6.1 Paths and line integrals

A line integral adds up the part of a vector field that points along a path, step by step. It is the work done by a force, it does not care how fast you walk, it changes sign if you walk backwards, and with arc length in place of the dot product it gives the mass and balance point of a wire.

You push a box along a floor with a force that changes from place to place. How much work did you do? If the force is constant and the path is straight, the answer is force times distance, as in high school. Here the force is a vector field (5.1) and the path bends. The line integral is the number that answers the question, and it is the tool that Green’s theorem, Stokes’ theorem and the whole second half of this unit are about.

Before you start

5.3: the rate of change of a scalar field along a curve is the gradient dotted with the velocity. 1.5: the integral of a derivative is the change in the function. 2.9: the length of a vector. 4.4: a path is a continuous map from an interval.

Quick checkLet

f

be a scalar field and

r

a path. What is

\dfrac{d}{dt} f(r(t))

Quick checkIf

g

has a continuous derivative on

[0, 1]

, what is

\displaystyle\int_0^1 g'(t)\,dt

Quick checkWhat is the length of the velocity vector

(-\sin t, \cos t, 1)

The picture: work accumulating along a path

Slide $u$ and watch the walker move from $A = (1, 0)$ to $B = (0, 1)$ . The red arrow is the field at the walker, the orange arrow is the velocity. The graph on the right is the work so far. It rises while the red arrow points along the orange one and falls while it points against it. Switch the field to $(y, -x)$ and compare the three paths: the same two endpoints, three different totals. Switch to $(y, x)$ and the three totals agree. Chapter 6.2 explains exactly when that happens.

The idea, in words

Cut the time interval into tiny steps of length $\Delta t$ . In one step the walker moves by about $r'(t)\,\Delta t$ , and the field barely changes, so it is about $F(r(t))$ . Only the part of the force that points along the motion does any pushing. That part, times the distance moved, is the dot product $F(r(t)) \cdot r'(t)\,\Delta t$ . Adding the steps gives a Riemann sum (1.5), and the limit of the sums is the integral

\int_a^b F(r(t)) \cdot r'(t)\,dt.

Why· Why the dot product, and why the velocity

A force at right angles to the motion does nothing: carrying a suitcase at constant height along a flat corridor needs no work in the physics sense, because the force you exert (up) is perpendicular to the motion (forward). The dot product $F \cdot r'$ is zero exactly when the two are perpendicular, positive when the force helps and negative when it resists. Using the velocity $r'$ rather than the position means the answer is built from how the path moves, and that is why it will turn out not to depend on how fast you walk, only on where you walk and in which direction.

Where this came from· Where the word 'work' came from

Engineers in the early 1800s wanted one number to rate what a machine does, and force alone or distance alone would not do. In 1829 Gaspard-Gustave Coriolis, in Du calcul de l’effet des machines, gave the name travail, ‘work’, its modern meaning: the force times the distance moved along the force, added up along the path of its point of application. Jean-Victor Poncelet, who was working on the same problem, later recalled that he had used the phrase ‘mechanical work’ in his lectures at Metz in the 1820s, and he acknowledged that the word came into general use through Coriolis. The modern definition, a dot product integrated along a curve, is the same idea written with vectors.

Paths, curves and corners

In 4.4 a path was a continuous map from $[0, 1]$ . In this unit a path is a continuous map $r: [a, b] \to \mathbb R^n$ on any closed interval, which only rescales the parameter. Think of $r(t)$ as the position at time $t$ . The set of points visited is the curve $C$ . Many different paths trace the same curve.

Definition T

Piecewise smooth path

A path $r: [a, b] \to \mathbb R^n$ is piecewise smooth if it has three properties.

The whole path is continuous on $[a, b]$ . The pieces must join.
There is a finite partition $a = t_0 < t_1 < \dots < t_m = b$ of the interval (1.5).
On each piece $[t_{j-1}, t_j]$ the derivative $r'$ exists (one-sided at the ends of the piece) and is continuous.

The official solution of the MATH 376 final accepts either closed pieces or open pieces $(t_{j-1}, t_j)$ , and accepts ‘smooth’ for ‘continuously differentiable’. It does not accept ‘differentiable’. (With open pieces the derivative may blow up at an end of a piece: $r(t) = (t, \sqrt t)$ on $[0, 1]$ is continuous, and on $(0, 1)$ its derivative is continuous but unbounded near $0$ . This chapter asks for the derivative to exist and be continuous on each closed piece, with one-sided derivatives at the ends, so that the integrand is bounded and the line integral is an ordinary Riemann integral.)

In one lineA continuous path $r: [a, b] \to \mathbb R^n$ for which there is a finite partition $a = t_0 < \dots < t_m = b$ such that on each piece $[t_{j-1}, t_j]$ the derivative $r'$ exists and is continuous.

Why it existsIt allows corners, like the boundary of a square, but keeps the velocity well behaved on every straight run between them, which is exactly what is needed for the integral of $F(r(t)) \cdot r'(t)$ to exist.

Each of the three requirements has a job, and each is easy to forget.

Continuous overall. If the pieces do not meet, there are two paths, not one path from $A$ to $B$ .
Finitely many pieces. The line integral will be a sum of one ordinary integral per piece. A sum has to be finite to be defined.
A continuous derivative, which is more than a derivative. The line integral is the ordinary Riemann integral of $F(r(t)) \cdot r'(t)$ , and a Riemann integrable function must be bounded. A derivative can exist everywhere and still be unbounded. The function $g(t) = t^2\sin(1/t^2)$ with $g(0) = 0$ has $g'(0) = 0$ , but for $t \ne 0$ we get $g'(t) = 2t\sin(1/t^2) - \frac2t\cos(1/t^2)$ , which is unbounded near $0$ (at $t = 1/\sqrt{2\pi k}$ it equals $-2\sqrt{2\pi k}$ ). A continuous derivative on a closed interval is bounded.

Worked example MATH 376 Final 1(b); see [A.1](/vol2/a1-what-you-lost-points-on/)

Write the definition of a piecewise smooth path, then decide whether (A)

r(t) = (t, |t|)

[-1, 1]

and (B)

r(t) = (t, 0)

for

0 \le t \le 1

r(t) = (t, 1)

for

1 < t \le 2

are piecewise smooth.

Solution.

Write the definition in three parts. A path $r: [a, b] \to \mathbb R^n$ is piecewise smooth if (1) $r$ is continuous on $[a, b]$ , (2) there is a partition $a = t_0 < \dots < t_m = b$ with $m$ a natural number, and (3) on each $[t_{j-1}, t_j]$ the derivative $r'$ exists and is continuous. This is the answer the exam wants. ‘Continuous and once-differentiable chunks, any amount of’ misses all three parts and lost a point.
(A) Continuity. Both coordinates $t$ and $|t|$ are continuous. Part (1).
(A) Choose the break point. Partition $-1 < 0 < 1$ . $|t|$ changes its formula at $0$ , so $0$ must be a break point. Two pieces is a finite number: part (2).
(A) Check each piece. On $[-1, 0]$ we have $r(t) = (t, -t)$ with $r' = (1, -1)$ . On $[0, 1]$ we have $r(t) = (t, t)$ with $r' = (1, 1)$ . Constant velocities are continuous. Part (3).
(B) Look at the joint. As $t \to 1$ from the left, $r(t) \to (1, 0)$ . From the right, $r(t) \to (1, 1)$ . The two formulas disagree at $t = 1$ .
(B) Conclude. $r$ is not continuous at $t = 1$ , so part (1) fails. Each piece is smooth. That is the ‘chunks’ picture, and it hides the missing global condition.

