Depends on

Used by

Gabriel's horn has unbounded truncated lateral area Example
The lateral area of a right circular cone is π R√R²+H² Example
The surface generated by rotating y=sin x on [0,π] has area 2π(√2+arsinh1) Example
FALSE: finite volume implies finite lateral surface area False statement
FALSE: surface area is the supremum of inscribed polyhedral areas False statement

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. E. Taylor, Introduction to Analysis in Several Variables, Exercise 17 (standard reference, not scraped)
APEX Calculus II, Section 7.4 (standard reference, not scraped)

The surface of revolution has area 2π∫abr(s)1+r′(s)2 ds2π∫ab​r(s)1+r′(s)2​ds