Learning Objectives
- Describe simple and closed curves; define connected and simply connected regions.
Curves and Regions
Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here.
We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve (Figure 1).
Definition
Curve [latex]C[/latex] is a closed curve if there is a parameterization [latex]{\bf{r}}(t)[/latex], [latex]a\leq{t}\leq{b}[/latex] of [latex]C[/latex] such that the parameterization traverses the curve exactly once and [latex]{\bf{r}}(a)={\bf{r}}(b)[/latex]. Curve [latex]C[/latex] is a simple curve if [latex]C[/latex] does not cross itself. That is, [latex]C[/latex] is simple if there exists a parameterization [latex]{\bf{r}}(t)[/latex], [latex]a\leq{t}\leq{b}[/latex] of [latex]C[/latex] such that [latex]{\bf{r}}[/latex] is one-to-one over [latex](a, b)[/latex]. It is possible for [latex]{\bf{r}}(a)={\bf{r}}(b)[/latex], meaning that the simple curve is also closed.
Figure 1. Types of curves that are simple or not simple and closed or not closed.
Example: determining whether a curve is simple and closed
Is the curve with parameterization [latex]{\bf{r}}(t)=\left\langle\cos{t},\frac{\sin{(2t)}}2\right\rangle, \ 0\leq{t}\leq2\pi[/latex] a simple closed curve?
try it
Is the curve given by parameterization [latex]{\bf{r}}(t)=\langle2\cos{t},3\sin{t}\rangle[/latex], [latex]0\leq{t}\leq6\pi[/latex] a simple closed curve?
Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region’s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.
definition
A region [latex]D[/latex] is a connected region if, for any two points [latex]P_1[/latex] and [latex]P_2[/latex], there is a path from [latex]P_1[/latex] to [latex]P_2[/latex] with a trace contained entirely inside [latex]D[/latex]. A region [latex]D[/latex] is a simply connected region if D is connected for any simple closed curve [latex]C[/latex] that lies inside [latex]D[/latex], and curve [latex]C[/latex] can be shrunk continuously to a point while staying entirely inside [latex]D[/latex]. In two dimensions, a region is simply connected if it is connected and has no holes.
All simply connected regions are connected, but not all connected regions are simply connected (Figure 3).
Figure 3. Not all connected regions are simply connected. (a) Simply connected regions have no holes. (b) Connected regions that are not simply connected may have holes but you can still find a path in the region between any two points. (c) A region that is not connected has some points that cannot be connected by a path in the region.
try it
Is the region in the below image connected? Is the region simply connected?
Try It
Fundamental Theorem for Line Integrals
Now that we understand some basic curves and regions, let’s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function [latex]f[/latex] has an antiderivative F, then the integral of [latex]f[/latex] from a to b depends only on the values of F at a and at b—that is,
[latex]\large{\displaystyle\int_a^bf(x) \ dx =F(b)-F(a)}[/latex].
If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.
Example: Evaluating a line integral and the Antiderivatives of the endpoints
Let [latex]{\bf{F}}(x,y)=\langle2x,4y\rangle[/latex]. Calculate [latex]\displaystyle\int_C{\bf{F}}. \ d{\bf{r}}[/latex], where [latex]C[/latex] is the line segment from [latex](0, 0)[/latex] to [latex](2, 2)[/latex] (Figure 4).
The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the Fundamental Theorem for Line Integrals.
theorem: the fundamental theorem for line integrals
Let [latex]C[/latex] be a piecewise smooth curve with parameterization [latex]{\bf{r}}(t)[/latex], [latex]a\leq{t}\leq{b}[/latex]. Let [latex]f[/latex] be a function of two or three variables with first-order partial derivatives that exist and are continuous on [latex]C[/latex]. Then,
[latex]\large{\displaystyle\int_C\nabla{f}. \ d{\bf{r}}=f({\bf{r}}(b))-f({\bf{r}}(a))}[/latex].
Proof
By [latex]\displaystyle\int_{C} {\bf{F}}\cdot{ds}=\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt[/latex],
[latex]\displaystyle\int_C\nabla{f}\cdot \ d{\bf{r}}=\displaystyle\int_a^b\nabla{f}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t) \ dt[/latex].