Answer. (A) is piecewise smooth, with a corner at $(0, 0)$ where $r'$ does not exist (allowed, since $0$ is a break point). (B) is not.

Quick checkWhich list contains all the requirements for a piecewise smooth path

r: [a, b] \to \mathbb R^n

The line integral

Definition T

Line integral

Let $r: [a, b] \to \mathbb R^n$ be piecewise smooth with curve $C$ , and let $F$ be a vector field that is continuous on a set containing $C$ . The line integral of $F$ along $C$ is

\int_C F \cdot dr = \int_a^b F(r(t)) \cdot r'(t)\,dt.

If $r$ has corners at $t_1, \dots, t_{m-1}$ , the right side means the sum of the integrals over the pieces $[t_{j-1}, t_j]$ . In the plane, with $F = (P, Q)$ and $r = (x(t), y(t))$ , the same integral is written $\int_C P\,dx + Q\,dy = \int_a^b \big(P\,x'(t) + Q\,y'(t)\big)\,dt$ . Apostol writes $\int_C F \cdot d\alpha$ and the lectures write $\int_\alpha f \cdot d\alpha$ .

In one lineFor a piecewise smooth path $r: [a, b] \to \mathbb R^n$ and a vector field $F$ continuous on the curve, $\int_C F \cdot dr = \int_a^b F(r(t)) \cdot r'(t)\,dt$ , a sum of one ordinary integral per smooth piece.

Why it existsIt is the limit of the sums $F \cdot (\text{displacement})$ over tiny steps, so it adds up the part of $F$ that points along the motion.

What each hypothesis does. $F$ continuous makes $F(r(t)) \cdot r'(t)$ piecewise continuous, hence Riemann integrable on each piece. A piecewise smooth path gives the velocity that is continuous on each piece. And $F$ must be defined at every point of the curve: if the curve passes through a point where $F$ is not defined, the integral has no meaning.

Definition T

Work

If $F$ is a force and $C$ is the path of a particle, $\int_C F \cdot dr$ is the work done by $F$ on the particle. It is positive where the force helps the motion and negative where it opposes it.

In one lineThe work done by a force field $F$ on a particle moving along $C$ is the line integral $\int_C F \cdot dr$ .

Why it existsIt turns 'force times distance' into a quantity that works when the force changes from place to place and the path bends.

The method

Parametrize $C$ in the stated direction. Reversing the direction changes the sign. Check the start and end points of your $r$ against the problem.
Differentiate: compute $r'(t)$ .
Substitute the path into the field: write $F(r(t))$ by putting $x(t), y(t)$ (and $z(t)$ ) into $F$ . You put the path into the field, never the field into the path.
Dot $F(r(t))$ with $r'(t)$ .
Integrate over the parameter interval.

Parametrizations used all the time:

Curve	Parametrization	Range
segment from point $P$ to point $Q$	$r(t) = P + t(Q - P)$	$0 \le t \le 1$
graph $y = g(x)$ from $x = a$ to $x = b$	$r(t) = (t, g(t))$	$a \le t \le b$
circle of radius $R$ about $(0, 0)$ , counterclockwise	$r(t) = (R\cos t, R\sin t)$	$0 \le t \le 2\pi$
ellipse $\frac{x^2}{p^2} + \frac{y^2}{q^2} = 1$ , counterclockwise	$r(t) = (p\cos t, q\sin t)$	$0 \le t \le 2\pi$

Worked example MATH 376 worksheet 3 problem 6

Give a parametrization of (a) the segment from

(2, -3)

(-4, 5)

; (b) the circle of radius

5

centred at

(0, 0, 4)

in the plane

z = 4

; (c) the segment from

(1, 0, 5)

(3, -2, 12)

; (d) the ellipse

\frac{x^2}{9} + \frac{y^2}{16} = 1

Solution.

(a) Segment formula. $Q - P = (-6, 8)$ , so $r(t) = (2 - 6t,\ -3 + 8t)$ for $0 \le t \le 1$ . At $t = 0$ this is $P$ , at $t = 1$ it is $Q$ , and in between it moves in a straight line.
(b) Circle in a plane. $r(t) = (5\cos t,\ 5\sin t,\ 4)$ for $0 \le t \le 2\pi$ . The first two coordinates are the circle formula with $R = 5$ , and the third is fixed at the height $4$ of the plane.
(c) Segment in space. $Q - P = (2, -2, 7)$ , so $r(t) = (1 + 2t,\ -2t,\ 5 + 7t)$ for $0 \le t \le 1$ . The same formula, with three coordinates.
(d) Ellipse. $r(t) = (3\cos t,\ 4\sin t)$ for $0 \le t \le 2\pi$ . Check: $\frac{9\cos^2 t}{9} + \frac{16\sin^2 t}{16} = \cos^2 t + \sin^2 t = 1$ .

Answer. As in steps 1 to 4. Any speed would do; these are the simplest.

Worked example MATH 376 worksheet 4 problem 2 (problem 1 is the same with a segment)

Compute

\int_C F \cdot dr

for

F = (y, x^2)

along

y = x^2

from

(0, 0)

(2, 4)

Solution.

Parametrize. The curve is a graph, so $r(t) = (t, t^2)$ for $0 \le t \le 2$ . The direction $(0, 0) \to (2, 4)$ means $t$ increases from $0$ to $2$ .
Differentiate. $r'(t) = (1, 2t)$ .
Substitute. $F(r(t)) = (t^2, t^2)$ . The first component of $F$ is $y = t^2$ and the second is $x^2 = t^2$ .
Dot. $F(r(t)) \cdot r'(t) = t^2 \cdot 1 + t^2 \cdot 2t = t^2 + 2t^3$ . Multiply matching components and add.
Integrate. $\displaystyle\int_0^2 (t^2 + 2t^3)\,dt = \Big[\frac{t^3}{3} + \frac{t^4}{2}\Big]_0^2 = \frac83 + 8 = \frac{32}{3}$ . Antiderivatives of $t^2$ and $2t^3$ , evaluated at $2$ and $0$ .

Answer. $\dfrac{32}{3}$ . (Worksheet 4 problem 1, $F = (x^2, -xy)$ along the segment from $(0, 0)$ to $(1, 2)$ , uses $r(t) = (t, 2t)$ and gives $\int_0^1 (t^2 - 4t^2)\,dt = -1$ .)

Worked example MATH 376 worksheet 4 problem 3

Let

F = (y, -x)

. Compute

\int F \cdot dr

from

(1, 0)

(0, 1)

along (i) the straight segment and (ii) the quarter of the unit circle. What does the result say?

