ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13

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A vector line integral around the vortex counts repeated traversals

Example

For the vortex field

$F (x, y) = (\frac{- y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}}),$

let $γ (t) = (\cos t, \sin t)$ and $β (t) = (\cos (2 t), \sin (2 t))$ on $[0, 2 π]$ . Then

$\int_{γ} F \cdot d r = 2 π, \int_{β} F \cdot d r = 4 π .$

Facts & Assumptions

Given: The field and paths in the Example.

[L1]

Vector line integrals integrate $⟨ F (η (t)), η^{'} (t) ⟩$ (Scalar line integrals with respect to arc length and vector-field line integrals).

[L2]

Sine and cosine have derivatives $\cos t$ and $- \sin t$ and satisfy $\sin^{2} t + \cos^{2} t = 1$ (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L3]

The integral of a constant $q$ on $[a, b]$ is $q (b - a)$ (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 $\int_{a}^{b} c = c (b - a)$ ).

[L4]

Parametrization invariance assumes a bijective oriented reparametrization (Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses).

Verification

technique · direct

1.1

By [L2], $F (γ (t)) = γ^{'} (t) = (- \sin t, \cos t)$ , so [L1], [L2], and [L3] give $\int_{γ} F \cdot d r = \int_{0}^{2 π} 1 d t = 2 π$ .

givenL1L2L3algebra

1.2

By [L2], $β^{'} (t) = 2 (- \sin (2 t), \cos (2 t)), F (β (t)) = (- \sin (2 t), \cos (2 t)) .$

givenL2algebra

2.1

Hence [L1], [L2], and [L3] give $\int_{β} F \cdot d r = \int_{0}^{2 π} 2 (\sin^{2} (2 t) + \cos^{2} (2 t)) d t = 4 π .$

step 1.2L1L2L3

3.1

The two paths have the same counterclockwise unit-circle image, but $t \mapsto 2 t$ on $[0, 2 π]$ is not a bijection onto $[0, 2 π]$ ; it covers the circle twice. Thus [L4] does not assert equality here.

givenL4algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 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

J. Lebl, Basic Analysis II, sections 9.2 and 9.3 (standard reference, not scraped)