By the chain rule,
[latex]\frac{d}{dt}(f({\bf{r}}(t))=\nabla{f}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t)[/latex].
Therefore, by the Fundamental Theorem of Calculus,
[latex]\begin{aligned} \displaystyle\int_C\nabla{f}\cdot \ d{\bf{r}}&=\displaystyle\int_a^b\nabla{f}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t) \ dt \\ &=\displaystyle\int_a^b\frac{d}{dt}(f({\bf{r}}(t)) \ dt \\ &=[{f}({\bf{r}}(t))]_{t=a}^{t=b} \\ &={f}({\bf{r}}(b))-{f}({\bf{r}}(a)) \end{aligned}[/latex].
[latex]_\blacksquare[/latex]
We know that if [latex]\bf{F}[/latex] is a conservative vector field, there are potential functions [latex]f[/latex] such that [latex]\nabla{f}={\bf{F}}[/latex]. Therefore [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_C\nabla{f}\cdot{d}{\bf{r}}={f}({\bf{r}}(b))-{f}({\bf{r}}(a))[/latex]. In other words, just as with the Fundamental Theorem of Calculus, computing the line integral [latex]\displaystyle\int_c{\bf{F}}\cdot{d}{\bf{r}}[/latex], where [latex]\bf{F}[/latex] is conservative, is a two-step process: (1) find a potential function (“antiderivative”) [latex]f[/latex] for [latex]\bf{F}[/latex] and (2) compute the value of [latex]f[/latex] at the endpoints of [latex]C[/latex] and calculate their difference [latex]{f}({\bf{r}}(b))-{f}({\bf{r}}(a))[/latex]. Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.
Example: applying the fundamental theorem
Calculate integral [latex]\displaystyle\int_c{\bf{f}}.d{\bf{r}}[/latex], where [latex]{\bf{F}}(x,y,z)=\left\langle2x\ln{y},\frac{x^2}y+z^2,2yz\right\rangle[/latex] and [latex]C[/latex] is a curve with parameterization [latex]{\bf{r}}(t)=\langle{t^2},t,t\rangle[/latex], [latex]1\leq{t}\leq1{e}[/latex]
- without using the Fundamental Theorem of Line Integrals and
- using the Fundamental Theorem of Line Integrals.
Therefore,
Thus,
- Given that [latex]f(x,y,z)=x^2\ln{y}+yz^2[/latex] is a potential function for [latex]{\bf{F}}[/latex], let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that
[latex]\hspace{6cm}\begin{align} \displaystyle\int_C{\bf{F}}.d{\bf{r}}&=\displaystyle\int_C\nabla{f}.d{\bf{r}} \\ &=f({\bf{r}}(e))-f({\bf{r}}(1)) \\ &=f(e^2,e,e)-f(1,1,1) \\ &=e^4+e^3-1 \end{align}[/latex].
This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.
Example “Applying the Fundamental Theorem” illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.
try it
Given that [latex]f(x,y)=(x-1)^2y+(y+1)^2x[/latex] is a potential function for [latex]{\bf{F}}=\langle2xy-2y+(y+1)^2,(x-1)^2+2yx+2x\rangle[/latex], calculate integral [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}[/latex], where [latex]C[/latex] is the lower half of the unit circle oriented counterclockwise.
Watch the following video to see the worked solution to the above Try It
The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if [latex]{\bf{F}}[/latex] is conservative and [latex]C[/latex] is a closed curve, then the circulation of [latex]{\bf{F}}[/latex] along [latex]C[/latex] is zero—that is, [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=0[/latex]. To see why this is true, let [latex]f[/latex] be a potential function for [latex]{\bf{F}}[/latex]. Since [latex]C[/latex] is a closed curve, the terminal point [latex]{\bf{r}}(b)[/latex] of [latex]C[/latex] is the same as the initial point [latex]{\bf{r}}(a)[/latex] of [latex]C[/latex]—that is, [latex]{\bf{r}}(a)={\bf{r}}(b)[/latex]. Therefore, by the Fundamental Theorem for Line Integrals,
[latex]\begin{aligned} \displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\oint_C\nabla{f}\cdot{d}{\bf{r}} \\ &=f({\bf{r}}(b))-f({\bf{r}}(a)) \\ &=f({\bf{r}}(b))-f({\bf{r}}(b)) \\ &=0 \end{aligned}[/latex].
