ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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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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Volume density in polar coordinates

Example

In a polar chart of the Euclidean plane, the density is $r ∣ d r d θ ∣$ and the positive-oriented volume form is $r d r \land d θ$ .

Facts & Assumptions

Given: $x = r \cos θ$ , $y = r \sin θ$ , with $r > 0$ and $θ$ in an open interval of length less than $2 π$ .

[F1]

Riemannian volume density: The Riemannian volume density is $μ_{g} = \sqrt{\det G_{x}} ∣ d x^{1} \dots d x^{n} ∣$ in coordinates. The matrix is that of prop-coordinate-criterion-for-a-riemannian-metric, so its determinant is positive and smooth. The density frames and their absolute-Jacobian law are def-density-bundle-and-smooth-density. In dimension zero take the empty determinant to be one, giving weight one at every point, independently of orientation. The compatibility of these local formulas is proved in lem-the-riemannian-volume-density-is-coordinate-independent.

[F2]

The riemannian volume density is coordinate independent: The local Riemannian volume densities glue to a positive smooth density independent of coordinates.

Verification

technique · direct

1.1F1given

Differentiation gives $d x = \cos θ d r - r \sin θ d θ$ and $d y = \sin θ d r + r \cos θ d θ$ . Expanding $d x^{2} + d y^{2}$ , the mixed terms cancel and the diagonal terms sum to $d r^{2} + r^{2} d θ^{2}$ . Thus $\det G = r^{2}$ , whose positive square root is $r$ on this domain, and the density is $r ∣ d r d θ ∣$ .

2.1F2step 1.1∎

The wedge expansion gives $d x \land d y = r (\cos^{2} θ + \sin^{2} θ) d r \land d θ = r d r \land d θ$ . The Jacobian is positive, so the chart has the standard orientation and this is its positive volume form. The absolute Jacobian density law agrees with step 1.1. At $r = 2$ , both coordinate coefficients are $2$ . The excluded value $r = 0$ is a failure of polar coordinates, not a zero of the Euclidean density.

Source locator

Lee, Example 13.12, p.332, polar metric; Proposition 15.31, p.390, coordinate volume formula; pp.430–431, Riemannian density.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)