Integrating zero-forms and one-forms

Section 5.1 Integrating zero-forms and one-forms

Before we move on to two-forms, for completeness we start by defining how zero-forms can be integrated over oriented points, and review how one-forms can be integrated over oriented curves. We highlight the main steps of the construction, setting up the stage for the development of surface integrals.

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Objectives

You should be able to:

Define the integral of a zero-form over oriented points.
Rephrase the Fundamental Theorem of line integrals in terms of oriented integrals of one- and zero-forms.

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Subsection 5.1.1 General strategy

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Let us start by summarizing step-by-step how we are developing our theory of integration. Suppose that we want to define integration of

k

-forms over

k

-dimensional spaces:

We define the orientation of a closed bounded region $D \subset R^{k}$ (such as a closed interval in $R$ ), and the notion of canonical orientation. We define the induced orientation on the boundary of the region $\partial D .$
We define the integral of a $k$ -form over an oriented region $D$ in terms of standard multiple integrals from calculus. To define the integral over a bounded region that has many components, we sum over the integrals on each components.
We define the notion of a parametric space, which maps a region $D$ to a $k$ -dimensional subspace $S \subset R^{n},$ with $n > k .$ We show that the parametrization induces an orientation on $S .$
To define the integral of a $k$ -form in $R^{n}$ on the subspace $S \subset R^{n}$ with a choice of orientation, we use the parametrization to pullback the $k$ -form to the region $D \subset R^{k},$ and then we integrate as in (2) using standard calculus.
We show that the integral is invariant under orientation-preserving reparametrizations and changes sign under orientation-reversing reparametrizations, thus ensuring that our theory is oriented and reparametrization-invariant. We conclude that the integral is defined geometrically in terms of the subspace $S$ and a choice of orientation.
As a last step, we study what happens in the case of an exact $k$ -form: this leads to Stokes' Theorem, which is the higher dimensional generalization of the Fundamental Theorem of Calculus and the Fundamental Theorem of line integrals.

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This summarizes the conceptual steps to construct our theory of integration. This is pretty much exactly what we did for line integrals in Chapter 3; we will review these steps below. But before we do that, let us apply this strategy to construct integrals of zero-forms.

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Subsection 5.1.2 Integrating zero-forms over oriented points

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We consider first the very simple and particular case of zero-forms. We would like to integrate a zero-form over a zero-dimensional space. What is a zero-dimensional space? It is just a point (or a union of points). But let us construct the theory step-by-step.

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STEP 1. We first need to define the orientation of a point.

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Definition 5.1.1. Oriented points.

Pick a point $a \in R^{n} .$ The orientation of a point $a \in R^{n}$ is given by a choice of $+$ or $- .$ We write an oriented point as $(a, +)$ or $(a, -) .$ The canonical orientation is $+ .$

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According to our recipe, we should talk about the induced orientation on the boundary, but a point has no boundary, so this step is meaningless in this case.

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STEP 2, 3, 4, 5. The next step is to define the “integral” of a zero-form on an oriented point in

R .

Then we would define “parametrizations for points in higher-dimensional spaces”, but this is all rather trivial here, so we can do steps 2 to 5 all at the same time and simply define the integral of a zero-form on an oriented point in

R^{n}

directly.

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Recall that a zero-form is just a function

f .

In this case, the integral is defined very simply by just evaluating the function at the point.

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Definition 5.1.2. Integral of a zero-form over an oriented point.

Let $f$ be a zero-form on an open subset $U \subseteq R^{n}$ and $(P, \pm)$ a point in $U$ with a choice of orientation. We define the integral of $f$ on $(P, \pm)$ by:

\int_{(P, \pm)} f = \pm f (P) .

In other words, we just evaluate the function at $P \in R^{n},$ and multiply by the sign corresponding to the chosen orientation.

To define the integral over a set of oriented points, we sum up the integrals over each point separately.

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For instance, in this language we can define the integral of a function

f

over two oriented points

{(P_{0}, -), (P_{1}, +)}

as:

\int_{{(P_{0}, -), (P_{1}, +)}} f = f (P_{1}) - f (P_{0}) .

