Unoriented line integrals

Section 7.1 Unoriented line integrals

We define unoriented line integrals of functions along parametric curves. As a particular case, we study how to calculate the arc length of a parametric curve.

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Objectives

You should be able to:

Determine the arc length of a parametrized curve in $R^{n}$ using an unoriented line integral.
Evaluate the unoriented integral of a function along a parametrized curve in $R^{n} .$

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Subsection 7.1.1 Unoriented line integrals

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In this course we developed a theory of integration along curves and surfaces using differential forms. By construction, our theory was oriented, as integrals of differential forms naturally depend on a choice of orientation on the space over which we are integrating.

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However, not all integrals should be oriented. Sometimes we want to calculate a quantity associated to a curve or a surface that should not depend on a choice of orientation. Typical examples would be the length of a curve or the area of a surface: such quantities should not depend on a choice of orientation. As integrals of differential forms are naturally oriented, it follows that integrals calculating arc lengths or surface areas cannot be represented as integrals of differential forms. We need to study unoriented line and surface integrals. In this section we look at unoriented line integrals.

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Before we define the concept of unoriented line integral of a function along a parametric curve, let us review how we defined oriented line integrals. Let

α : [a, b] \to R^{n}

be a parametric curve, with

α (t) = (x_{1} (t), \dots, x_{n} (t)) .

We defined the oriented line integral of a one-form

ω

along the parametric curve

α

via pullback (Definition 3.3.2). In terms of the vector field

F = (f_{1}, \dots, f_{n})

associated to the one-form

ω,

by evaluating the pullback the oriented line integral can be rewritten as (Lemma 3.3.7):

\int_{a}^{b} (F (α (t)) \cdot T) d t,

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where

T

is the tangent vector

T (t) = (x_{1}^{'} (t), \dots, x_{n}^{'} (t)) .

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From the point of view of vector fields, the orientation of the integral is encapsulated in the choice of tangent vector

T .

However, the way it is formulated, the tangent vector includes more information than just the orientation, as it also has a non-trivial norm specified by the parametrization. To isolate the oriented nature of the integral, we normalize the tangent vector, and define the unit tangent vector

\hat{T} = \frac{T}{| T |}, | T | = \sqrt{(x_{1}^{'} (t))^{2} + \dots + (x_{n}^{'} (t))^{2}} .

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We can then rewrite the oriented line integral as

\int_{a}^{b} (F (α (t)) \cdot \hat{T}) | T | d t = \int_{a}^{b} (F (α (t)) \cdot \hat{T}) d s,

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where we defined the “line element”

d s = | T | d t = \sqrt{(x_{1}^{'} (t))^{2} + \dots + (x_{n}^{'} (t))^{2}} d t .

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With this formulation, we see that the choice of orientation is completely encapsulated in the expression

F (α (t)) \cdot \hat{T} (t),

which is function of

t

which depends on the choice of direction on the parametric curve.

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We can now see how unoriented line integrals can be naturally defined: we simply replace the function

F (α (t)) \cdot \hat{T} (t),

constructed out of a vector field and a choice of orientation on the curve, by an arbitrary function

f (α (t))

that does not depend on a choice of orientation. This leads to the following definition.

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Definition 7.1.1. Unoriented line integrals.

Let $α : [a, b] \to R^{n}$ be a parametric curve, with image curve $C = α ([a, b]) \subset R^{n},$ and let $f : C \to R$ be a continuous function. We define the unoriented line integral of $f$ along the curve $C$ to be

\int_{C} f d s = \int_{a}^{b} f (α (t)) | T (t) | d t = \int_{a}^{b} f (α (t)) \sqrt{(x_{1}^{'} (t))^{2} + \dots + (x_{n}^{'} (t))^{2}} d t .

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A similar calculation as in the proof of Lemma 3.3.5 shows that unoriented line integrals are invariant under reparametrizations, regardless of whether the reparametrization preserves the orientation or not (the integrals are unoriented). What this means is that the line integral does not depend on the choice of parametrization, but only on the image curve

C .

This is why we wrote

\int_{C} f d s

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for the unoriented line integral of the function

f

along the curve

C \subset R^{n},

without specifying the parametrization

α,

since the integral is independent of the choice of parametrization.

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Remark 7.1.2.

We note that even though the notation “ $d s$ ” is similar to the notation we used for one-forms, the line element is not a one-form. For instance,

\int_{C_{+}} d s = \int_{C_{-}} d s,

i.e. the integral remains the same if we change the orientation of the curve, which would not be the case if $d s$ was a one-form.

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Example 7.1.3. An example of an unoriented line integral.

