Skip to main content

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.

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: ω=df. Conversely, its associated vector field F can be written as the gradient of a function, F=f; we say that F is conservative and that f is its associated potential.

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:

αdf=[a,b]α(df).

Next, we can use one of the fundamental properties of the pullback, which is that α(df)=d(αf). So we can write:

αdf=[a,b]d(αf).

If we introduce a parameter t for the parametric curve, i.e. α(t)=(x1(t),,xn(t)), then αf(t)=f(α(t)), and we can write the integral as:

αdf=abddt(f(α(t)) dt.

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(α(b))f(α(a)). We thus get:

αdf=f(α(b))f(α(a)).

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!

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 α(b)=α(a) (so that the image curve is a “loop”), and so the right-hand-side in Theorem 3.4.1 vanishes.

Suppose that you want to integrate the one-form ω=y2z dx+2xyz dy+xy2 dz over the line segment joining the origin to the point (1,1,1)R3. 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 ω is exact! Indeed, if you pick the function f(x,y,z)=xy2z, then

df=fx dx+fy dy+fz dz=y2z dx+2xyz dy+xy2 dz,

which is ω. Thus we can use the Fundamental Theorem of line integrals to evaluate the line integral. Let α be any parametrization of the line segment joining the origin to the point (1,1,1). We get:

αω=f(1,1,1)f(0,0,0)=10=1.

What is neat as well is that you know that the line integral of ω 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 β:[0,1]R3 with β(t)=(t,t2,t3), whose image curve joins the origin to (1,1,1). Let us show that this works. By definition of line integrals,

βω=01((t2)2t3(1)+2(t)(t2)(t3)(2t)+t(t2)2(3t2)) dt=01(t7+4t7+3t7) dt=801t7 dt=t8|01=1.

Neat!

One thing that we did not explain however here: how did we know that ω 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.

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 ω=(y+zex) dx+(x+eysinz) dy+(z+ex+eycosz) dz on R3. Show that ω is an exact form, and use this fact to evaluate the integral of ω along the parametric curve α:[0,π]R3 with α(t)=(t,et,sint).

Solution.

To show that it is exact, we simply find a function f(x,y,z) such that ω=df 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)=xy+zex+eysinz+12z2, and check that it works. Its differential is:

df=(y+zex) dx+(x+eysinz) dy+(z+ex+eycosz) dz,

which is indeed ω. So our guess is correct, and we have shown that ω is exact.

Using the Fundamental Theorem of line integrals, we can integrate ω directly along α:

αω=f(α(π))f(α(0))=f(π,eπ,0)f(0,1,0)=πeπ.

2.

Recall from Example 2.2.13 (see also Example 3.6.5) that the one-form ω=yx2+y2 dx+xx2+y2 dy on R2{(0,0)} is closed. However, we said that it was not exact. Use the integral of ω along one turn counterclockwise around the unit circle to show that ω cannot be exact.

Solution.

We parametrize the unit circle as usual by α:[0,2π]R2 with α(t)=(cost,sint). The pullback one-form is

αω=(sintcos2t+sin2t(sint)+costcos2t+sin2t(cost)) dt=(sin2t+cos2t) dt=dt.

The line integral thus simply becomes:

αω=02πdt=2π.

In particular, it is non-zero. This proves that ω cannot be exact on R2{(0,0)}, since if it was exact its line integral along a closed curve would have to vanish.

3.

Suppose that F is a conservative vector field in R2 and that its integral from point (1,0) to (1,0) along the upper half of the unit circle is 5. What should the integral from (1,0) to (1,0) but along the lower half of the circle be?

Solution.

It should be 5! Indeed, since 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 ω=df on R2 with f(x,y)=sin(x+y). Find a parametric curve α that is not closed but such that

αω=0.
Solution.

The one-form ω=df is obviously exact. By the Fundamental Theorem for line integrals, we know that

αω=f(α(b))f(α(a)).

If we write α(t)=(x(t),y(t)), then this becomes

αω=sin(x(b)+y(b))sin(x(a)+y(a)).

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

sin(x(b)+y(b))=sin(x(a)+y(a)).

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

α(a)=(0,0),α(b)=(π,0).

Then

sin(x(a)+y(a))=sin(0)=0,sin(x(b)+y(b))=sin(π)=0.

Thus

αω=0

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

5.

Let ω be a one-form that is defined on all of R2. Let P0,P1,P0,P1 be any four points in R2. Suppose that

C1ω=C2ω

for any two curves C1 and C2 that start at P0 and end at P1. Show that it implies that

C1ω=C2ω

for any two curves C1 and C2 that start at P0 and end at P1.

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 P0,P1R2, and pick any two other points P0,P1R2. Let D0 be a fixed curve from P0 to P0, and D1 a fixed curve from P1 to P1. Suppose that C1 and C2 are two curves from P0 to P1.

On the one hand, the curve C1=D0C1D1 is curve from P0 to P1. The line integral of ω along C1 is

C1ω=D0ω+C1ω+D1ω.

On the other hand, the curve C2=D0C2D1 is also a curve from P0 to P1. The line integral of ω along C2 is

C2ω=D0ω+C2ω+D1ω.

But we know that

C1ω=C2ω.

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

C1ω=C2ω.

Since this must be true for any points P0 and P1, and any curves C1 and C2 from P0 to P1, we conclude that the line integral of ω is path independent everywhere.