Solution.

(i) Parametrize the segment. $r_1(t) = (1 - t, t)$ for $0 \le t \le 1$ , with $r_1' = (-1, 1)$ . Segment formula with $P = (1, 0)$ and $Q - P = (-1, 1)$ .
(i) Substitute and dot. $F(r_1(t)) = (t, -(1 - t))$ , so $F \cdot r_1' = -t - (1 - t) = -1$ . The terms $-t$ and $+t$ cancel.
(i) Integrate. $\int_0^1 (-1)\,dt = -1$ .
(ii) Parametrize the arc. $r_2(t) = (\cos t, \sin t)$ for $0 \le t \le \pi/2$ , with $r_2' = (-\sin t, \cos t)$ . It starts at $(1, 0)$ when $t = 0$ and ends at $(0, 1)$ when $t = \pi/2$ , going counterclockwise.
(ii) Substitute and dot. $F(r_2(t)) = (\sin t, -\cos t)$ , so $F \cdot r_2' = -\sin^2 t - \cos^2 t = -1$ .
(ii) Integrate. $\int_0^{\pi/2} (-1)\,dt = -\dfrac\pi2$ .

Answer. $-1$ and $-\frac\pi2$ . Two paths with the same start and the same end give different answers, so the integral of this field depends on the path. By 6.2, such a field cannot be a gradient field. This is the red-arrow picture at the top of the chapter.

Worked example MATH 376 worksheet 10 problem 1

Compute

\int_C F \cdot dr

for

F = (y, -x, z)

along the helix

r(t) = (\cos t, \sin t, t)

0 \le t \le 2\pi

Solution.

Differentiate. $r'(t) = (-\sin t, \cos t, 1)$ .
Substitute. $F(r(t)) = (\sin t, -\cos t, t)$ . Put $x = \cos t$ , $y = \sin t$ , $z = t$ into $F = (y, -x, z)$ .
Dot. $F \cdot r' = -\sin^2 t - \cos^2 t + t = t - 1$ . $\sin^2 + \cos^2 = 1$ .
Integrate. $\displaystyle\int_0^{2\pi} (t - 1)\,dt = \Big[\frac{t^2}{2} - t\Big]_0^{2\pi} = 2\pi^2 - 2\pi$ .

Answer. $2\pi^2 - 2\pi$ . The method is the same in three variables: only the number of components grows.

Worked example MATH 376 homework 2 problem 1

Compute the line integral of

f(x, y) = (x^2 - 2xy,\ y^2 - 2x)

along the parabola

y = x^2

from

(-1, 1)

(1, 1)

Solution.

Parametrize. $r(t) = (t, t^2)$ for $-1 \le t \le 1$ , with $r'(t) = (1, 2t)$ . At $t = -1$ the point is $(-1, 1)$ and at $t = 1$ it is $(1, 1)$ , matching the direction.
Substitute. $f(r(t)) = (t^2 - 2t^3,\ t^4 - 2t)$ .
Dot. $f \cdot r' = (t^2 - 2t^3) + (t^4 - 2t)(2t) = 2t^5 - 2t^3 - 3t^2$ . Expanding $(t^4 - 2t)(2t) = 2t^5 - 4t^2$ and combining $t^2 - 4t^2 = -3t^2$ .
Use symmetry. $2t^5 - 2t^3$ is an odd function of $t$ , so its integral over $[-1, 1]$ is $0$ . The part over $[-1, 0]$ cancels the part over $[0, 1]$ .
Integrate the rest. $\int_{-1}^1 (-3t^2)\,dt = [-t^3]_{-1}^1 = -1 - 1 = -2$ .

Answer. $-2$ . Worksheet 27 problem 1 ( $F = (x^2y, xy^2)$ along $(t, t^2)$ , $0 \le t \le 1$ , integrand $t^4 + 2t^6$ ) gives $\frac15 + \frac27 = \frac{17}{35}$ , and homework 2 problem 2 ( $f = (y^2 - z^2, 2yz, x^2)$ along $(t, t^2, t^3)$ , integrand $4t^4 + 3t^6$ ) gives $\frac{43}{35}$ .

Worked example MATH 376 Midterm 1 problem 3; see [A.2](/vol2/a2-midterm-1/). The sibling on the practice exam (a segment from $(0, 0)$ to $(1, 1)$ with $u = (\frac23, y^2)$) gives $1$.

The curve

C

\alpha(t) = (t, t^2)

for

t \in [0, 1]

. Compute the line integral of

u(x, y) = (0, y)

along

C

Solution.

Differentiate. $\alpha'(t) = (1, 2t)$ .
Substitute. $u(\alpha(t)) = (0, t^2)$ . Put $y = t^2$ into $u = (0, y)$ . The first component is the constant $0$ .
Dot. $u(\alpha(t)) \cdot \alpha'(t) = 0 \cdot 1 + t^2 \cdot 2t = 2t^3$ .
Integrate. $\int_0^1 2t^3\,dt = \big[\frac{t^4}{2}\big]_0^1 = \frac12$ .

Answer. $\dfrac12$ . A check that needs no integral, previewed from 6.2: $u = \nabla(y^2/2)$ , so the answer is the change in $\frac{y^2}{2}$ from $(0, 0)$ to $(1, 1)$ , namely $\frac12 - 0$ .

Your answerCompute

\int_C F \cdot dr

for

F = (xy,\ x + y)

along

r(t) = (t, t^2)

0 \le t \le 1

Worked example MATH 376 worksheet 4 problem 4

A force

F

is perpendicular to the velocity

r'(t)

at every point of the path. What is the work? Test it with

F = (-y, x)

along the ray

r(t) = (t, t)

0 \le t \le 3

Solution.

State the condition. $F(r(t)) \cdot r'(t) = 0$ for every $t$ . Two vectors are perpendicular exactly when their dot product is $0$ (2.9).
Integrate. $\int_C F \cdot dr = \int_a^b 0\,dt = 0$ . The integrand is the zero function.
Test. $F(r(t)) = (-t, t)$ and $r' = (1, 1)$ , so $F \cdot r' = -t + t = 0$ , and the integral over $[0, 3]$ is $0$ .

Answer. The work is $0$ . Examples in physics: the magnetic force on a moving charge ( $qv \times B$ is perpendicular to $v$ ), and the tension in a pendulum string.

Quick checkA satellite moves in a circle at constant speed, pulled by a force that always points at the centre. What is the work done by that force in one lap?

Linearity and additivity

Proposition T MATH 376 homework 2 problem 3; Apostol 10.4

Line integrals are linear and additive

Let $F, G$ be continuous on the curve, and $\lambda, \mu$ real numbers. Then the first equation holds. If $C_1$ is traced by $r$ on $[a, b]$ and $C_2$ by $r$ on $[b, c]$ , so that the end of $C_1$ is the start of $C_2$ , and $C$ is traced by $r$ on $[a, c]$ , then the second equation holds.

