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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Surface area and flux on a sphere, with scalar integrals on a hemisphere

Example

For $R > 0$ , the sphere of radius $R$ has area $4 π R^{2}$ . With outward orientation, the field $F (x) = x$ has flux $4 π R^{3}$ . On the northern hemisphere, the scalar integral of the height coordinate is $π R^{3}$ .

Facts & Assumptions

Given: The parametrization $φ (ϕ, θ) = R (\sin ϕ \cos θ, \sin ϕ \sin θ, \cos ϕ)$ on $[0, π] \times [0, 2 π]$ .

[L1]

The sine and cosine derivative, sign, range, Pythagorean, and endpoint-value formulas hold (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine, Quarter-turn values and shifts by pi/2 and pi).

[L2]

Regular-patch area and scalar integrals are $\int_{D} J_{φ}$ and $\int_{D} (q \circ φ) J_{φ}$ , and flux in the orientation induced by $φ$ is $\int_{D} (F \circ φ) \cdot (φ_{u} \times φ_{v})$ (Regular parametrized surface patches on compact Jordan parameter regions, Surface area and scalar surface integrals on a regular patch, Unit normal fields, orientations, and flux through a regular surface patch); Jordan-Fubini and the fundamental theorem evaluate the rectangular integrals (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[L3]

The area density satisfies $J_{φ} = ∥ φ_{u} \times φ_{v} ∥_{2}$ at every parameter point, with the cross product given by the coordinate formula (The surface area density is the norm of the cross product of the parameter tangents, The cross product in $R^{3}$ ).

Verification

technique · direct

1.1

Direct differentiation using [L1] and the coordinate formula in [L3] gives $φ_{ϕ} \times φ_{θ} = R \sin ϕ φ$ , the outward area vector, whose norm is $R^{2} \sin ϕ$ ; by [L3] this is the area density $J_{φ}$ . It is nonzero in the parameter interior; the longitude seam and poles lie on the boundary, so [L2] gives a regular patch.

givenL1L2L3algebra

2.1

By [L2], the area is $\int_{0}^{2 π} \int_{0}^{π} R^{2} \sin ϕ d ϕ d θ = 4 π R^{2}$ .

step 1.1L2algebra

2.2

Since $F (φ) = φ$ , its dot product with the outward area vector is $R^{3} \sin ϕ$ , whose integral is $4 π R^{3}$ .

step 1.1L2algebra

2.3

On the northern hemisphere $0 \leq ϕ \leq π / 2$ , the height is $R \cos ϕ$ , so its scalar integral is $\int_{0}^{2 π} \int_{0}^{π / 2} R^{3} \sin ϕ \cos ϕ d ϕ d θ = π R^{3}$ .

step 1.1L1L2algebra

3.1

Steps 2.1, 2.2, and 2.3 establish the area, outward flux, and hemisphere scalar integral with the orientation stated.

step 2.1step 2.2step 2.3∎

Depends on

Used by

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Dependency tree · two levels

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Sources

University of Toronto MAT237 notes, Section 5.3, Example 1 and Basic Problem 3 (standard reference, not scraped)
M. E. Taylor, Introduction to Analysis in Several Variables, Section 3.2 (standard reference, not scraped)