Fundamental Theorem of line integrals

Section 3.4 Fundamental Theorem of line integrals

In Section 2.2 we studied an important class of one-forms called exact, which arise as differentials of functions. Their associated vector fields are called conservative, and can be expressed as the gradient of a potential function. In this section we see that line integrals of such one-forms are very nice and satisfy beautiful properties. This leads us to the Fundamental Theorem of line integrals, which is a natural generalization of the Fundamental Theorem of Calculus.

Objectives

You should be able to:

State the Fundamental Theorem of line integrals for line integrals of exact one-forms, and use it to evaluate line integrals.
Show that the Fundamental Theorem of line integrals implies that line integrals of exact one-forms only depend on the starting and ending points of the curve.
Show that the Fundamental Theorem of line integrals implies that line integrals of exact one-forms over closed curves vanish.
State these results in terms of conservative vector fields and their associated potential functions.
Use the Fundamental Theorem of line integrals and its consequences to show that a given one-form cannot be exact.

Subsection 3.4.1 The Fundamental Theorem of line integrals

Recall from Section 2.2 that an exact one-form is a one-form that can be written as the differential of a function: \(\omega = d f\text{.}\) Conversely, its associated vector field \(\mathbf{F}\) can be written as the gradient of a function, \(\mathbf{F} = \boldsymbol{\nabla} f\text{;}\) we say that \(\mathbf{F}\) is conservative and that \(f\) is its associated potential.

Theorem 3.4.1. The Fundamental Theorem of line integrals.

Let \(\omega= d f\) be an exact one-form on an open subset \(U \subseteq \mathbb{R}^n\text{,}\) and \(\alpha:[a,b] \to \mathbb{R}^n\) be a parametric curve whose image \(\alpha([a,b]) = C \subset U\text{.}\) Then:

\begin{equation*} \int_\alpha \omega = \int_\alpha d f = f(\alpha(b)) - f(\alpha(a)). \end{equation*}

The integral thus only depends on the starting and ending points of the image curve \(C\text{.}\)

Proof.

You have probably noticed that this theorem is similar in flavour to the Fundamental Theorem of Calculus for definite integrals; in fact it follows from it, as we will see.

First, by the definition of line integrals, we have:

\begin{equation*} \int_\alpha df = \int_{[a,b]} \alpha^*(df ). \end{equation*}

Next, we can use one of the fundamental properties of the pullback, which is that \(\alpha^*(d f) = d(\alpha^* f)\text{.}\) So we can write:

\begin{equation*} \int_\alpha df = \int_{[a,b]} d(\alpha^* f). \end{equation*}

If we introduce a parameter \(t\) for the parametric curve, i.e. \(\alpha(t) = (x_1(t), \ldots, x_n(t))\text{,}\) then \(\alpha^* f(t) = f(\alpha(t))\text{,}\) and we can write the integral as:

\begin{equation*} \int_\alpha df = \int_a^b \frac{d}{dt} ( f(\alpha(t) )\ dt. \end{equation*}

But then, the right-hand-side is just a standard definite integral of the derivative of a function. By the Fundamental Theorem of Calculus (part 2), we know that the right-hand-side is simply equal to \(f(\alpha(b)) - f(\alpha(a))\text{.}\) We thus get:

\begin{equation*} \int_\alpha df = f(\alpha(b)) - f(\alpha(a)). \end{equation*}

This result makes it very easy to evaluate line integrals for exact one-forms. But it also has deeper implications. Since the integral only depends on the starting and ending points on the image curve, this means that it does not actually depend on the choice of curve itself! Pick any two parametric curves whose images start and end at the same place: the integral will be the same. This is rather striking!

Corollary 3.4.2. The line integrals of an exact form along two curves that start and end at the same points are equal.

If \(\alpha\) and \(\beta\) are two parametric curves whose image share the same starting and ending points, and \(\omega = d f\) is exact, then:

\begin{equation*} \int_\alpha \omega = \int_\beta \omega. \end{equation*}

Another direct consequence of the Fundamental Theorem of line integrals is that the integral of an exact one-form over a closed curve always vanishes! Indeed, the curve is closed if \(\alpha(b) = \alpha(a)\) (so that the image curve is a “loop”), and so the right-hand-side in Theorem 3.4.1 vanishes.