In one line $\int_C (\lambda F + \mu G) \cdot dr = \lambda\int_C F \cdot dr + \mu\int_C G \cdot dr$ , and if $C$ is $C_1$ followed by $C_2$ then $\int_C F \cdot dr = \int_{C_1} F \cdot dr + \int_{C_2} F \cdot dr$ .

Why it is trueIt lets you split a hard field into easy fields, and a hard path into easy pieces, which is how almost every computation in the rest of this unit is done.

ProofMATH 376 homework 2 problem 3

The idea: Each statement is the corresponding fact about ordinary integrals, once the definition is unwrapped: the dot product is linear in its first factor, and an integral over [a, c] splits at b.

\int_C (\lambda F + \mu G) \cdot dr = \int_a^b \big(\lambda F(r(t)) + \mu G(r(t))\big) \cdot r'(t)\,dt

Definition of the line integral.

= \int_a^b \big(\lambda\,F(r(t)) \cdot r'(t) + \mu\,G(r(t)) \cdot r'(t)\big)\,dt

The dot product is linear in its first factor.

= \lambda\int_a^b F(r(t)) \cdot r'(t)\,dt + \mu\int_a^b G(r(t)) \cdot r'(t)\,dt = \lambda\int_C F \cdot dr + \mu\int_C G \cdot dr

An ordinary integral is linear: constants come out and a sum splits (1.5).

For additivity,

\int_C F \cdot dr = \int_a^c F(r(t)) \cdot r'(t)\,dt

Definition on

[a, c]

= \int_a^b F(r(t)) \cdot r'(t)\,dt + \int_b^c F(r(t)) \cdot r'(t)\,dt

An ordinary integral over

[a, c]

is the sum of the integrals over

[a, b]

and

[b, c]

(additivity over intervals in 1.5).

= \int_{C_1} F \cdot dr + \int_{C_2} F \cdot dr

\blacksquare

Definition on each piece.

Speed does not matter, direction does

A curve can be traced by many paths. Two things can change: how fast you move along it, and which way you go.

Definition T

Reparametrization

Let $\varphi: [c, d] \to [a, b]$ be continuously differentiable. If $\varphi' > 0$ everywhere, with $\varphi(c) = a$ and $\varphi(d) = b$ , then $\rho(s) = r(\varphi(s))$ traces the same curve from the same start to the same end, at a different speed: a reparametrization that keeps the orientation. If instead $\varphi' < 0$ everywhere with $\varphi(c) = b$ and $\varphi(d) = a$ , then $\rho$ traces the curve backwards: it reverses the orientation. A curve together with a chosen direction is an oriented curve.

In one lineIf $r: [a, b] \to \mathbb R^n$ is a path and $\varphi: [c, d] \to [a, b]$ is continuously differentiable with $\varphi(c) = a$ and $\varphi(d) = b$ (and $\varphi' > 0$ ), then $\rho = r \circ \varphi$ is a reparametrization of $r$ .

Why it existsIt changes the speed at which the curve is traced without changing the curve, the starting point or the end point.

Theorem T MATH 376 lecture notes, week 2; Apostol 10.4

Independence of parametrization

Let $r$ be a piecewise smooth path, and $\rho = r \circ \varphi$ with $\varphi$ continuously differentiable and monotone as above. Then

\int_\rho F \cdot d\rho = \int_r F \cdot dr \ \text{ if } \varphi' > 0, \qquad \int_\rho F \cdot d\rho = -\int_r F \cdot dr \ \text{ if } \varphi' < 0.

In one lineReparametrizing a path with $\varphi' > 0$ does not change the line integral, and reversing the direction ( $\varphi' < 0$ ) changes its sign.

Why it is trueIt is what lets us write $\int_C F \cdot dr$ for a curve with a direction, without naming a particular path, and it shows that only the curve and the direction matter.

ProofMATH 376 lecture notes, week 2

The idea: Differentiate the composite with the chain rule: the velocity gets multiplied by the scalar phi-prime. Then the substitution rule for ordinary integrals absorbs that factor into the limits, and a swap of limits is the sign.

\rho'(s) = r'(\varphi(s))\,\varphi'(s)

Chain rule for

r \circ \varphi

, applied to each coordinate. We do one smooth piece of

r

. For a path with corners, the corners of

\rho

are the points

s

with

\varphi(s)

a corner of

r

, finitely many because

\varphi

is monotone, and we add the pieces.

\int_\rho F \cdot d\rho = \int_c^d F(r(\varphi(s))) \cdot r'(\varphi(s))\,\varphi'(s)\,ds

Definition of the line integral. The number

\varphi'(s)

is a scalar, so it sits beside the dot product.

= \int_{\varphi(c)}^{\varphi(d)} F(r(u)) \cdot r'(u)\,du

Substitution rule (1.5) with

u = \varphi(s)

and

du = \varphi'(s)\,ds

, valid because

\varphi

is continuously differentiable.This is the step where 'continuously differentiable' for

\varphi

is used.

In that case this is

\int_a^b F(r(u)) \cdot r'(u)\,du = \int_r F \cdot dr

\varphi' > 0

, the limits are

\varphi(c) = a

and

\varphi(d) = b

, the limits of the original definition.

In that case it is

\int_b^a F(r(u)) \cdot r'(u)\,du = -\int_a^b F(r(u)) \cdot r'(u)\,du = -\int_r F \cdot dr

\blacksquare

\varphi' < 0

, the limits are

\varphi(c) = b

and

\varphi(d) = a

, and swapping the limits of an ordinary integral changes its sign.

Proof workshopGive the reason for each line: independence of parametrizationMATH 376 week 2

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\rho'(s) = r'(\varphi(s))\,\varphi'(s)$ .
$\int_\rho F \cdot d\rho = \int_c^d F(r(\varphi(s))) \cdot r'(\varphi(s))\,\varphi'(s)\,ds$ .
$= \int_{\varphi(c)}^{\varphi(d)} F(r(u)) \cdot r'(u)\,du$ .
If $\varphi(c) = a$ and $\varphi(d) = b$ this is $\int_r F \cdot dr$ .
If $\varphi(c) = b$ and $\varphi(d) = a$ it is $-\int_r F \cdot dr$ .

What fails without each hypothesis. If $\varphi$ is not monotone, the new path goes forward and then back, and the answer changes. Take $r(t) = (t, 0)$ on $[0, 1]$ , the field $F = (1, 0)$ , and $\varphi(s) = \sin^2(\pi s/2)$ on $[0, 2]$ . This $\varphi$ is continuously differentiable and goes from $0$ up to $1$ and back to $0$ . Then $\int_r F \cdot dr = \int_0^1 1\,dt = 1$ but $\int_\rho F \cdot d\rho = \int_0^2 \varphi'(s)\,ds = \varphi(2) - \varphi(0) = 0$ . The curve is the same, the integrals differ, because the segment is covered once forward and once backward. And if $\varphi$ is not continuously differentiable, the chain rule and the substitution rule have nothing to work with.

Quick checkYou parametrize a curve from

A

B

and get

\int = 5

. A friend parametrizes the same curve from

B

A

, running twice as fast. What does the friend get?