Recall that the reason a conservative vector field [latex]{\bf{F}}[/latex] is called “conservative” is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field [latex]{\bf{F}}[/latex] in integral [latex]\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}[/latex] as a gravitational field, then the equation [latex]\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}=0[/latex] follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.
The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path—meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.
definition
Let [latex]{\bf{F}}[/latex] be a vector field with domain [latex]D[/latex]. The vector field [latex]{\bf{F}}[/latex] is independent of path (or path independent) if [latex]\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}[/latex] for any paths [latex]C_1[/latex] and [latex]C_2[/latex] in [latex]D[/latex] with the same initial and terminal points.
The second consequence is stated formally in the following theorem.
theorem: path independence of conservative fields
If [latex]{\bf{F}}[/latex] is a conservative vector field, then [latex]{\bf{F}}[/latex] is independent of path.
Proof
Let [latex]D[/latex] denote the domain of [latex]{\bf{F}}[/latex] and let [latex]C_1[/latex] and [latex]C_2[/latex] be two paths in [latex]D[/latex] with the same initial and terminal points (Figure 5). Call the initial point [latex]P_1[/latex] and the terminal point [latex]P_2[/latex]. Since [latex]{\bf{F}}[/latex] is conservative, there is a potential function [latex]f[/latex] for [latex]{\bf{F}}[/latex]. By the Fundamental Theorem for Line Integrals,
[latex]\large{\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=f({\bf{P}}_2)-f({\bf{P}}_1)=\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}}[/latex].
Therefore, [latex]\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}[/latex] and [latex]{\bf{F}}[/latex] is independent of path.
[latex]_\blacksquare[/latex]
![<img src="/apps/archive/20220422.171947/resources/db3b7b07dc4a4243dcbf4dd01addbc41609ec1eb" data-media-type="image/jpeg" alt="A vector field in two dimensions. The arrows are shorter the closer to the x axis and line x=1.5 they become. The arrows point up, converging around x=1.5 in quadrant 1. That line is approached from the left and from the right. Below, in quadrant 4, the arrows in the rough interval [1,2.5] curve out, away from the given line x=1.5, but do turn back in and converge to x=1.5 above the x axis. Outside of that interval, the arrows go to the left and right horizontally for x values less than 1 and greater than 2.5, respectively. A line is drawn from P_1 at the origin to P_2 at (3,.75) and labeled C_2. C_1 is a simple curve that connects the given endpoints above C_2, C_3 is a simple curve that connects the given endpoints below C_2." id="19">](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/5667/2021/11/31223535/6.29.jpg)
Figure 5. The vector field is conservative, and therefore independent of path.
To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.
We have shown that if [latex]{\bf{F}}[/latex] is conservative, then [latex]{\bf{F}}[/latex] is independent of path. It turns out that if the domain of [latex]{\bf{F}}[/latex] is open and connected, then the converse is also true. That is, if [latex]{\bf{F}}[/latex] is independent of path and the domain of [latex]{\bf{F}}[/latex] is open and connected, then [latex]{\bf{F}}[/latex] is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.
theorem: The path independence test of conservative fields
If [latex]{\bf{F}}[/latex] is a continuous vector field that is independent of path and the domain [latex]D[/latex] of [latex]{\bf{F}}[/latex] is open and connected, then [latex]{\bf{F}}[/latex] is conservative.
