Line Integrals and Path Parametrisation

Integrating a vector field along a route rather than over an interval: how a parametrisation turns the line integral into an ordinary single-variable integral, why the value is a property of the oriented curve, and why the answer generally depends on which route was taken.

Definition

A vector field on a region of the plane assigns a vector $F (x, y) = (P (x, y), Q (x, y))$ to each point.

The line integral of $F$ along a curve $C$ parametrised by $r (t) = (x (t), y (t))$ for $a \leq t \leq b$ is

\int_{C} P d x + Q d y = \int_{a}^{b} [P (r (t)) x^{'} (t) + Q (r (t)) y^{'} (t)] d t,

a single-variable integral once the substitution is made. Its value is unchanged by reparametrising $C$ in the same direction, and reverses sign if the direction reverses.

Assumptions and scope

The value of a line integral is unchanged by reparametrising the curve in the same direction and changes sign under reversal, so an orientation must be stated before a value means anything.
These statements are for the plane. The analogous three-dimensional condition involves the curl rather than a single cross-partial comparison, and the corresponding theorems are those of Stokes and the divergence theorem, which these units do not cover.
All figures in this unit come from exact evaluation of the stated integrals, with each conservative case checked twice (once by direct parametrisation along every stated path, and once as a difference of potential values), and with both sides of Green's theorem computed independently rather than one being inferred from the other.

Worked material

Example

Four routes, parametrised and evaluated

One field, $F = (- y, x)$ , and four routes. The arithmetic is short in every case; what varies is how the route is described.

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1. A straight segment: $(0, 0)$ to $(1, 1)$ . Parametrise $r (t) = (t, t)$ for $0 \leq t \leq 1$ , so $d x = d t$ and $d y = d t$ :

\int_{0}^{1} [(- t) (1) + (t) (1)] d t = \int_{0}^{1} 0 d t = 0 .

The field is perpendicular to this route at every point, so nothing accumulates.

2. A polygonal route: $(0, 0) \to (1, 0) \to (1, 1)$ . Do each leg separately.

Leg	Constant	Vanishing term	Contribution
$(0, 0) \to (1, 0)$	$y = 0$ , so $d y = 0$ and $P = - y = 0$	both	$0$
$(1, 0) \to (1, 1)$	$x = 1$ , so $d x = 0$	the $P d x$ term	$\int_{0}^{1} 1 d y = 1$

Total: $1$ . On a leg where one coordinate is constant, that differential is zero and one term drops, often the whole integrand. Recognising this before parametrising saves most of the work.

3. A circular arc: the unit circle, counterclockwise, from $(1, 0)$ to $(0, 1)$ . Parametrise $r (t) = (\cos t, \sin t)$ for $0 \leq t \leq π / 2$ , so $d x = - \sin t d t$ and $d y = \cos t d t$ :

\int_{0}^{π / 2} [(- \sin t) (- \sin t) + (\cos t) (\cos t)] d t = \int_{0}^{π / 2} (\sin^{2} t + \cos^{2} t) d t = \int_{0}^{π / 2} 1 d t = \frac{π}{2} .

The Pythagorean identity collapses the integrand to a constant, which is characteristic of this field on circles centred at the origin: it runs exactly along them.

4. The same arc, traversed the other way. From $(0, 1)$ to $(1, 0)$ , take $t$ from $π / 2$ down to $0$ , or equivalently negate:

\int_{π / 2}^{0} 1 d t = - \frac{π}{2} .

Same set of points, opposite sign. The integral is a property of the oriented curve, so a value quoted without a direction is ambiguous by a factor of $- 1$ .

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A reparametrisation changes nothing. Redo route 1 as $r (s) = (s^{2}, s^{2})$ for $0 \leq s \leq 1$ , which traces the same segment in the same direction at a different speed. Then $d x = d y = 2 s d s$ and the integrand is $(- s^{2}) (2 s) + (s^{2}) (2 s) = 0$ , giving $0$ again. The substitution introduces a derivative factor that cancels against the change of variable, which is why the parametrisation is a device for computing and not part of the answer.

What the four show. Routes 1 and 2 share endpoints and give $0$ and $1$ : the value depends on the path. Routes 3 and 4 are the same points in opposite directions and differ in sign. And route 1 under two parametrisations gives one answer. So a line integral is determined by the field, the point set, and the direction, and by nothing else about how the route was written down.

Common errors

Common misconception

That a line integral depends only on its endpoints, as a single-variable integral depends only on its limits. For a general vector field it depends on the entire route. Take $F = (- y, x)$ and three paths from $(0, 0)$ to $(1, 1)$ . Along the straight line $(t, t)$ the integrand is $- t + t = 0$ , so the integral is $0$ . Along the route through $(1, 0)$ the horizontal leg contributes $0$ because $y = 0$ there, and the vertical leg contributes $\int_{0}^{1} 1 d y = 1$ , so the total is $1$ . Along the route through $(0, 1)$ the vertical leg contributes $0$ because $x = 0$ , and the horizontal leg contributes $\int_{0}^{1} - 1 d x = - 1$ , so the total is $- 1$ . Same endpoints, three values: $0$ , $1$ and $- 1$ . The cross-partials explain it, $\partial (- y) / \partial y = - 1$ while $\partial (x) / \partial x = 1$ , so the field is not conservative and no potential exists. Contrast $F = (2 x y, x^{2})$ , whose cross-partials are both $2 x$ : all three of the same paths give $1$ , matching $f (1, 1) - f (0, 0)$ for the potential $f = x^{2} y$ . Endpoint-only evaluation is a privilege of conservative fields, not a general property of line integrals, and assuming it silently converts a route-dependent quantity into a number that was never well defined.

Line Integrals and Path Parametrisation

Assumptions and scope

Worked material

Four routes, parametrised and evaluated

Common errors

Related units

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