Math Core

An ordinary integral adds up a quantity along a straight interval of the $x$ -axis. A line integral adds up a quantity along a curve: the mass of a bent wire whose density varies, or the work a force field does on a particle that travels along a winding path. The trick is always the same: parametrize the curve, and the line integral becomes an ordinary integral in the parameter $t$ .

Line integrals of functions

Let $C$ be a smooth curve given by $\mathbf{r}(t) = \langle x(t), y(t)\rangle$ for $a \le t \le b$ , and let $f(x, y)$ be a function defined on $C$ . Chop $C$ into tiny pieces of arc length $\Delta s$ , multiply the value of $f$ on each piece by its length, and add. In the limit you get the line integral of $f$ with respect to arc length.

Recall from arc length that a small change $dt$ in the parameter moves you a distance $ds = \lvert\mathbf{r}'(t)\rvert\,dt$ along the curve. That gives the formula you actually compute with.

Definition

Line integral with respect to arc length

\int_C f(x, y)\,ds = \int_a^b f\big(x(t), y(t)\big)\,\lvert\mathbf{r}'(t)\rvert\,dt, \qquad \lvert\mathbf{r}'(t)\rvert = \sqrt{x'(t)^2 + y'(t)^2}.

In space, add a $z$ -component: $\lvert\mathbf{r}'(t)\rvert = \sqrt{x'^2 + y'^2 + z'^2}$ .

If $f = 1$ , the integral is just the length of $C$ . If $f$ is the density of a wire (mass per unit length), the integral is the wire's mass. The value does not depend on which parametrization you choose or which way you traverse the curve, because $ds$ is always a positive length.

Worked example: A straight segment

Evaluate $\displaystyle\int_C (x + y)\,ds$ , where $C$ is the segment from $(0, 0)$ to $(3, 4)$ .

Parametrize: $\mathbf{r}(t) = \langle 3t, 4t\rangle$ , $0 \le t \le 1$ . Then $\mathbf{r}'(t) = \langle 3, 4\rangle$ and $\lvert\mathbf{r}'(t)\rvert = 5$ . On the curve, $x + y = 7t$ . So

\int_C (x + y)\,ds = \int_0^1 7t \cdot 5\,dt = 35\cdot\frac{1}{2} = \frac{35}{2}.

Worked example: A quarter circle

Evaluate $\displaystyle\int_C xy\,ds$ , where $C$ is the quarter of the circle $x^2 + y^2 = 4$ in the first quadrant.

Parametrize: $\mathbf{r}(t) = \langle 2\cos t, 2\sin t\rangle$ , $0 \le t \le \tfrac{\pi}{2}$ . Then $\lvert\mathbf{r}'(t)\rvert = \sqrt{4\sin^2 t + 4\cos^2 t} = 2$ , and

\int_C xy\,ds = \int_0^{\pi/2} (2\cos t)(2\sin t)\cdot 2\,dt = 8\int_0^{\pi/2}\sin t\cos t\,dt = 8\cdot\frac{1}{2} = 4.

Line integrals of vector fields

Now let $\mathbf{F}$ be a force field and let a particle move along $C$ . On a tiny piece of the curve, the displacement is $d\mathbf{r} = \mathbf{r}'(t)\,dt$ , and the work done is the component of force along the motion times the distance: $\mathbf{F}\cdot d\mathbf{r}$ . Adding these up gives the work.

Line integral of a vector field

\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t)\,dt.

If $\mathbf{F} = \langle P, Q\rangle$ , this is also written $\displaystyle\int_C P\,dx + Q\,dy$ , since $dx = x'(t)\,dt$ and $dy = y'(t)\,dt$ . In space, $\displaystyle\int_C P\,dx + Q\,dy + R\,dz$ .

Equivalently, $\mathbf{F}\cdot d\mathbf{r} = (\mathbf{F}\cdot\mathbf{T})\,ds$ , where $\mathbf{T}$ is the unit tangent vector. So the vector line integral is the scalar line integral of the tangential component of $\mathbf{F}$ . Force perpendicular to the path does no work.

The path y = x² from (0, 0) to (1, 1) used in the work example.Open in grapher →

Worked example: Work along a parabola

Find the work done by $\mathbf{F}(x, y) = \langle y, x^2\rangle$ on a particle moving along $y = x^2$ from $(0, 0)$ to $(1, 1)$ .

Parametrize with $x = t$ : $\mathbf{r}(t) = \langle t, t^2\rangle$ , $0 \le t \le 1$ , so $\mathbf{r}'(t) = \langle 1, 2t\rangle$ . On the curve $\mathbf{F} = \langle t^2, t^2\rangle$ . Then

W = \int_0^1 \langle t^2, t^2\rangle\cdot\langle 1, 2t\rangle\,dt = \int_0^1 \left(t^2 + 2t^3\right)dt = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}.

Worked example: A helix in space

Evaluate $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle z, x, y\rangle$ along one turn of the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle$ , $0 \le t \le 2\pi$ .

$\mathbf{r}'(t) = \langle -\sin t, \cos t, 1\rangle$ and $\mathbf{F}(\mathbf{r}(t)) = \langle t, \cos t, \sin t\rangle$ . The dot product is $-t\sin t + \cos^2 t + \sin t$ . Integrate each piece over $[0, 2\pi]$ :

\int_0^{2\pi} -t\sin t\,dt = \Big[t\cos t - \sin t\Big]_0^{2\pi} = 2\pi, \qquad \int_0^{2\pi}\cos^2 t\,dt = \pi, \qquad \int_0^{2\pi}\sin t\,dt = 0.

The total is $3\pi$ .

Orientation matters

Traversing $C$ in the opposite direction (written $-C$ ) flips every $d\mathbf{r}$ , so

\int_{-C}\mathbf{F}\cdot d\mathbf{r} = -\int_C \mathbf{F}\cdot d\mathbf{r}, \qquad\text{but}\qquad \int_{-C} f\,ds = \int_C f\,ds.

A curve made of several pieces, like the sides of a triangle, is handled one piece at a time; add the results.

Common mistake

Don't forget the factor $\lvert\mathbf{r}'(t)\rvert$ in $\displaystyle\int_C f\,ds$ , and don't insert it into $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ . The vector integral uses $\mathbf{r}'(t)$ itself, dotted with $\mathbf{F}$ , with no square root anywhere.

Tip

To parametrize the segment from $P$ to $Q$ , use $\mathbf{r}(t) = (1 - t)P + tQ$ for $0 \le t \le 1$ . Then $\mathbf{r}'(t) = Q - P$ , a constant vector.