Quick checkFor

r(t) = (t, 0)

[0, 1]

and

F = (1, 0)

, the path

\rho(s) = r(\sin^2(\pi s / 2))

for

0 \le s \le 2

gives a different integral from

r

. Why does the independence theorem not apply?

Worked example new (reversal, MATH 376 week 2)

Reverse a path. If the line integral of

F

along

C

from

(1, 0)

(0, 1)

-1

(the segment of the earlier example), what is the integral along the same segment from

(0, 1)

(1, 0)

? Check it.

Solution.

Use the theorem. Going the other way is a reparametrization with $\varphi(s) = a + b - s$ , so $\varphi' = -1 < 0$ . The new path is $r(a + b - s)$ , which starts at $r(b)$ and ends at $r(a)$ .
Conclude. The integral is $-(-1) = +1$ .
Check directly. Take $\rho(t) = (t, 1 - t)$ for $0 \le t \le 1$ , so $\rho' = (1, -1)$ and $F(\rho(t)) = (1 - t, -t)$ for $F = (y, -x)$ . Then $F \cdot \rho' = (1 - t) + t = 1$ , and $\int_0^1 1\,dt = 1$ . Same field, same segment, opposite direction.

Answer. $+1$ .

Closed paths and circulation

Definition T

Closed path

A path is closed if $r(a) = r(b)$ . A closed path is simple if it does not cross or touch itself, apart from the start-end point. For a closed curve the line integral is written $\oint_C F \cdot dr$ .

In one lineA path $r: [a, b] \to \mathbb R^n$ with $r(a) = r(b)$ : it ends where it starts.

Why it existsIntegrals around closed paths are the test for gradient fields in 6.2 and the left side of Green's theorem in 6.4.

Definition T

Circulation

The circulation of $F$ around the closed curve $C$ is $\oint_C F \cdot dr$ . For a velocity field it measures how much the fluid swirls around the loop. For a force field, circulation and work in one lap are the same number.

In one lineThe circulation of a field $F$ around a closed curve $C$ is $\oint_C F \cdot dr$ .

Why it existsFor a velocity field of a fluid it measures how much the fluid goes around the loop; for a force field it is the work done in one lap.

Worked example MATH 376 worksheet 27 problem 3 (done there with $R = 1$, directly and by Green)

Find the circulation of

F = (-y, x)

counterclockwise around the circle of radius

R

centred at the origin.

Solution.

Parametrize. $r(t) = (R\cos t, R\sin t)$ for $0 \le t \le 2\pi$ , with $r' = (-R\sin t, R\cos t)$ . The circle formula, counterclockwise.
Substitute. $F(r(t)) = (-R\sin t, R\cos t)$ . $F$ is the same vector as $r'$ here: the field is the velocity of a rotation.
Dot. $F \cdot r' = R^2\sin^2 t + R^2\cos^2 t = R^2$ .
Integrate. $\oint F \cdot dr = \int_0^{2\pi} R^2\,dt = 2\pi R^2$ .

Answer. $2\pi R^2$ , which is twice the area $\pi R^2$ of the disc. 6.4 explains why.

Your answerWhat is the circulation of

F = (-y, x)

counterclockwise around the circle of radius

3

centred at the origin?

Line integrals with respect to arc length

Beyond the course: for understanding· Apostol 10.7 to 10.9 (label A): the course did not teach this

The dot product $F \cdot r'$ pairs a vector field with the direction of motion. Sometimes the thing being added up along a curve is a scalar: the density of a wire. The line integral of a scalar field is a different integral, and it does not care about direction.

Definition A

Line integral with respect to arc length

The arc length function of a path $r$ with continuous $r'$ is $s(t) = \int_a^t \lVert r'(u) \rVert\,du$ , so $s'(t) = \lVert r'(t) \rVert$ , the speed. For a scalar field $\varphi$ defined and bounded on $C$ ,

\int_C \varphi\,ds = \int_a^b \varphi(r(t))\,s'(t)\,dt = \int_a^b \varphi(r(t))\,\lVert r'(t) \rVert\,dt.

With $\varphi = 1$ this is the length of the curve.

In one lineFor a scalar field $\varphi$ and a path $r$ with continuous $r'$ , $\int_C \varphi\,ds = \int_a^b \varphi(r(t))\,\lVert r'(t) \rVert\,dt$ .

Why it existsThe length of the tiny piece of curve traced in time $\Delta t$ is about the speed $\lVert r'(t) \rVert$ times $\Delta t$ , so this adds up $\varphi$ times length.

This integral is unchanged if you reverse the direction (the speed is the same in both directions) and unchanged by any monotone reparametrization, by the same proof as above. The two kinds of line integral are related. With the unit tangent $T = r'/\lVert r' \rVert$ , we have $F \cdot r' = (F \cdot T)\,\lVert r' \rVert$ , so

\int_C F \cdot dr = \int_C (F \cdot T)\,ds.

The vector integral adds up the tangential component of $F$ . When $F$ is a velocity field this is the flow integral, and around a closed curve it is the circulation.

Worked example Apostol 10.7 and 10.9 problem 7

Find the length of one turn of the helix

r(t) = (\cos t, \sin t, t)

0 \le t \le 2\pi

, and the integral

\int_T (x + y)\,ds

over the triangle with vertices

(0, 0)

(1, 0)

(0, 1)

, traversed counterclockwise.

Solution.

Helix, speed. $r' = (-\sin t, \cos t, 1)$ , so $\lVert r' \rVert = \sqrt2$ . $\sin^2 + \cos^2 + 1 = 2$ .
Helix, length. $\int_0^{2\pi} \sqrt2\,dt = 2\pi\sqrt2$ . The integral of $1\,ds$ is the length.
Triangle, bottom. $r = (t, 0)$ , $0 \le t \le 1$ , speed $1$ : $\int_0^1 t\,dt = \frac12$ . $x + y = t$ .
Triangle, hypotenuse. $r = (1 - t, t)$ , $0 \le t \le 1$ , speed $\sqrt2$ , and $x + y = 1$ : $\int_0^1 1 \cdot \sqrt2\,dt = \sqrt2$ . The speed is $\lVert(-1, 1)\rVert = \sqrt 2$ .
Triangle, left side. $r = (0, 1 - t)$ , speed $1$ , $x + y = 1 - t$ : $\int_0^1 (1 - t)\,dt = \frac12$ . Run it from $(0, 1)$ down to $(0, 0)$ as the counterclockwise order requires. The direction does not change an arc length integral.
Add the three sides. $\frac12 + \sqrt2 + \frac12 = 1 + \sqrt2$ . Additivity over pieces.

Answer. The length is $2\pi\sqrt2$ and the integral over the triangle is $1 + \sqrt2$ .

Your answerWhat is the length of one turn of the helix

r(t) = (\cos t, \sin t, t)

0 \le t \le 2\pi

Mass, balance point and moment of inertia

Think of a curve $C$ in space as a thin wire, and let $\varphi(x, y, z)$ be its density in mass per unit length. A wire of constant density is uniform.