Evaluate the unoriented line integral

\int_{C} x y^{6} d s,

where $C$ is the right half of the circle $x^{2} + y^{2} = 4.$

First, we parametrize the curve as $α : [- π / 2, π / 2] \to R^{2}$ with

α (θ) = (2 \cos (θ), 2 \sin (θ)) .

As $θ \in [- π / 2, π / 2],$ we are parametrizing the right half of the circle, as required. We do not need to check here whether the parametrization induces the right orientation on the curve, as we do not care about the orientation whatsoever: the integral is unoriented.

To evaluate the unoriented line integral, we need the line element $d s .$ We calculate:

\begin{aligned} d s = & \sqrt{(x^{'} (θ))^{2} + (y^{'} (θ))^{2}} \\ = & \sqrt{4 \sin^{2} (θ) + 4 \cos^{2} (θ)} \\ = & 2. \end{aligned}

Using this parametrization, the unoriented line integral becomes:

\begin{aligned} \int_{C} x y^{6} d s = & \int_{- π / 2}^{π / 2} (2 \cos (θ)) (2 \sin (θ))^{6} (2) d θ \\ = & 256 \int_{- π / 2}^{π / 2} \cos (θ) \sin^{6} (θ) d θ \\ = & 256 \int_{- 1}^{1} u^{6} d u \\ = & \frac{512}{7} . \end{aligned}

Note that we used the substitution $u = \sin (θ)$ to evaluate the trigonometric integral.

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Subsection 7.1.2 Arc length of a curve

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A particularly important example of an unoriented line integral calculates the arc length of a curve

C .

This is the most trivial example, where we choose the function that we are integrating to be the constant function

f = 1 .

More precisely:

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Definition 7.1.4. Arc length of a curve.

Let $α : [a, b] \to R^{n}$ be a parametric curve, with image curve $C = α ([a, b]) \subset R^{n} .$ The arc length of $C$ is given by the unoriented line integral

\int_{C} d s = \int_{a}^{b} \sqrt{(x_{1}^{'} (t))^{2} + \dots + (x_{n}^{'} (t))^{2}} d t .

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You may have seen this formula before for the arc length, at least in

R^{2}

R^{3} .

It is straightforward to justify that it calculates the arc length of the curve, using the standard slicing argument from integral calculus. Consider a small curve segment between two points

P (α (t))

and

Q (α (t + d t)) .

The length

d s

of this curve segment can be approximated by the length of the line joining the two points, which can be written as

d s ≃ | α (t + d t) - α (t) | ≃ | α^{'} (t) | d t,

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where on the right-hand-side we kept only terms of first-order in

d t .

α^{'} (t) = T (t),

we recover the formula above for the line element. Finally, we sum over line elements and take the limit of an infinite number of line element of infinitesimal size, which turns the sum into the definite integral

\int_{a}^{b} | T (t) | d t = \int_{a}^{b} \sqrt{(x_{1}^{'} (t))^{2} + \dots + (x_{n}^{'} (t))^{2}} d t .

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Note that this also justifies our definition of unoriented line integrals in general above; it is constructed via the same slicing process, but where we also introduce a function

f

evaluated the point

α (t)

where we calculate the line element. We then sum over slices and take the limit of infinite number of infinitesimal slices as usual, and the integral of the function over the parametric curve becomes Definition 7.1.1.

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Example 7.1.5. Calculating the arc length of a parametric curve.

Find the length of the parametric curve $α : [0, 2] \to R^{3}$ with

α (t) = (1, t^{2}, t^{3}) .

We first calculate the line element $d s :$

\begin{aligned} d s = & \sqrt{(x^{'} (t))^{2} + (y^{'} (t))^{2} + (z^{'} (t))^{2}} d t \\ = & \sqrt{4 t^{2} + 9 t^{4}} d t \\ = & t \sqrt{4 + 9 t^{2}} d t, \end{aligned}

where in the last line we used $\sqrt{t^{2}} = t$ since we know that $t \in [0, 2]$ and hence it is positive. The arc length is thus given by

\begin{aligned} \int_{C} d s = & \int_{0}^{2} t \sqrt{4 + 9 t^{2}} d t \\ = & \frac{1}{18} \int_{4}^{40} \sqrt{u} d u \\ = & \frac{1}{27} (40^{3 / 2} - 4^{3 / 2}) \\ = & \frac{8}{27} (10^{3 / 2} - 1), \end{aligned}

where we used the substitution $u = 4 + 9 t^{2},$ $d u = 18 t d t .$

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Exercises 7.1.3 Exercises

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1.

Find the arc length of the circular helix $α : [0, 3] \to R^{3}$ with

α (t) = (t, 2 \cos (t), 2 \sin (t)) .