Proof
We prove the theorem for vector fields in [latex]\mathbb{R}^2[/latex]. The proof for vector fields in [latex]\mathbb{R}^3[/latex] is similar. To show that [latex]{\bf{F}}=\langle{P},Q\rangle[/latex] is conservative, we must find a potential function [latex]f[/latex] for [latex]{\bf{F}}[/latex]. To that end, let [latex]X[/latex] be a fixed point in [latex]D[/latex]. For any point [latex](x, y)[/latex] in [latex]D[/latex], let [latex]C[/latex] be a path from [latex]X[/latex] to [latex](x, y)[/latex]. Define [latex]f(x, y)[/latex] by [latex]f(x,y)=\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}[/latex]. (Note that this definition of [latex]f[/latex] makes sense only because [latex]{\bf{F}}[/latex] is independent of path. If [latex]{\bf{F}}[/latex] was not independent of path, then it might be possible to find another path [latex]C'[/latex] from [latex]X[/latex] to [latex](x, y)[/latex] such that [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}\ne\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}[/latex], and in such a case [latex]f(x, y)[/latex] would not be a function.) We want to show that [latex]f[/latex] has the property [latex]\nabla{f}={\bf{F}}[/latex].
Since domain [latex]D[/latex] is open, it is possible to find a disk centered at [latex](x, y)[/latex] such that the disk is contained entirely inside [latex]D[/latex]. Let [latex](a, y)[/latex] with [latex]a<x[/latex] be a point in that disk. Let [latex]C[/latex] be a path from [latex]X[/latex] to [latex](x, y)[/latex] that consists of two pieces: [latex]C_1[/latex] and [latex]C_2[/latex]. The first piece, [latex]C_1[/latex], is any path from [latex]X[/latex] to [latex](a, y)[/latex] that stays inside [latex]D[/latex]; [latex]C_2[/latex] is the horizontal line segment from [latex](a, y)[/latex] to [latex](x ,y)[/latex] (Figure 6). Then
[latex]\large{f(x,y)=\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}}[/latex].
The first integral does not depend on [latex]x[/latex], so
[latex]\large{f_x=\frac{\partial}{\partial{x}}\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}}[/latex].
If we parameterize [latex]C_2[/latex] by [latex]{\bf{r}}(t)=\langle{t},y\rangle[/latex], [latex]a\leq{t}\leq{x}[/latex], then
[latex]\begin{aligned} f_x&=\frac\partial{\partial{x}}\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^x{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t) \ dt \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^x{\bf{F}}({\bf{r}}(t))\cdot\frac{d}{dt}(\langle{t},y\rangle \ dt \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^x{\bf{F}}({\bf{r}}(t))\cdot\langle1,0\rangle \ dt \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^xP(t,y) \ dt \end{aligned}[/latex].
By the Fundamental Theorem of Calculus (part 1),
[latex]\large{f_x=\frac\partial{\partial{x}}\displaystyle\int_a^xP(t,y) \ dt=P(x,y)}[/latex].
Figure 6. Here, [latex]C_1[/latex] is any path from [latex]X[/latex]; to [latex](a, y)[/latex] that stays inside [latex]D[/latex], and [latex]C_2[/latex] is the horizontal line segment from [latex](a, y)[/latex] to [latex](x, y)[/latex].
A similar argument using a vertical line segment rather than a horizontal line segment shows that [latex]f_y= Q(x, y)[/latex].
Therefore [latex]\nabla{f}={\bf{F}}[/latex] and [latex]{\bf{F}}[/latex] is conservative.
[latex]_\blacksquare[/latex]
We have spent a lot of time discussing and proving Path Independence of Conservative Fields Theorem and The Path Independence Test for Conservative Fields Theorem, but we can summarize them simply: a vector field [latex]{\bf{F}}[/latex] on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.
Example: showing that a vector field is not conservative
Use path independence to show that vector field [latex]{\bf{F}}(x,y)=\langle{x^2}y,y+5\rangle[/latex] is not conservative.
try it
Show that [latex]{\bf{F}}(x,y)=\langle{x}y,x^2y^2\rangle[/latex] is not path independent by considering the line segment from [latex](0, 0)[/latex] to [latex](2, 2)[/latex] and the piece of the graph of [latex]y=\frac{x^2}2[/latex] that goes from [latex](0, 0)[/latex] to [latex](2, 2)[/latex].
