Scalar Line Integrals, multivariable and vector calculus lesson

Picture a bent piece of wire whose thickness changes along its length, so some stretches weigh more per centimeter than others. To find its total mass you cut it into short pieces, multiply each piece’s length by the density there, and add. The scalar line integral is the limit of that sum, and it works for any quantity spread along a curve, such as charge on a wire or temperature averaged along a path.

Key terms

Regular and piecewise regular curves

Definition A parametrization $\mathbf{r} : [a, b] \to \mathbb{R}^n$ with $a < b$ is regular (or smooth) if $\mathbf{r}'$ exists and is continuous on $[a, b]$ , using one-sided derivatives at the endpoints, and

\mathbf{r}'(t) \ne \mathbf{0} \quad \text{for every } t \text{ in } [a, b].

A curve with a regular parametrization that is one-to-one, except possibly that $\mathbf{r}(a) = \mathbf{r}(b)$ for a closed curve, is a regular curve. A curve $C$ is piecewise regular if it is a union $C = C_1 \cup C_2 \cup \cdots \cup C_m$ of finitely many regular curves, where the end point of each $C_i$ is the starting point of $C_{i+1}$ .

In plain words A regular curve has a tangent direction at every point that changes continuously, and a point moving along it never stops. A piecewise regular curve may have finitely many corners, where you start a new piece.

Example The graph of $y = \lvert x \rvert$ for $-1 \le x \le 1$ has the parametrization $\mathbf{r}(t) = \langle t, \lvert t \rvert \rangle$ , which has no derivative at $t = 0$ . Split it at the corner. The pieces $\langle t, -t \rangle$ for $-1 \le t \le 0$ and $\langle t, t \rangle$ for $0 \le t \le 1$ both have constant nonzero velocity, so the V shape is piecewise regular.

The arc length element

Definition Let $C$ be a regular curve with parametrization $\mathbf{r}(t)$ , $a \le t \le b$ . The arc length function measured from $\mathbf{r}(a)$ is

s(t) = \int_a^t \lvert \mathbf{r}'(u) \rvert \, du,

so by the fundamental theorem of calculus $s'(t) = \lvert \mathbf{r}'(t) \rvert$ . The arc length element is

ds = \lvert \mathbf{r}'(t) \rvert \, dt.

In plain words During a short time $\Delta t$ the point covers a distance of about speed times time, $\lvert \mathbf{r}'(t) \rvert \Delta t$ . That speed factor converts a step in the parameter into a step in length.

Example For the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ , the velocity is $\langle -\sin t, \cos t, 1 \rangle$ and the speed is

\begin{aligned} \lvert \mathbf{r}'(t) \rvert &= \sqrt{\sin^2 t + \cos^2 t + 1} \\[0.4em] &= \sqrt{2}. \end{aligned}

So $ds = \sqrt{2} \, dt$ , and one full turn, $0 \le t \le 2\pi$ , has length $2\sqrt{2}\,\pi \approx 8.89$ .

Integration with respect to arc length

Definition Let $C$ be a regular curve in $\mathbb{R}^n$ and $f$ a function defined on $C$ . Cut $C$ into $N$ arcs of lengths $\Delta s_1, \dots, \Delta s_N$ and pick a point $P_i^*$ on the $i$ th arc. The scalar line integral of $f$ over $C$ is

\int_C f \, ds = \lim_{\max \Delta s_i \to 0} \sum_{i=1}^{N} f(P_i^*) \, \Delta s_i,

when this limit exists and is the same for every choice of cuts and points. If $f$ is continuous on $C$ and $\mathbf{r}(t)$ , $a \le t \le b$ , is a regular parametrization that traces $C$ exactly once (one-to-one, except possibly $\mathbf{r}(a) = \mathbf{r}(b)$ for a closed curve), the limit exists and

\int_C f \, ds = \int_a^b f(\mathbf{r}(t)) \, \lvert \mathbf{r}'(t) \rvert \, dt.

With $f = 1$ the integral is the length of $C$ .

In plain words Each small piece contributes two separate factors, the value of $f$ there and the length of the piece. To compute, substitute the parametrization into $f$ , multiply by the speed, and do an ordinary integral in $t$ .