Definition A

Centre of mass

The mass of the wire is $M = \int_C \varphi\,ds$ . The centre of mass is the point $(\bar x, \bar y, \bar z)$ determined by

\bar x\,M = \int_C x\,\varphi(x, y, z)\,ds, \qquad \bar y\,M = \int_C y\,\varphi\,ds, \qquad \bar z\,M = \int_C z\,\varphi\,ds.

For a uniform wire the centre of mass is also called the centroid.

In one lineFor a wire with density $\varphi$ , the mass is $M = \int_C \varphi\,ds$ , and the centre of mass is the point $(\bar x, \bar y, \bar z)$ with $\bar x\,M = \int_C x\,\varphi\,ds$ , and likewise for $y$ and $z$ .

Why it existsIt is the average position weighted by mass: the single point where all the mass could be placed without changing the first moments, and the balance point.

Definition A

Moment of inertia

If $\delta(x, y, z)$ is the perpendicular distance from the point $(x, y, z)$ of the wire to the axis $L$ , the moment of inertia about $L$ is $I_L = \int_C \delta^2\,\varphi\,ds$ . For the coordinate axes, $I_z = \int_C (x^2 + y^2)\,\varphi\,ds$ , and so on. Chapter 6.3 uses the same idea for flat plates and solids.

In one lineThe moment of inertia of a wire about an axis $L$ is $I_L = \int_C \delta^2\,\varphi\,ds$ , where $\delta$ is the distance from a point of the wire to $L$ .

Why it existsIt measures how hard the wire is to spin about $L$ : mass far from the axis counts with the square of its distance.

Worked example Apostol 10.9 problem 11

A uniform semicircular wire of radius

a

lies in the upper half plane. Find its centroid and its moment of inertia about the diameter through its endpoints.

Solution.

Parametrize. $r(t) = (a\cos t, a\sin t)$ for $0 \le t \le \pi$ , with $\lVert r' \rVert = a$ . Let the constant density be $\varphi$ . Half of the circle formula; the speed of a circle of radius $a$ at unit angular speed is $a$ .
Mass. $M = \int_0^\pi \varphi\,a\,dt = \varphi\pi a$ .
Symmetry. By symmetry in the $y$ -axis, $\bar x = 0$ . The reflection $x \mapsto -x$ maps the wire to itself.
The height. $\bar y\,M = \int_0^\pi a\sin t\cdot\varphi\,a\,dt = 2\varphi a^2$ , so $\bar y = \dfrac{2\varphi a^2}{\varphi\pi a} = \dfrac{2a}{\pi}$ . $\int_0^\pi \sin t\,dt = 2$ .
Moment of inertia. About the $x$ -axis (the diameter through the endpoints), $\delta = y = a\sin t$ , so $I = \int_0^\pi a^2\sin^2 t\cdot\varphi\,a\,dt = \varphi a^3\cdot\dfrac\pi2$ . $\int_0^\pi \sin^2 t\,dt = \frac\pi2$ .
Compare with the mass. $I = \frac12 (\varphi\pi a)\,a^2 = \frac12 M a^2$ .

Answer. The centroid is $(0, 2a/\pi)$ , about $0.64a$ above the centre. The moment of inertia is $\frac12 Ma^2$ .

Worked example Apostol 10.8, examples 1 and 2

Find the mass of one coil of the spring

r(t) = (a\cos t, a\sin t, bt)

0 \le t \le 2\pi

, if the density at

(x, y, z)

x^2 + y^2 + z^2

, and its moment of inertia about the

z

-axis.

Solution.

Speed. $r' = (-a\sin t, a\cos t, b)$ , so $\lVert r' \rVert = \sqrt{a^2 + b^2}$ , a constant.
Density along the wire. $x^2 + y^2 + z^2 = a^2 + b^2t^2$ . $a^2\cos^2 t + a^2\sin^2 t = a^2$ .
Mass. $M = \sqrt{a^2 + b^2}\int_0^{2\pi}(a^2 + b^2t^2)\,dt = \sqrt{a^2 + b^2}\,\Big(2\pi a^2 + \dfrac{8\pi^3b^2}{3}\Big)$ . $\int_0^{2\pi} t^2\,dt = \frac{(2\pi)^3}{3}$ .
Distance to the axis. For the $z$ -axis, $\delta^2 = x^2 + y^2 = a^2$ along the whole wire. The wire stays at distance $a$ from the axis.
Moment of inertia. $I_z = \int_C a^2\,\varphi\,ds = a^2M$ .

Answer. $M = \sqrt{a^2 + b^2}\,\big(2\pi a^2 + \frac83\pi^3b^2\big)$ and $I_z = a^2M$ .

Your answerA semicircular wire of radius

0.1

m has density

0.2(1 + \sin\theta)

kg per metre at the point at angle

\theta

, for

0 \le \theta \le \pi

. What is its mass in kilograms?

Python labPath dependence, reparametrization and going back and forth, numerically

import numpy as np
c, s = np.cos, np.sin

def work(F, path, n):
    r, dr, a, b = path
    t = np.linspace(a, b, n + 1); m = (t[:-1] + t[1:]) / 2     # midpoints
    (x, y), (u, v) = r(m), dr(m); P, Q = F(x, y)
    return np.sum((P * u + Q * v) * np.diff(t))               # sum of F . r' dt

def circle(th, dth, a, b):                                      # angle th(t) on the unit circle
    return (lambda t: (c(th(t)), s(th(t))), lambda t: (-s(th(t)) * dth(t), c(th(t)) * dth(t)), a, b)

paths = {
  "segment":          (lambda t: (1 - t, t), lambda t: (-1 + 0 * t, 1 + 0 * t), 0, 1),
  "quarter circle":   circle(lambda t: np.pi / 2 * t, lambda t: np.pi / 2 + 0 * t, 0, 1),
  "circle, speeding": circle(lambda t: np.pi / 2 * t**2, lambda t: np.pi * t, 0, 1),
  "circle, out+back": circle(lambda t: np.pi / 2 * s(np.pi * t / 2)**2,
                             lambda t: np.pi**2 / 4 * s(np.pi * t / 2) * c(np.pi * t / 2), 0, 2),
}
F = lambda x, y: (y, -x)                                        # try (y, x) next
for name, p in paths.items():
    print(f"{name:18s}", "  ".join(f"{work(F, p, n):9.5f}" for n in (10, 100, 1000)))

Try this. Run it. The columns are the Riemann sums with 10, 100 and 1000 steps. For F = (y, -x) the segment gives -1 and the quarter circle gives -1.5708 (that is -pi/2): same endpoints, different answers. The circle traced at a changing speed (angle = (pi/2) t^2) still gives -1.5708, because only the curve and the direction matter. The out-and-back path (the angle goes 0 to pi/2 and then back to 0) gives 0, as the independence theorem warns for a map that is not monotone. Now change the field to (y, x): every path gives 0, because that field is the gradient of x*y and the endpoints (1, 0) and (0, 1) have the same value of x*y. Try your own field, for example (x, x*y), and your own path.

