Line Integrals and Path Parametrisation

Integrate the vector field $\mathbf{F}=(-y,x)$ from $(0,0)$ to $(1,1)$.

Along the straight diagonal, the answer is $0$.

Along the route through $(1,0)$, right, then up, the answer is $1$.

Along the route through $(0,1)$, up, then right, the answer is $-1$.

Same field, same endpoints, three values. Nothing has gone wrong: this integral genuinely depends on the route, and asking for "the integral from $(0,0)$ to $(1,1)$" is not a well-posed question until a path is named.

That is unfamiliar, because an ordinary integral ${\int }_a^{b}fdx$ has only one way of getting from $a$ to $b$. Introduce a second dimension and the route becomes a choice, so the first question about any line integral is whether the choice matters.

This unit covers evaluating such integrals: parametrising a route, substituting, and reducing the whole thing to an ordinary single-variable integral. Whether the route can be avoided altogether is the next unit's question.

The canonical statement above gives the definitions and the two theorems. What follows is what each condition in them is load-bearing for, since a hypothesis that seems decorative is the one that later produces a wrong answer.

The parametrisation is a device, not part of the answer. Two parametrisations of the same curve in the same direction give the same integral. The substitution $t\mapsto s$ cancels against the derivative it introduces. Reversing direction reverses the sign, because $dx$ and $dy$ change sign while the field does not. So a line integral is a property of the oriented curve, and stating a value without an orientation states half a result.

Treat $\mathbf{F}$ as a current and ${\int }_C\mathbf{F}\cdot d\mathbf{r}$ as the work done travelling along $C$: with the current counts positively, against it negatively, across it not at all.

That one reading predicts every value in the figure, and the three disagreeing answers between the same two points are what path dependence means.

The picture misleads in one respect. It suggests the value should depend on how far you travel, and it does not: reparametrising a route changes the speed and leaves the integral alone. Direction relative to the field is what counts, not pace.

One field, $\mathbf{F}=(-y,x)$, 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 $\mathbf{r}(t)=(t,t)$ for $0\le t\le 1$, so $dx=dt$ and $dy=dt$:

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.

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 $\mathbf{r}(t)=(\cos t,\sin t)$ for $0\le t\le \pi /2$, so $dx=-\sin tdt$ and $dy=\cos tdt$:

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 $\pi /2$ down to $0$, or equivalently negate:

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 $\mathbf{r}(s)=(s^2,s^2)$ for $0\le s\le 1$, which traces the same segment in the same direction at a different speed. Then $dx=dy=2sds$ and the integrand is $(-s^2)(2s)+(s^2)(2s)=0$, 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.

To evaluate a line integral directly.

1. Parametrise the curve as $\mathbf{r}(t)=(x(t),y(t))$ with a stated range, in the direction the problem specifies.
2. Differentiate to get $dx=x^{\prime }(t)dt$ and $dy=y^{\prime }(t)dt$.
3. Substitute everything. The $x$ and $y$ inside $P$ and $Q$ as well as the differentials. A common slip is substituting the differentials and leaving $x$ and $y$ in the integrand.
4. Evaluate the resulting single-variable integral in $t$.
5. For a polygonal path, do each leg separately and add. On a leg where one coordinate is constant, that differential is zero and one term drops, often the whole integrand.

Checks. Verify an orientation by checking one leg's sign explicitly: traversing the top of a rectangle counterclockwise means integrating from larger $x$ to smaller. And confirm both the coordinates and the differentials were substituted, since leaving $x$ or $y$ in the integrand produces an expression that cannot be integrated in $t$.

Part 1: a field whose integral depends on the route.

Take $\mathbf{F}=(-y,x)$, so $P=-y$ and $Q=x$, and integrate from $(0,0)$ to $(1,1)$.

Test first. ${\displaystyle \frac{\partial P}{\partial y}}=-1$ and ${\displaystyle \frac{\partial Q}{\partial x}}=1$. They differ, so no potential exists and path dependence is expected. One derivative each, and the question is settled before any integration.

Path A. The diagonal. Parametrise $\mathbf{r}(t)=(t,t)$, $0\le t\le 1$, so $dx=dt$ and $dy=dt$:

Path B, right then up. On the horizontal leg $y=0$ so $dy=0$, and $P=-y=0$: the leg contributes $0$. On the vertical leg $x=1$ so $dx=0$, and $Q=x=1$:

Total: $0+1=1$.

Path C, up then right. On the vertical leg $x=0$ so $dx=0$ and $Q=0$: contributes $0$. On the horizontal leg $y=1$ so $dy=0$ and $P=-1$:

Total: $0+(-1)=-1$.

Three paths, three values: $0$, $1$, $-1$. The endpoints alone do not determine the answer.

Quoting a line integral without an orientation. Reversing direction reverses the sign, so a value stated without a direction is ambiguous by a factor of $-1$. On the unit square, $(-y,x)$ gives $2$ counterclockwise and $-2$ clockwise.

Substituting the differentials but not the coordinates. Replacing $dx$ by $x^{\prime }(t)dt$ while leaving $x$ and $y$ in the integrand produces an expression in mixed variables that cannot be integrated in $t$, and the resulting number, if one is forced out, means nothing.

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Two that follow from the one-dimensional habit.

Assuming endpoints determine the value. They do for a conservative field and not otherwise. Three paths between the same endpoints give $0$, $1$ and $-1$ for $(-y,x)$, so "the integral from $(0,0)$ to $(1,1)$" is not a well-posed quantity for this field.

Expecting a closed path to give zero. That is a property of gradient fields, not of closed paths. A circulating field gives $2$ around the unit square, and the punctured-plane field gives $2\pi$ around the unit circle.

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And one about the word itself. A vector field is not a field in the algebraic sense. A set with addition and multiplication satisfying the field axioms, which is what the term means in linear algebra. Nothing connects the two uses. Where both readings are available, write vector field in full.