Example Take $f(x, y) = xy$ on the quarter circle of radius $2$ from $(2, 0)$ to $(0, 2)$ . With $\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$ , $0 \le t \le \pi/2$ , the speed is $2$ and $f(\mathbf{r}(t)) = 4\cos t \sin t$ . So

\begin{aligned} \int_C xy \, ds &= \int_0^{\pi/2} 4\cos t \sin t \cdot 2 \, dt \\[0.4em] &= 4\sin^2 t \, \Big|_0^{\pi/2} = 4. \end{aligned}

Integrals over piecewise regular curves

Definition If $C = C_1 \cup \cdots \cup C_m$ is piecewise regular and $f$ is continuous on $C$ , then

\begin{aligned} \int_C f \, ds &= \int_{C_1} f \, ds + \int_{C_2} f \, ds \\[0.4em] &\quad + \cdots + \int_{C_m} f \, ds, \end{aligned}

where each piece is computed with its own regular parametrization.

In plain words Break the curve at its corners, integrate along each smooth piece, and add the results.

Example Let $f(x, y) = x + y^2$ and let $C$ go from $(0, 0)$ to $(1, 0)$ and then up to $(1, 1)$ . On the first piece take $\mathbf{r}_1(t) = \langle t, 0 \rangle$ with speed $1$ , so $f = t$ . On the second take $\mathbf{r}_2(t) = \langle 1, t \rangle$ with speed $1$ , so $f = 1 + t^2$ . Both run over $0 \le t \le 1$ , and

\begin{aligned} \int_C f \, ds &= \int_0^1 t \, dt + \int_0^1 (1 + t^2) \, dt \\[0.4em] &= \frac{1}{2} + \frac{4}{3} = \frac{11}{6}. \end{aligned}

Density on a wire, mass and balance point

Definition A thin wire in the shape of a piecewise regular curve $C$ with continuous linear density $\delta \ge 0$ , measured in mass per unit length, has mass

m = \int_C \delta \, ds.

If $m > 0$ , its center of mass $(\bar{x}, \bar{y}, \bar{z})$ is given by

\begin{gathered} \bar{x} = \frac{1}{m} \int_C x \, \delta \, ds, \\[0.4em] \bar{y} = \frac{1}{m} \int_C y \, \delta \, ds, \\[0.4em] \bar{z} = \frac{1}{m} \int_C z \, \delta \, ds, \end{gathered}

with only the first two for a wire in the plane.

In plain words Mass is density times length, added up along the wire. The balance point is the average position, where each piece is weighted by its mass.

Example A wire lies on the upper half of the unit circle, $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ , $0 \le t \le \pi$ , with density $\delta(x, y) = y$ , so it gets heavier toward the top. The speed is $1$ . The mass is

m = \int_0^\pi \sin t \, dt = 2.

The integral of $x\delta$ is the integral of $\cos t \sin t$ over $[0, \pi]$ , which is $0$ , so $\bar{x} = 0$ as the symmetry suggests. For the other coordinate,

\begin{aligned} \bar{y} &= \frac{1}{2} \int_0^\pi \sin^2 t \, dt \\[0.4em] &= \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \approx 0.785. \end{aligned}

The balance point $(0, \pi/4)$ lies below the wire, in empty space, which is normal for a curved wire.

Parametrization independence

Definition Let $\mathbf{r}(t)$ , $a \le t \le b$ , be a regular parametrization of $C$ and let $\varphi : [c, d] \to [a, b]$ be a one-to-one and onto function with continuous derivative and $\varphi'(u) \ne 0$ for all $u$ . Then $\mathbf{q}(u) = \mathbf{r}(\varphi(u))$ is another regular parametrization of $C$ , and for every $f$ continuous on $C$

\begin{aligned} &\int_c^d f(\mathbf{q}(u)) \, \lvert \mathbf{q}'(u) \rvert \, du \\[0.4em] &\quad = \int_a^b f(\mathbf{r}(t)) \, \lvert \mathbf{r}'(t) \rvert \, dt. \end{aligned}

The reason is the chain rule, $\mathbf{q}'(u) = \mathbf{r}'(\varphi(u)) \, \varphi'(u)$ , followed by the substitution $t = \varphi(u)$ , where the absolute value $\lvert \varphi'(u) \rvert$ is exactly what the substitution needs whether $\varphi$ increases or decreases. Any two regular parametrizations that trace $C$ exactly once are related by such a $\varphi$ , after splitting $C$ at the two starting points when $C$ is closed and they start at different points, so the value of $\int_C f \, ds$ does not depend on which one you use. The once condition matters, since a parametrization that runs around a circle twice gives twice the integral.