Where it is used

Used inside maths· Inside maths: the integral behind Green, Stokes and complex analysis

Everything in the rest of this unit is a statement about line integrals. The question of 6.2 is when $\int_C F \cdot dr$ depends only on the endpoints. Green’s theorem (6.4) turns $\oint_C F \cdot dr$ into an integral over the region inside. In complex analysis, a contour integral $\int_C f(z)\,dz$ with $f = u + iv$ and $dz = dx + i\,dy$ equals $\int_C (u\,dx - v\,dy) + i\int_C (v\,dx + u\,dy)$ , two real line integrals; Cauchy’s theorem says both vanish around a closed curve when $f$ is analytic, and the proof is Green’s theorem. In analysis, the bound $\big|\int_C F \cdot dr\big| \le \max_C\lVert F \rVert \cdot \text{length}(C)$ (a problem below) is the ‘ML inequality’ used to show that integrals over big circles go to zero.

Used in the real world· Used in the real world: work against a wind, and the mass of a curved wire

A drone circles a mast along the circle of radius $R = 100$ m at constant height. A rotating wind pushes with force $F = (-0.02y,\ 0.02x)$ newtons on the drone at position $(x, y)$ metres, so at the drone’s position the push has size $0.02R = 2$ N and points around the circle. With $r(t) = (100\cos t, 100\sin t)$ the dot product $F \cdot r'$ is $0.02 \cdot 100^2 = 200$ N·m per unit of $t$ , constant, so one counterclockwise lap gives $\oint F \cdot dr = 200 \cdot 2\pi = 400\pi \approx 1257$ joules. The wind does $1257$ J of work on the drone in one lap, even though the drone ends where it started: this force is not a gradient field (6.2). Flying clockwise the wind does $-1257$ J and the drone’s motors have to supply the difference.

A bent copper rod in the shape of a semicircle of radius $0.1$ m has a density that varies along it, because it was cast with more alloy at one end. If the density is $0.2(1 + \sin\theta)$ kg/m at angle $\theta$ , the mass is $0.02(\pi + 2) = 0.103$ kg (the check above), and the balance point is found with the same integrals weighted by $x$ and $y$ . A designer who needs the rod to balance on a pivot must compute $\bar x$ and $\bar y$ from exactly the formulas of this chapter; for a uniform rod the answer $2a/\pi = 6.4$ cm above the centre is the one in the example.

Your answerA drone flies one counterclockwise lap of the circle of radius

100

m in the wind force

F = (-0.02y, 0.02x)

newtons. How many joules of work does the wind do on it? Give the exact answer.

Traps

Watch out· The definition of a piecewise smooth path has three parts

On the 376 final, a student wrote, in problem 1(b), that a piecewise smooth curve ‘can be broken down into discrete continuous and once-differentiable chunks’, with ‘any amount of’ written above ‘curve’. The rubric item applied was ‘Correct, aside from a conceptual mistake’ (1 of 2 points), and their regrade request was declined with the reply that each piece ‘should be more than differentiable; its derivative should also be continuous.’ Three things were missing: continuity of the whole path (two continuous chunks may not meet), a finite partition (‘any amount of’ is the wrong word), and a continuous derivative on each piece. Write all three, in order, as in the worked example above.

Watch out· Substitute the path into the field, then dot with the velocity

The five steps of the method are in a fixed order. A common slip is to dot $F(x, y)$ with $r'(t)$ before replacing $x$ and $y$ by functions of $t$ , which leaves a mix of variables. Another is to leave the answer containing $t$ : the value of a line integral is a number. If $t$ appears in your boxed answer, something was not integrated.

Watch out· Direction matters, speed does not

The words ‘from $(0, 0)$ to $(2, 4)$ ’ or ‘counterclockwise’ fix the direction, and a wrong direction flips the sign. Check that your $r$ starts and ends at the right points. On the other hand, you cannot change the answer by walking faster; if you got a different answer from two parametrizations of the same oriented curve, one of them has an error, or one of them is not monotone.

Watch out· Line integral of a vector field and of a scalar field are different integrals

$\int_C F \cdot dr$ (a vector field, a dot product, sign depends on direction) is not $\int_C \varphi\,ds$ (a scalar field, a factor $\lVert r' \rVert$ , no sign). Do not put $\lVert r' \rVert$ into a work integral, and do not drop it from a mass integral.

Problems

checknew (MATH 376 Final 1(b) style)

Is $r(t) = (t^3, |t|^3)$ on $[-1, 1]$ a smooth path (continuous velocity everywhere)? Its image is the corner $y = |x|$ for $-1 \le x \le 1$ .

Solution

Yes. Both coordinates have derivatives $3t^2$ and $3t|t|$ (for $t \ge 0$ this is $3t^2$ , for $t \le 0$ it is $-3t^2$ , and at $0$ the one-sided derivatives both equal $0$ ). So $r'(t) = (3t^2, 3t|t|)$ is continuous on $[-1, 1]$ , and $r'(0) = (0, 0)$ . The image has a corner, but the path is smooth, because the walker stops for an instant at the corner. Smoothness is a property of the path, not of the curve.

checknew (MATH 376 worksheet 3 problem 6 style)

Parametrize (a) the segment from $(1, 2)$ to $(4, -2)$ ; (b) the circle of radius $3$ centred at $(1, -1)$ , counterclockwise.

Solution

(a) $Q - P = (3, -4)$ , so $r(t) = (1 + 3t,\ 2 - 4t)$ , $0 \le t \le 1$ . (b) $r(t) = (1 + 3\cos t,\ -1 + 3\sin t)$ , $0 \le t \le 2\pi$ : the centre plus the circle of radius $3$ about the origin.

check

A line integral along a curve from $A$ to $B$ equals $\frac{32}{3}$ . What is the integral along the same curve from $B$ to $A$ , parametrized at half the speed?

Solution

$-\frac{32}{3}$ . Reversing the direction flips the sign; the speed has no effect.

usenew (compare MATH 376 worksheet 5 exercise 4)

Compute $\int F \cdot dr$ for $F = (y^2, x)$ from $(0, 0)$ to $(1, 1)$ along $y = x^2$ .

Solution

$r(t) = (t, t^2)$ , $r' = (1, 2t)$ , $F(r) = (t^4, t)$ , $F \cdot r' = t^4 + 2t^2$ , and $\int_0^1 (t^4 + 2t^2)\,dt = \frac15 + \frac23 = \frac{13}{15}$ . (Along the line $y = x$ the same field gives $\frac56$ ; see 6.2.)

use

Compute $\int F \cdot dr$ for $F = (yz,\ xz,\ xy)$ along $r(t) = (t, t^2, t^3)$ , $0 \le t \le 1$ .

Solution

$F(r(t)) = (t^5, t^4, t^3)$ and $r' = (1, 2t, 3t^2)$ , so $F \cdot r' = t^5 + 2t^5 + 3t^5 = 6t^5$ , and $\int_0^1 6t^5\,dt = 1$ . (This $F$ is the gradient of $xyz$ , and $xyz$ goes from $0$ to $1$ : see 6.2.)

usenew (circulation, compare MATH 376 worksheet 27 problem 3)

Compute the circulation of $F = (-y, x)$ counterclockwise around the boundary of the square $[0, 2]^2$ , edge by edge.