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And that's basically it, as far as zero-forms go. There's no question of parametrization here, and the integral is clearly oriented by definition. There's no step 6, as there is no such thing as an exact zero-form.

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Example 5.1.3. Integral of a zero-form at points.

Consider the function $f (x, y, z) = x y + z$ on $R^{3} .$ Pick the two points $P_{0} = (0, 0, 0)$ and $P_{1} = (1, 1, 1) .$ Let's give $P_{0}$ a negative orientation, and $P_{1}$ a positive orientation. Then

\int_{{(P_{0}, -), (P_{1}, +)}} f = f (P_{1}) - f (P_{0}) = f (1, 1, 1) - f (0, 0, 0) = 2.

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Subsection 5.1.3 Integrating one-forms over oriented curves

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We move to the case of one-forms, which was already covered in Chapter 3. Here we review the construction to highlight how it fits within our general strategy.

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STEP 1. We start by considering a closed interval

[a, b] \in R .

We define its orientation as in Definition 3.1.3. We also define an induced orientation on the boundary.

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Definition 5.1.4. The orientation of an interval.

We define the orientation of an interval $[a, b] \subset R$ to be a choice of direction. By $[a, b]_{+},$ we mean the interval $[a, b]$ with the orientation of increasing real numbers (from $a$ to $b$ ), and by $[a, b]_{-}$ we denote the same interval but with the orientation of decreasing real numbers (from $b$ to $a$ ). We define the canonical orientation to be the orientation of increasing real numbers.

The boundary of $[a, b]$ is the set of points ${a, b} .$ To the interval $[a, b]$ in canonical orientation, we define the induced orientation on its boundary to be ${(a, -), (b, +)},$ and vice-versa for the interval $[a, b]_{-}$ in the reverse orientation.

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STEP 2. The next step is to define the integral of a one-form over an interval

[a, b]

with a choice of orientation. This is what we did in Definition 3.1.1.

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Definition 5.1.5. The integral of a one-form over an oriented interval $[a, b]_{\pm}$ .

Let $ω$ be a one-form on $U \subseteq R,$ with $[a, b] \subset U .$ We define the integral of $ω$ over the oriented interval $[a, b]_{\pm}$ as:

\int_{[a, b]_{\pm}} ω = \pm \int_{a}^{b} f (x) d x,

where on the right-hand-side we use the standard definition of definite integrals from calculus.

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STEP 3. Next step: introduce parametric curves

α : [a, b] \to R^{n} .

This was done in Definition 3.2.1. We do not repeat the definition here, but simply restate that it induces an orientation on the image curve

C = α ([a, b]) \subset R^{n},

and on its boundary

\partial C = {(α (a), -), (α (b), +)}

if it is not closed. ¹ We will write

\partial α

to denote the boundary of the parametric curve with its induced orientation.

This is clear since the domain of a parametric curve is always the interval

[a, b]

with its canonical orientation. Since the canonical orientation induces the orientation

{(a, -), (b, +)}

on the boundary of the interval, it induces the orientation

{(α (a), -), (α (b), +)}

on the boundary of the parametric curve.

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STEP 4. We can then define the integral over a parametric curve

α

using the pullback. This is Definition 3.3.2.

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Definition 5.1.6. (Oriented) line integrals.

Let $ω$ be a one-form on an open subset $U$ of $R^{n},$ and let $α : [a, b] \to R^{n}$ be a parametric curve whose image $C = α ([a, b]) \subset U .$ We define the integral of $ω$ along $α$ as follows:

\int_{α} ω = \int_{[a, b]} α^{*} ω,

where the integral one the right-hand-side is defined in Definition 5.1.5.

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STEP 5. Then, in Lemma 3.3.5, we showed that the integral above is invariant under orientation-preserving reparametrizations, and changes sign under orientation-reversing reparametrizations. Thus we can think of the integral as being defined intrinsically in terms of the image curve

C = α ([a, b])

and a choice of orientation.

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STEP 6. Finally we consider the case of an exact one-form

ω = d f .

