Introduction to Vector Integral Calculus

Work = 0.000

We know how to integrate functions over intervals. But what does it mean to integrate along a curve?

Three questions, one idea

Forget definitions. Here are three problems.

Problem 1. A wire bends into a curve $C$. How long is it?

$$\text{Length} = \int_C \mathrm{d}s$$

Problem 2. The wire has varying density $f(x,y)$ - thicker in some places, thinner in others. What is its total mass?

$$\text{Mass} = \int_C f(x,y)\,\mathrm{d}s$$

Problem 3. A force field $\mathbf{F}$ pushes a particle along the wire. How much work does it do?

$$\text{Work} = \int_C \mathbf{F} \cdot \mathrm{d}\mathbf{r}$$

All three involve the same idea: add up tiny contributions along a curve. That's a line integral. The only difference is what we're adding up.

Arc length - the simplest line integral

How do you measure the length of a curve? The same way you'd measure a winding road with a ruler: chop it into tiny straight segments and add them up.

Each tiny segment is approximately straight. Its length comes from the Pythagorean theorem:

$$\Delta s_k \;\approx\; \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2}$$

Sum them all up:

$$\text{Length} \;\approx\; \sum_{k=1}^{n} \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2}$$

Take $n \to \infty$ and the sum becomes an integral. If the curve is parametrized by $\mathbf{r}(t) = (x(t), y(t))$, each $\Delta x_k \approx x'(t)\,\Delta t$ and $\Delta y_k \approx y'(t)\,\Delta t$, so:

$$\int_C \mathrm{d}s \;=\; \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;\mathrm{d}t \;=\; \int_a^b |\mathbf{r}'(t)|\,\mathrm{d}t$$

We're summing tiny Pythagorean hypotenuses. That's all arc length is.

Use the slider to increase $n$. Watch the chord approximation converge to the true length.

Work - force times displacement

Constant force $\mathbf{F}$ on a straight segment from $A$ to $B$:

$$W \;=\; \underbrace{\mathbf{F} \cdot \Delta\mathbf{r}}_{\text{vector dot} \,\to\, \text{scalar}} \;=\; \underbrace{(|\mathbf{F}|\cos\alpha)\,|\Delta\mathbf{r}|}_{\text{scalar}\,\times\,\text{scalar}}.$$

Force-along-motion (scalar) times distance (scalar).

On a real path, $\mathbf{F}$ varies and the path curves - so neither assumption holds. Fix: zoom in. On a piece short enough,

the piece is straight - replace it with a vector $\Delta\mathbf{r}_k$,
$\mathbf{F}$ is constant - evaluate once at $\mathbf{F}(\mathbf{r}_k)$.

Our old formula works on that one piece:

$$\Delta W_k \;\approx\; \mathbf{F}(\mathbf{r}_k)\cdot \Delta\mathbf{r}_k \qquad \text{(vector dot)}$$

Sum and shrink:

$$W \;\approx\; \sum_{k=1}^{n} \mathbf{F}(\mathbf{r}_k) \cdot \Delta\mathbf{r}_k \;\;\xrightarrow{\;n\to\infty\;}\;\; \int_C \mathbf{F} \cdot \mathrm{d}\mathbf{r}.$$

Trap. $W = \mathbf{F}\cdot(\mathbf{r}_B - \mathbf{r}_A)$ is only valid when $\mathbf{F}$ is constant and $C$ is straight. Otherwise path shape matters: different paths, different work.

Tangential form. The tangent vector satisfies $\Delta\mathbf{r}_k \approx \mathbf{T}_k\,\Delta s_k$ (unit tangent vector times arc-length scalar). Plug in:

$$\underbrace{\mathbf{F}\cdot\Delta\mathbf{r}_k}_{\text{vector dot}} \;\approx\; \underbrace{(\mathbf{F}\cdot\mathbf{T}_k)}_{\text{scalar } F_T} \;\underbrace{\Delta s_k}_{\text{scalar}}.$$

So the work integral can be read two equivalent ways:

$$\underbrace{\int_C \mathbf{F} \cdot \mathrm{d}\mathbf{r}}_{\text{vectors}} \;=\; \underbrace{\int_C F_T\,\mathrm{d}s}_{\text{scalars}}.$$

Only the component of force along motion does work. Perpendicular force: zero (pushing down on a sliding book). Opposing force: negative (braking).

Component form. With $\mathbf{F} = (F_x, F_y)$ and $\mathrm{d}\mathbf{r} = (\mathrm{d}x, \mathrm{d}y)$, the vector dot product expands to a sum of scalar products:

$$\int_C \mathbf{F} \cdot \mathrm{d}\mathbf{r} \;=\; \int_C \underbrace{F_x\,\mathrm{d}x}_{\text{scalar}} + \underbrace{F_y\,\mathrm{d}y}_{\text{scalar}}.$$

The pieces $\int_C P\,\mathrm{d}x$ and $\int_C Q\,\mathrm{d}y$ aren't new definitions - they fall out of the dot product.

Here's the dot product in action. A constant rightward force $\mathbf{F} = (1, 0)$ acts on a particle as it traces an ellipse counterclockwise. The red arrow is $\mathbf{F}$ - same at every point. The small slate chevron on the path shows which way we're headed at the current position.

On the bottom half, we move rightward - with the wind - so $\mathbf{F}\cdot\mathbf{T} > 0$ and the path is green. On top, we move leftward against the wind and the path is red. At the left and right extremes motion is purely vertical, perpendicular to $\mathbf{F}$, and $\mathbf{F}\cdot\mathbf{T} = 0$.

Aiding motion Opposing motion

We've previewed the work integral here as one of the three types of line integrals (§5.1's job). The dedicated treatment - parametric reduction, the differential-form notation, properties, conservative-field shortcuts, and worked Kaplan problems - lives on the next page: §5.4 Line Integrals as Integrals of Vectors.

The three types

We've arrived at three kinds of line integral:

$$\int_C f(x,y)\,\mathrm{d}s, \qquad \int_C P(x,y)\,\mathrm{d}x, \qquad \int_C Q(x,y)\,\mathrm{d}y$$

The first - $\int_C f\,\mathrm{d}s$ - weights by arc length. It doesn't care which direction you walk along the curve. Reverse the path and the mass of the wire stays the same.

The second and third - $\int_C P\,\mathrm{d}x$ and $\int_C Q\,\mathrm{d}y$ - weight by displacement components. Reverse the path and the sign flips. Walk against the force field, and work becomes negative.

These three types are the atoms. Every line integral in this chapter is built from them.

What's ahead: In §5.2, we set up the precise definitions. In §5.3, we learn to compute these integrals by parametrization. In §5.5, Green's theorem reveals a stunning connection: a line integral around a closed curve equals a double integral over the region inside. Line integrals and area integrals are secretly the same thing.

Practice Problems - §5.1

From Kaplan, problems after §5.3

From Kaplan, problems after §5.3

1(a) $\displaystyle\int_{(0,0)}^{(1,1)} y^2\,\mathrm{d}x$ along the line from $(0,0)$ to $(1,1)$

Evaluate $\displaystyle\int_C y^2\,\mathrm{d}x$ along the straight line from $(0,0)$ to $(1,1)$.

Step 1: Parametrize.
The straight line from $(0,0)$ to $(1,1)$: let $x = t$, $y = t$, $0 \le t \le 1$. So $\mathrm{d}x = \mathrm{d}t$.

Step 2: Substitute and integrate.
$$\int_C y^2\,\mathrm{d}x = \int_0^1 t^2\,\mathrm{d}t = \left[\frac{t^3}{3}\right]_0^1 = \frac{1}{3}$$

1(b) $\displaystyle\int_{(0,0)}^{(1,1)} y\,\mathrm{d}x$ along the line from $(0,0)$ to $(1,1)$

Evaluate $\displaystyle\int_C y\,\mathrm{d}x$ along the straight line from $(0,0)$ to $(1,1)$.

Step 1: Parametrize.
Same line: $x = t$, $y = t$, $0 \le t \le 1$. So $\mathrm{d}x = \mathrm{d}t$.

Step 2: Substitute and integrate.
$$\int_C y\,\mathrm{d}x = \int_0^1 t\,\mathrm{d}t = \left[\frac{t^2}{2}\right]_0^1 = \frac{1}{2}$$

1(c) $\displaystyle\int_{(0,0)}^{(2,1)} x\,\mathrm{d}y$ along the line from $(0,0)$ to $(2,1)$

Evaluate $\displaystyle\int_C x\,\mathrm{d}y$ along the straight line from $(0,0)$ to $(2,1)$.

Step 1: Parametrize.
The line from $(0,0)$ to $(2,1)$: let $x = 2t$, $y = t$, $0 \le t \le 1$. So $\mathrm{d}y = \mathrm{d}t$.

Step 2: Substitute and integrate.
$$\int_C x\,\mathrm{d}y = \int_0^1 2t\,\mathrm{d}t = \left[t^2\right]_0^1 = 1$$