Solution.

To find the arc length, we first calculate the line element $d s .$ The tangent vector is

T (t) = (1, - 2 \sin (t), 2 \cos (t)) .

Its norm is

| T (t) | = \sqrt{1 + 4 \sin^{2} (t) + 4 \cos^{(} t)} = \sqrt{5} .

Thus the line element is

d s = \sqrt{5} d t .

We then calculate the arc length:

\begin{aligned} \int_{α} d s = & \int_{0}^{3} \sqrt{5} d t \\ = & 3 \sqrt{5} . \end{aligned}

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2.

Show that the arc length of the curve $C$ at the intersection of the surfaces $x^{2} = 2 y$ and $3 z = x y$ between the origin and the point $(6, 18, 36)$ is the answer to the ultimate question of life, the universe, and everything.

Solution.

We first parametrize the curve as $α : [0, 6] \to R^{3}$ with

α (t) = (t, \frac{t^{2}}{2}, \frac{t^{3}}{6}) .

The tangent vector is

T (t) = (1, t, \frac{t^{2}}{2}) .

Its norm is

| T (t) | = \sqrt{1 + t^{2} + \frac{t^{4}}{4}} = \sqrt{{(1 + \frac{t^{2}}{2})}^{2}} = 1 + \frac{t^{2}}{2},

since $1 + \frac{t^{2}}{2} > 0 .$ The line element is then

d s = (1 + \frac{t^{2}}{2}) d t,

and the arc length is

\begin{aligned} \int_{C} d s = & \int_{0}^{6} (1 + \frac{t^{2}}{2}) d t \\ = & 6 + \frac{6^{3}}{6} \\ = & 42, \end{aligned}

which is of course the answer to the ultimate question of life, the univers, and everything! :-)

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3.

Evaluate the unoriented line integral

\int_{C} (x z + e^{- y}) d s,

where $C$ is the line segment between the origin and the point $(1, 2, 3) .$

Solution.

We parametrize $C$ as $α : [0, 1] \to R^{3}$ with

α (t) = (t, 2 t, 3 t) .

The tangent vector is

T (t) = (1, 2, 3),

with norm

| T (t) | = \sqrt{1 + 2^{2} + 3^{2}} = \sqrt{14} .

The line element is

d s = \sqrt{14} d t,

and the unoriented line integral can be evaluated:

\begin{aligned} \int_{C} (x z + e^{- y}) d s = & \int_{0}^{1} ((t) (3 t) + e^{- 2 t}) \sqrt{14} d t \\ = & \sqrt{14} \int_{0}^{1} (3 t^{2} + e^{- 2 t}) d t \\ = & \sqrt{14} (1 - \frac{e^{- 2}}{2} + \frac{1}{2}) \\ = & \frac{\sqrt{14}}{2} (3 - e^{- 2}) . \end{aligned}

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4.

Evaluate the unoriented line integral

\int_{C} x^{3} y d s,

where $C$ is the circular helix parametrized by $α : [0, π / 2] \to R^{3}$ with

α (t) = (\cos (2 t), \sin (2 t), t) .

Solution.

The tangent vector to the parametric curve is

T (t) = (- 2 \sin (2 t), 2 \cos (2 t), 1),

with norm

| T (t) | = \sqrt{4 \sin^{2} (2 t) + 4 \cos^{2} (2 t) + 1} = \sqrt{5} .

The line element is

d s = \sqrt{5} d t .

The unoriented line integral becomes:

\begin{aligned} \int_{C} x^{3} y d s = & \sqrt{5} \int_{0}^{π / 2} \cos^{3} (2 t) \sin (2 t) d t \\ = & - \frac{\sqrt{5}}{2} \int_{1}^{- 1} u^{3} d u \\ = & 0, \end{aligned}

where we did the substitution $u = \cos (2 t) .$

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5.

In single-variable calculus, you saw that the length of the curve $y = f (x)$ in $R^{2},$ with $x \in [a, b],$ is given by the definite integral

\int_{a}^{b} \sqrt{1 + (f^{'} (x))^{2}} d x .

Show that this is consistent with our definition of arc length in this section.

Solution.

From our point of view, we realize the curve as the parametric curve $α : [a, b] \to R^{2}$ with

α (t) = (t, f (t)) .

Then the tangent vector is

T (t) = (1, f^{'} (t)),

with norm

| T (t) | = \sqrt{1 + (f^{'} (t))^{2}} .

So our arc length formula is

\int_{C} d s = \int_{a}^{b} | T (t) | d t = \int_{a}^{b} \sqrt{1 + (f^{'} (t))^{2}} d t,

which is indeed the formula that you obtained in single-variable calculus.