In this case, the integral simplifies drastically. Let us first look at what happens when we integrate an exact one-form over an interval

[a, b] \subset R .

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Theorem 5.1.7. The Fundamental Theorem of Calculus.

Let $f$ be a zero-form on $U \subseteq R,$ and $[a, b] \subset U$ with its canonical orientation. Let $\partial ([a, b]) = {(a, -), (b, +)}$ be the boundary of the interval with its induced orientation. Then

\int_{[a, b]} d f = \int_{\partial ([a, b])} f,

which is the Fundamental Theorem of Calculus (part II).

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It may not seem obvious that this is the Fundamental Theorem of Calculus, but it is! On the left-hand-side, from our definition of integration, we have:

\int_{[a, b]} d f = \int_{a}^{b} \frac{d f}{d x} d x .

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On the right-hand-side, from our definition of integration of zero-forms over oriented points (Definition 5.1.2), we get:

\int_{\partial ([a, b])} f = f (b) - f (a) .

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So the theorem above is

\int_{a}^{b} \frac{d f}{d x} d x = f (b) - f (a),

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which is the Fundamental Theorem of Calculus (part II). Cool!

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Now let us move on to integration of exact one-forms over parametric curves.

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Theorem 5.1.8. The Fundamental Theorem of line integrals.

Let $f$ be a zero-form on $U \subseteq R^{n},$ and $α : [a, b] \to R^{n}$ be a parametric curve whose image $C = α ([a, b]) \subset U .$ The boundary of the parametric curve, with its induced orientation, consists in the two oriented points $\partial α = {(α (a), -), (α (b), +)}$ if the image curve is not closed; otherwise it is the empty set. Then:

\int_{α} d f = \int_{\partial α} f .

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Again, the statement above may look different from Theorem 3.4.1, but it is really the same thing. Indeed, the integral on the right-hand-side should be understood as the integral of a zero-form as in Definition 5.1.2. We can thus write:

\int_{\partial α} f = \int_{{(α (a), -), (α (b), +)}} f = f (α (b)) - f (α (a)),

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which is the statement in Theorem 3.4.1. In particular, if the curve is closed,

\partial α

is the empty set, and the right-hand-side vanishes, as in Corollary 3.4.3.

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Exercises 5.1.4 Exercises

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1.

Evaluate the integral of the zero-form

f (x, y, z) = \sin (x) + \cos (x y z) + e^{x y}

at the set of oriented points

S = {(p, +), (q, -), (r, +)}

with

p = (0, 0, 5), q = (\frac{π}{2}, \frac{2}{π}, π), r = (π, 1, 1) .

Solution.

By definition, the integral is

\begin{aligned} \int_{S} f = & f (p) - f (q) + f (r) \\ = & f (0, 0, 5) - f (\frac{π}{2}, \frac{2}{π}, π) + f (π, 1, 1) \\ = & (0 + 1 + 1) - (1 - 1 + e) + (0 - 1 + e^{π}) \\ = & 1 + e^{π} - e . \end{aligned}

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2.

Let $f$ be a zero-form on $R^{n},$ and suppose that $a \in R^{n}$ is a zero of the function $f .$ Show that the integral of $f$ at the point $a$ does not depend on the orientation of the point.

Solution.

Let $(a, \pm)$ be the point $a \in R^{n}$ with the positive and negative orientations. The integral of $f$ at $(a, \pm)$ is:

\begin{aligned} \int_{(a, \pm)} f = & \pm f (a) \\ = & 0, \end{aligned}

since $a$ is a zero of $f .$ Since the result is the same regardless of what orientation the point $a$ has, we conclude that the integral does not depend on the orientation of the point. (That's of course only because the point $a$ is a zero of $f,$ that wouldn't be true for an arbitrary point).

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3.

Suppose that $f$ is a smooth function on $R^{n},$ and let $p, q \in R^{n}$ be two distinct points such that $f (p) = f (q) .$ Show that the line integral of $d f$ along any curve starting at $p$ and ending at $q$ is zero.

Solution.