Solution

Bottom ( $y = 0$ , $r = (t, 0)$ , $0 \le t \le 2$ ): $F = (0, t)$ , $r' = (1, 0)$ , so $0$ . Right ( $x = 2$ , $r = (2, t)$ ): $F = (-t, 2)$ , $r' = (0, 1)$ , so $\int_0^2 2\,dt = 4$ . Top, from $(2, 2)$ to $(0, 2)$ : $r = (2 - t, 2)$ , $F = (-2, 2 - t)$ , $r' = (-1, 0)$ , so $\int_0^2 2\,dt = 4$ . Left, from $(0, 2)$ to $(0, 0)$ : $r = (0, 2 - t)$ , $F = (-(2-t), 0)$ , $r' = (0, -1)$ , so $0$ . Total $8$ , twice the area $4$ of the square, as for the circle.

useApostol 10.9 problem 8 style

Find $\int_C x\,ds$ over the quarter of the circle of radius $2$ in the first quadrant.

Solution

$r(t) = (2\cos t, 2\sin t)$ , $0 \le t \le \frac\pi2$ , $\lVert r' \rVert = 2$ , so $\int_0^{\pi/2} 2\cos t \cdot 2\,dt = 4[\sin t]_0^{\pi/2} = 4$ .

proveApostol 10.6 example 1

Let $F$ be the constant field $F = c$ . Show that the work done in moving a particle from $p$ to $q$ along any piecewise smooth path is $c \cdot (q - p)$ .

Solution

Let $r$ run from $p$ to $q$ . On each smooth piece, $\int_{t_{j-1}}^{t_j} c \cdot r'(t)\,dt = \sum_k c_k\int_{t_{j-1}}^{t_j} r_k'(t)\,dt = \sum_k c_k\,[r_k(t_j) - r_k(t_{j-1})] = c \cdot [r(t_j) - r(t_{j-1})]$ , by the fundamental theorem of calculus in each coordinate. Adding over the pieces, the sum telescopes, because $r$ is continuous at the break points, to $c \cdot [r(b) - r(a)] = c \cdot (q - p)$ .

provenew (the ML inequality)

Let $r$ be a path with continuous $r'$ on $[a, b]$ , of length $\ell = \int_a^b \lVert r'(t) \rVert\,dt$ , and let $F$ be continuous with $\lVert F \rVert \le M$ on the curve. Prove that $\big|\int_C F \cdot dr\big| \le M\ell$ .

Solution

By the Cauchy–Schwarz inequality (2.9), $|F(r(t)) \cdot r'(t)| \le \lVert F(r(t)) \rVert\,\lVert r'(t) \rVert \le M\lVert r'(t) \rVert$ . The absolute value of an integral is at most the integral of the absolute value, so $\big|\int_a^b F \cdot r'\,dt\big| \le \int_a^b M\lVert r'(t) \rVert\,dt = M\ell$ . For a path with corners apply this on each piece and add the lengths.

proveMATH 376 homework 2 problem 3 / lecture, applied to arc length

Prove that $\int_C \varphi\,ds$ is unchanged when the path is reversed: if $\rho(s) = r(a + b - s)$ on $[a, b]$ then $\int_\rho \varphi\,ds = \int_r \varphi\,ds$ .

Solution

$\rho'(s) = -r'(a + b - s)$ , so $\lVert \rho'(s) \rVert = \lVert r'(a + b - s) \rVert$ . Then $\int_a^b \varphi(r(a + b - s))\,\lVert r'(a + b - s) \rVert\,ds$ ; substitute $u = a + b - s$ , $du = -ds$ , and the limits $s = a, b$ become $u = b, a$ : the integral becomes $-\int_b^a \varphi(r(u))\,\lVert r'(u) \rVert\,du = \int_a^b \varphi(r(u))\,\lVert r'(u) \rVert\,du$ . Here the minus sign from $du = -ds$ cancels the minus sign from swapping the limits, because the factor $\lVert\cdot\rVert$ does not change sign, unlike the velocity in the vector case.

stretchApostol 10.9 problem 8

Compute $\int_C y^2\,ds$ for one arch of the cycloid $r(t) = a(t - \sin t,\ 1 - \cos t)$ , $0 \le t \le 2\pi$ .

Solution

$r' = a(1 - \cos t,\ \sin t)$ , so $\lVert r' \rVert = a\sqrt{2 - 2\cos t} = 2a\sin(t/2)$ for $0 \le t \le 2\pi$ . Also $y^2 = a^2(1 - \cos t)^2 = 4a^2\sin^4(t/2)$ . The integral is $\int_0^{2\pi} 4a^2\sin^4(t/2)\cdot2a\sin(t/2)\,dt = 8a^3\int_0^{2\pi}\sin^5(t/2)\,dt = 16a^3\int_0^\pi \sin^5 u\,du = 16a^3\cdot\frac{16}{15} = \frac{256a^3}{15}$ , using $u = t/2$ and $\int_0^\pi \sin^5 u\,du = \frac{16}{15}$ .

Summary

A path is a continuous map $r: [a, b] \to \mathbb R^n$ . It is piecewise smooth if it is continuous on all of $[a, b]$ , and there is a finite partition with $r'$ existing and continuous on each piece. All three parts are needed (‘differentiable’ is not enough).
The line integral of a vector field is $\int_C F \cdot dr = \int_a^b F(r(t)) \cdot r'(t)\,dt$ : parametrize in the stated direction, differentiate, substitute the path into the field, dot, integrate. It is the work of a force and, around a closed path, the circulation.
It is linear in $F$ and additive over joined paths. It is unchanged by a reparametrization with $\varphi' > 0$ (speed does not matter) and changes sign if $\varphi' < 0$ (direction does). A non-monotone reparametrization can change it.
A force perpendicular to the motion does no work. Two paths with the same endpoints can give different integrals ( $F = (y, -x)$ gives $-1$ and $-\pi/2$ ), which is the starting point of 6.2.
The line integral of a scalar field with respect to arc length is $\int_C \varphi\,ds = \int_a^b \varphi(r(t))\,\lVert r'(t) \rVert\,dt$ ; it ignores direction, gives lengths, masses ( $\varphi$ = density), centres of mass and moments of inertia of wires (label A).
$\int_C F \cdot dr = \int_C (F \cdot T)\,ds$ , and $\big|\int_C F \cdot dr\big| \le (\max\lVert F \rVert) \cdot \text{length}$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

The picture: work accumulating along a path

The idea, in words

Paths, curves and corners

Piecewise smooth path

The line integral

Line integral

Work

The method

Linearity and additivity

Line integrals are linear and additive

Speed does not matter, direction does

Reparametrization

Independence of parametrization

Closed paths and circulation

Closed path

Circulation

Line integrals with respect to arc length

Line integral with respect to arc length

Mass, balance point and moment of inertia

Centre of mass

Moment of inertia

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere