ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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Cavalieri computes the area of the unit disc from its sections

Example

Let $D : = {(x, y) \in R^{2} : x^{2} + y^{2} < 1} .$ Then $λ_{2} (D) = \int_{- 1}^{1} 2 \sqrt{1 - x^{2}} d x = π .$

Facts & Assumptions

Given: The unit disc $D \subseteq R^{2}$ .

[L1]

Tonelli computes the area of a measurable set from the lengths of its sections. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[L2]

The unit ball in $R^{2}$ has area $π$ . (The closed form for the volume of the unit $n$ -ball)

Verification

technique · direct

1.1

For $x \in [- 1, 1]$ , the vertical section of $D$ is $D_{x} = (- \sqrt{1 - x^{2}}, \sqrt{1 - x^{2}}),$ so $λ_{1} (D_{x}) = 2 \sqrt{1 - x^{2}}$ ; outside $[- 1, 1]$ the section is empty. Applying [L1] to $1_{D}$ therefore gives $λ_{2} (D) = \int_{- 1}^{1} 2 \sqrt{1 - x^{2}} d x .$

2.1

The same set $D$ is the unit ball in $R^{2}$ , so [L2] gives $\int_{- 1}^{1} 2 \sqrt{1 - x^{2}} d x = λ_{2} (D) = π .$

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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

Terence Tao, An Introduction to Measure Theory, Exercise 1.7.24 (standard reference, not scraped)