Corollary 3.4.3. The line integral of an exact one-form along a closed curve vanishes.

Let \(\omega= d f\) be an exact one-form on an open subset \(U \subseteq \mathbb{R}^n\text{,}\) and \(\alpha\) be any closed parametric curve whose image \(C \subset U\text{.}\) Then

\begin{equation*} \int_\alpha \omega = 0. \end{equation*}

We sometimes write \(\oint_\alpha \omega\) for the line integral of a one-form along a closed parametric curve.

Example 3.4.4. An example of a line integral of an exact one-form.

Suppose that you want to integrate the one-form \(\omega = y^2 z\ dx + 2 x y z \ dy + x y^2\ dz\) over the line segment joining the origin to the point \((1,1,1) \in \mathbb{R}^3\text{.}\) In principle, to evaluate the line integral, you would need to find a parametrization for the line, and use the definition of line integrals Definition 3.3.2 to evaluate the integral. However, we notice here that \(\omega\) is exact! Indeed, if you pick the function \(f(x,y,z) = x y^2 z\text{,}\) then

\begin{equation*} d f = \frac{\partial f}{\partial x}\ dx + \frac{\partial f}{\partial y}\ dy + \frac{\partial f}{\partial z}\ dz = y^2 z \ dx + 2 x y z \ dy + x y^2 \ dz, \end{equation*}

which is \(\omega\text{.}\) Thus we can use the Fundamental Theorem of line integrals to evaluate the line integral. Let \(\alpha\) be any parametrization of the line segment joining the origin to the point \((1,1,1)\text{.}\) We get:

\begin{equation*} \int_\alpha \omega = f(1,1,1) - f(0,0,0) = 1-0 = 1. \end{equation*}

What is neat as well is that you know that the line integral of \(\omega\) along any curve joining the origin to the point \((1,1,1)\) will be equal to 1! The curve does not have to be a line. It could be a parabola, the arc of a circle, anything! For instance, just for fun let us pick the following parametric curve \(\beta:[0,1] \to \mathbb{R}^3\) with \(\beta(t) = (t, t^2, t^3)\text{,}\) whose image curve joins the origin to \((1,1,1)\text{.}\) Let us show that this works. By definition of line integrals,

\begin{align*} \int_\beta \omega =\amp \int_0^1 \left( (t^2)^2 t^3 (1) + 2 (t)(t^2)(t^3) (2t) + t (t^2)^2 (3 t^2) \right)\ dt\\ =\amp \int_0^1 \left( t^7 + 4 t^7 + 3 t^7 \right)\ dt \\ = \amp 8 \int_0^1 t^7\ dt\\ = \amp t^8 \Big|_0^1 \\ = \amp 1. \end{align*}

Neat!

One thing that we did not explain however here: how did we know that \(\omega\) was exact? This is not always so easy to figure out. We will discuss this further in Section 3.6.

Subsection 3.4.2 The Fundamental Theorem of line integrals for vector fields

To end this section, let us rewrite the Fundamental Theorem for line integrals in terms of the associated vector fields.

Theorem 3.4.5. The Fundamental Theorem of line integrals for vector fields.

Let \(\mathbf{F} = \boldsymbol{\nabla} f\) be a conservative vector field on an open subset \(U \subseteq \mathbb{R}^n\text{,}\) and \(\alpha:[a,b] \to \mathbb{R}^n\) a parametric curve whose image \(\alpha([a,b]) = C \subset U\text{.}\) Then:

\begin{equation*} \int_a^b \mathbf{F}(\alpha(t)) \cdot \mathbf{T}(t) \ dt = \int_a^b \boldsymbol{\nabla}f(\alpha(t)) \cdot \mathbf{T}(t) \ dt = f(\alpha(b)) - f(\alpha(a)). \end{equation*}

In the notation introduced in Remark 3.3.8, we can rewrite this integral as

\begin{equation*} \int_C \mathbf{F} \cdot d \mathbf{r} = \int_C \boldsymbol{\nabla} f \cdot d \mathbf{r} = f(P_1) - f(P_0), \end{equation*}

where \(P_0 = \mathbf{r}(a)\) and \(P_1 = \mathbf{r}(b)\) are the starting and ending points respectively of the curve \(C\) with direction of travel specified by the position function \(\mathbf{r}\text{.}\)

Note that from Corollary 3.4.2 and Corollary 3.4.3, we know that:

the line integral of a conservative vector field does not depend on the path chosen between two points;
the line integral of a conservative vector field along a closed curve is always zero.

Exercises 3.4.3 Exercises

1.

Consider the one-form \(\omega = (y + z e^x)\ dx + (x + e^y \sin z)\ dy + (z + e^x + e^y \cos z)\ dz\) on \(\mathbb{R}^3\text{.}\) Show that \(\omega\) is an exact form, and use this fact to evaluate the integral of \(\omega\) along the parametric curve \(\alpha:[ 0,\pi] \to \mathbb{R}^3\) with \(\alpha(t) = (t, e^t, \sin t)\text{.}\)

Solution.

To show that it is exact, we simply find a function \(f(x,y,z)\) such that \(\omega = d f\) by inspection (we can also do that by integrating the partial derivatives as we did a number of times already). We guess that \(f(x,y,z) = x y + z e^x + e^y \sin z + \frac{1}{2} z^2\text{,}\) and check that it works. Its differential is:

\begin{equation*} d f = (y + z e^x)\ dx + (x + e^y \sin z)\ dy + (z + e^x + e^y \cos z)\ dz, \end{equation*}

which is indeed \(\omega\text{.}\) So our guess is correct, and we have shown that \(\omega\) is exact.

Using the Fundamental Theorem of line integrals, we can integrate \(\omega\) directly along \(\alpha\text{:}\)

\begin{align*} \int_\alpha \omega =\amp f(\alpha(\pi)) - f(\alpha(0)) \\ =\amp f(\pi, e^\pi, 0) - f(0, 1, 0)\\ =\amp \pi e^\pi. \end{align*}

2.

Recall from Example 2.2.13 (see also Example 3.6.5) that the one-form \(\omega = -\frac{y}{x^2+y^2}\ dx + \frac{x}{x^2+y^2}\ dy\) on \(\mathbb{R}^2 \setminus \{(0,0) \}\) is closed. However, we said that it was not exact. Use the integral of \(\omega\) along one turn counterclockwise around the unit circle to show that \(\omega\) cannot be exact.

Solution.

We parametrize the unit circle as usual by \(\alpha:[0,2 \pi] \to \mathbb{R}^2\) with \(\alpha(t) = (\cos t, \sin t)\text{.}\) The pullback one-form is

\begin{align*} \alpha^* \omega =\amp \left( - \frac{\sin t}{\cos^2 t + \sin^2 t} (- \sin t) + \frac{\cos t}{\cos^2 t + \sin^2 t} (\cos t) \right) \ dt\\ =\amp (\sin^2 t + \cos^2 t)\ dt \\ =\amp dt. \end{align*}

The line integral thus simply becomes:

\begin{equation*} \int_\alpha \omega = \int_0^{2 \pi} dt = 2 \pi. \end{equation*}

In particular, it is non-zero. This proves that \(\omega\) cannot be exact on \(\mathbb{R}^2 \setminus \{(0,0)\}\text{,}\) since if it was exact its line integral along a closed curve would have to vanish.

3.

Suppose that \(\mathbf{F}\) is a conservative vector field in \(\mathbb{R}^2\) and that its integral from point \((1,0)\) to \((-1,0)\) along the upper half of the unit circle is \(5\text{.}\) What should the integral from \((1,0)\) to \((-1,0)\) but along the lower half of the circle be?

Solution.

It should be \(5\text{!}\) Indeed, since \(\mathbf{F}\) is conservative, we know that its line integral does not depend on the path chosen between two points. Since both paths here start and end at the same points, the line integrals along these paths must be equal.

4.

