Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-26 (gpt-6-sol)
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The inverse-square field is divergence free, and its flux through the sphere bounding the translated unit ball vanishes

Example

On U=R3∖{0} let F(x,y,z)=(x,y,z)(x2+y2+z2)3/2. Then div⁡F=0 on U. Consequently the outward flux of F through the sphere bounding the translated unit ball B={(x,y,z):(x2+y2+(z−2)2)≤1} is 0.

Facts & Assumptions

Given: The field F on U=R3∖{0}, and the translated closed unit ball B={(x,y,z):(x2+y2+(z−2)2)≤1}.

[F1]

The divergence of a field is the sum of its coordinate partial derivatives (Divergence and curl of a C1 vector field).

[L6]

For every real α, the function s↦sα is continuous and differentiable on (0,∞), with derivative αsα−1 (Continuity and derivatives of positive-base real powers).

[F2]

The Jacobian matrix records the coordinate partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F4]

Flux is computed against the oriented area vector of a patch (Unit normal fields, orientations, and flux through a regular surface patch).

[F5]

A subset of a metric space is open when every one of its points contains an open metric ball lying in the subset; the Euclidean metric on R3 is induced by ∥⋅∥2 (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it).

[L8]

A regular surface patch may identify parameters only on its boundary; its induced area vector gives the outward flux integral (Regular parametrized surface patches on compact Jordan parameter regions, Unit normal fields, orientations, and flux through a regular surface patch).

[L10]

Sine and cosine have the usual derivatives, sin⁡2u+cos⁡2u=1, sin⁡u>0 for 0<u<π, cosine is strictly decreasing on [0,π], and cos⁡0=1, cos⁡π=−1 (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).

[L11]

The unit-circle parametrization θ↦(cos⁡θ,sin⁡θ) is injective on [0,2π) (t↦(cos⁡t,sin⁡t) is a bijection from [0,2π) onto the real unit circle); the cross product has its coordinate determinant formula (The cross product in R3).

Verification

technique · direct
1.1L3L4L6L7F2F3given

Put s(x,y,z)=x2+y2+z2, which is positive and continuous on U by [L7]. The ith component of F is Fi=xis−3/2. By the product and chain rules [L3, L4], the positive-base power rule [L6], and the coordinate interpretation of partial derivatives [F2], every coordinate partial derivative is ∂jFi={s−3/2−3xi2s−5/2,j=i,−3xixjs−5/2,j≠i. The coordinate projections are continuous by [L7], so s is continuous; because s>0 on U, [L6] and [L7] make every function in the displayed formulas continuous there. Hence F is C1 on U.

2.1step 1.1F1

Summing the three diagonal formulas of step 1.1 gives div⁡F=3s−3/2−3s s−5/2=0 on U by [F1].

2.2step 1.1F3F5given

Every point of B has distance at least 1 from the origin, so B⊆U. The set U is open: if p∈U, then ∥p∥2>0 and the ball B(p,∥p∥2/2) cannot contain the deleted origin. Hence F is continuous on a neighbourhood of the sphere.

3.1L8L10L11step 2.2construct

Parametrize ∂B by ψ(ϕ,θ)=(sin⁡ϕcos⁡θ,sin⁡ϕsin⁡θ,2+cos⁡ϕ) on D=[0,π]×[0,2π]. Direct differentiation and the cross-product formula give ψϕ×ψθ=sin⁡ϕ(sin⁡ϕcos⁡θ,sin⁡ϕsin⁡θ,cos⁡ϕ). On D∘ this is nonzero and points outward. Strict monotonicity of cosine on [0,π] and [L11] make the parametrization injective on its interior; its only repeated boundary images lie at the seam or poles. Thus ψ is one regular patch covering the sphere in the sense of [L8].

4.1F3F4L8L9L10step 3.1

On this patch, ∣ψ∣2=5+4cos⁡ϕ≥1 and ψ⋅(ψϕ×ψθ)=sin⁡ϕ(1+2cos⁡ϕ). Thus [F4] and [L9] give the outward flux as 2π∫0π(1+2cos⁡ϕ)sin⁡ϕ (5+4cos⁡ϕ)−3/2 dϕ.

5.1L3L4L6L9L10step 4.1∎

Put H(u)=14(5+4u+3/5+4u) for −1≤u≤1. The power and chain rules give H′(u)=(1+2u)(5+4u)−3/2, so the integrand in step 4.1 is −ddϕH(cos⁡ϕ). Since H(1)=H(−1)=1, the fundamental theorem in [L9] makes the flux −2π[H(cos⁡ϕ)]0π=0.

Remarks

  • The translation moves the sphere away from the singular origin. The direct patch calculation proves its zero flux without requiring an elementary-solid presentation of the ball.

Depends on

Used by

Dependency tree · two levels

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Sources