Let $α$ be any parametric curve whose image starts at $p$ and ends at $q .$ By the Fundamental Theorem of line integrals, we know that

\int_{α} d f = \int_{\partial α} f = \int_{{(p, -), (q, +)}} f = f (q) - f (p) = 0,

where the last equality follows since we assume that $f (p) = f (q) .$ Therefore, the line integral of $d f$ along any such parametric curve $α$ is zero.

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4.

Let $f$ be a zero-form on $R^{n},$ and $S = {(a, +), (- a, -)}$ for some point $a \in R^{n}$ ( $- a$ denotes the point in $R^{n}$ whose coordinates are minus those of $a$ ). Show that

\int_{S} f = {\begin{cases} 0 & if f is even, \\ 2 f (a) & if f is odd. \end{cases}

Solution.

The integral of the zero-form is

\int_{S} f = \int_{{(a, +), (- a, -)}} f = f (a) - f (- a) .

If $f$ is even, $f (- a) = f (a),$ and hence

\int_{S} f = f (a) - f (- a) = f (a) - f (a) = 0.

If $f$ is odd, $f (- a) = - f (a),$ and hence

\int_{S} f = f (a) - f (- a) = f (a) - (- f (a)) = 2 f (a) .

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5.

Let $ω = 2 x y d x + x^{2} d y$ be a one-form on $R^{2} .$ Suppose that

\int_{α} ω = 5

for some parametric curve $α : [a, b] \to R^{2} .$ Show that the image curve $C = α ([a, b])$ is not a closed curve, i.e. it must have boundary points.

Solution.

First, we notice that $ω$ is exact. Indeed, let $f (x, y) = x^{2} y .$ Then

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = 2 x y d x + x^{2} d y = ω .

But then, by the Fundamental Theorem of line integrals, we know that

\int_{α} ω = \int_{α} d f = \int_{\partial α} f .

In particular, if the image curve is closed, then the boundary set is empty, i.e. $\partial α = \emptyset,$ and the right-hand-side is zero. But the question states that it is non-zero; it is equal to $5 .$ Therefore, the image curve cannot be closed.

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6.

You want to impress your calculus teacher, and you tell her that “integration by parts” can be rewritten as the “simple” statement that

\int_{[a, b]} d (f g) = \int_{\partial ([a, b])} f g,

i.e. it is just the Fundamental Theorem of Calculus (part II) for a product of functions ( $f, g$ are differentiable functions on $R$ ).

Explain why this is equivalent to integration by parts for definite integrals.

Solution.

First, using the graded product rule for the exterior derivative, we know that $d (f g) = g d f + f d g .$ So we can write the left-hand-side as

\int_{[a, b]} d (f g) = \int_{[a, b]} (g d f + f d g) = \int_{a}^{b} g d f + \int_{a}^{b} f d g .

As for the right-hand-side, the boundary of the interval is $\partial ([a, b]) = {(a, -), (b, +)} .$ So it can be rewritten as

\int_{\partial ([a, b])} f g = f (b) g (b) - f (a) g (a) .

Putting this together and rearranging a bit, we get

\int_{a}^{b} f d g = f g |_{a}^{b} - \int_{a}^{b} g d f,

which is the statement of integration by parts for definite integrals.

Section 5.1 Integrating zero-forms and one-forms

Objectives

Subsection 5.1.1 General strategy

Subsection 5.1.2 Integrating zero-forms over oriented points

Definition 5.1.1. Oriented points.

Definition 5.1.2. Integral of a zero-form over an oriented point.

Example 5.1.3. Integral of a zero-form at points.

Subsection 5.1.3 Integrating one-forms over oriented curves

Definition 5.1.4. The orientation of an interval.

Definition 5.1.5. The integral of a one-form over an oriented interval [a,b]±.

Definition 5.1.6. (Oriented) line integrals.

Theorem 5.1.7. The Fundamental Theorem of Calculus.

Theorem 5.1.8. The Fundamental Theorem of line integrals.

Exercises 5.1.4 Exercises

1.

2.

3.

4.

5.

6.

Definition 5.1.5. The integral of a one-form over an oriented interval $[a, b]_{\pm}$ .