Consider the one-form \(\omega = d f\) on \(\mathbb{R}^2\) with \(f(x,y) = \sin(x+y)\text{.}\) Find a parametric curve \(\alpha\) that is not closed but such that

\begin{equation*} \int_\alpha \omega = 0. \end{equation*}

Solution.

The one-form \(\omega = d f\) is obviously exact. By the Fundamental Theorem for line integrals, we know that

\begin{equation*} \int_\alpha \omega = f(\alpha(b)) - f(\alpha(a)). \end{equation*}

If we write \(\alpha(t) = (x(t), y(t))\text{,}\) then this becomes

\begin{equation*} \int_\alpha \omega = \sin(x(b)+y(b)) - \sin(x(a)+y(a)). \end{equation*}

Now we want this integral to be zero. Thus we want

\begin{equation*} \sin(x(b) + y(b) ) = \sin (x(a) + y(a)). \end{equation*}

But we don't want a closed curve, so we must choose our curve such that \(\alpha(b) \neq \alpha(a)\text{.}\) There are of course many possible choices. Here is one example:

\begin{equation*} \alpha(a) = (0,0), \qquad \alpha(b) = (\pi,0). \end{equation*}

Then

\begin{equation*} \sin(x(a)+y(a)) = \sin(0) = 0, \qquad \sin(x(b)+y(b)) = \sin(\pi) = 0. \end{equation*}

Thus

\begin{equation*} \int_\alpha \omega = 0 \end{equation*}

for any parametric curve that starts at \((0,0)\) and ends at \((\pi,0)\text{.}\) For instance, we could pick a straight line between the two points.

5.

Let \(\omega\) be a one-form that is defined on all of \(\mathbb{R}^2\text{.}\) Let \(P_0, P_1, P_0', P_1'\) be any four points in \(\mathbb{R}^2\text{.}\) Suppose that

\begin{equation*} \int_{C_1} \omega = \int_{C_2} \omega \end{equation*}

for any two curves \(C_1\) and \(C_2\) that start at \(P_0\) and end at \(P_1\text{.}\) Show that it implies that

\begin{equation*} \int_{C_1'}\omega = \int_{C_2'} \omega \end{equation*}

for any two curves \(C_1'\) and \(C_2'\) that start at \(P_0'\) and end at \(P_1'\text{.}\)

In other words, if the line integral of a one-form between two given points is path independent, then it is path independent everywhere.

Solution.

The proof is fairly intuitive. Fix \(P_0, P_1 \in \mathbb{R}^2\text{,}\) and pick any two other points \(P_0', P_1' \in \mathbb{R}^2\text{.}\) Let \(D_0\) be a fixed curve from \(P_0\) to \(P_0'\text{,}\) and \(D_1\) a fixed curve from \(P_1'\) to \(P_1\text{.}\) Suppose that \(C_1'\) and \(C_2'\) are two curves from \(P_0'\) to \(P_1'\text{.}\)

On the one hand, the curve \(C_1 = D_0 \cup C_1' \cup D_1\) is curve from \(P_0\) to \(P_1\text{.}\) The line integral of \(\omega\) along \(C_1\) is

\begin{equation*} \int_{C_1} \omega = \int_{D_0} \omega + \int_{C_1'} \omega + \int_{D_1} \omega. \end{equation*}

On the other hand, the curve \(C_2 = D_0 \cup C_2' \cup D_1\) is also a curve from \(P_0\) to \(P_1\text{.}\) The line integral of \(\omega\) along \(C_2\) is

\begin{equation*} \int_{C_2} \omega = \int_{D_0} \omega + \int_{C_2'} \omega + \int_{D_1} \omega. \end{equation*}

But we know that

\begin{equation*} \int_{C_1} \omega = \int_{C_2} \omega. \end{equation*}

Equating the two expressions for these line integrals, and simplifying, we end up with the statement that

\begin{equation*} \int_{C_1'} \omega = \int_{C_2'} \omega. \end{equation*}

Since this must be true for any points \(P_0'\) and \(P_1'\text{,}\) and any curves \(C_1'\) and \(C_2'\) from \(P_0'\) to \(P_1'\text{,}\) we conclude that the line integral of \(\omega\) is path independent everywhere.