In plain words The integral depends only on the curve and on $f$ , not on how fast or in which way you trace it. A faster parametrization covers each piece in less parameter time, and the larger speed factor makes up for it exactly.

Example Let $C$ be the segment from $(0, 0)$ to $(3, 4)$ and $f(x, y) = x + y$ . With $\mathbf{r}(t) = \langle 3t, 4t \rangle$ , $0 \le t \le 1$ , the speed is $5$ and

\int_0^1 7t \cdot 5 \, dt = \frac{35}{2}.

With the slower $\mathbf{q}(u) = \langle 3u/2, 2u \rangle$ , $0 \le u \le 2$ , the speed is $5/2$ and

\int_0^2 \frac{7u}{2} \cdot \frac{5}{2} \, du = \frac{35}{4} \cdot 2 = \frac{35}{2}.

Orientation invariance

Definition If $\mathbf{r}(t)$ , $a \le t \le b$ , traces $C$ in one direction, then $\mathbf{r}^-(t) = \mathbf{r}(a + b - t)$ , $a \le t \le b$ , traces the same points in the opposite direction. Its speed is $\lvert \mathbf{r}'(a + b - t) \rvert$ , and for $f$ continuous on $C$

\int_{-C} f \, ds = \int_C f \, ds,

where $-C$ is $C$ with the reversed direction. This is the special case $\varphi(u) = a + b - u$ of parametrization independence. In particular the length of a curve does not depend on its direction.

In plain words The element $ds$ is a length, and lengths are never negative, so walking the wire backwards gives the same mass. This is different from the integrals $\int_C f \, dx$ and $\int_C \mathbf{F} \cdot d\mathbf{r}$ of the next lesson, which change sign when the direction is reversed.

Example Trace the quarter circle from the earlier example the other way, from $(0, 2)$ to $(2, 0)$ , with $\mathbf{r}^-(t) = \langle 2\sin t, 2\cos t \rangle$ , $0 \le t \le \pi/2$ . The speed is still $2$ and the integrand is still $4\sin t \cos t \cdot 2$ , so

\int_{-C} xy \, ds = 4 = \int_C xy \, ds.

Check yourself

Find the integral of $f(x, y) = x$ with respect to arc length along the segment from $(1, 2)$ to $(4, 6)$ . Does the answer change if the segment is traced from $(4, 6)$ to $(1, 2)$ ?

Show the answer

Use $\mathbf{r}(t) = \langle 1 + 3t, 2 + 4t \rangle$ , $0 \le t \le 1$ , with speed $5$ . Then

\begin{aligned} \int_C x \, ds &= \int_0^1 (1 + 3t) \cdot 5 \, dt \\[0.4em] &= 5 \left( 1 + \frac{3}{2} \right) = \frac{25}{2}. \end{aligned}

The answer stays $25/2$ for the opposite direction, because scalar line integrals do not depend on orientation.

A wire has the shape of one turn of the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ , $0 \le t \le 2\pi$ , and its density is $\delta(x, y, z) = z$ . Find its mass.

Show the answer

The speed is $\sqrt{2}$ and $\delta(\mathbf{r}(t)) = t$ , so

\begin{aligned} m &= \int_0^{2\pi} t \sqrt{2} \, dt = \sqrt{2} \cdot \frac{(2\pi)^2}{2} \\[0.4em] &= 2\sqrt{2}\,\pi^2 \approx 27.9. \end{aligned}

A wire of constant density lies on the quarter circle $x^2 + y^2 = a^2$ in the first quadrant, with $a > 0$ . Find its balance point.

Show the answer

Let the density be $k > 0$ and use $\mathbf{r}(t) = \langle a\cos t, a\sin t \rangle$ , $0 \le t \le \pi/2$ , with speed $a$ . The mass is $k a \pi / 2$ and

\begin{aligned} \int_C x \, k \, ds &= \int_0^{\pi/2} k a \cos t \cdot a \, dt \\[0.4em] &= k a^2, \end{aligned}

so $\bar{x} = 2a/\pi$ . By symmetry about the line $y = x$ , also $\bar{y} = 2a/\pi$ . The balance point is $(2a/\pi, 2a/\pi)$ , about $(0.64a, 0.64a)$ .

Scalar Line Integrals

Key terms

Regular and piecewise regular curves

The arc length element

Integration with respect to arc length

Integrals over piecewise regular curves

Density on a wire, mass and balance point

Parametrization independence

Orientation invariance

Check yourself

How this lesson connects