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

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Curve $C$ is a closed curve if there is a parameterization ${\bf{r}}(t)$, $a\leq{t}\leq{b}$ of $C$ such that the parameterization traverses the curve exactly once and ${\bf{r}}(a)={\bf{r}}(b)$. Curve $C$ is a simple curve if $C$ does not cross itself. That is, $C$ is simple if there exists a parameterization ${\bf{r}}(t)$, $a\leq{t}\leq{b}$ of $C$ such that ${\bf{r}}$ is one-to-one over $(a, b)$. It is possible for ${\bf{r}}(a)={\bf{r}}(b)$, 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 ${\bf{r}}(t)=\left\langle\cos{t},\frac{\sin{(2t)}}2\right\rangle, \ 0\leq{t}\leq2\pi$ a simple closed curve?

Show Solution

Note that ${\bf{r}}(0)=\langle0,1\rangle$; therefore, the curve is closed. The curve is not simple, however. To see this, note that ${\bf{r}}\left(\frac{\pi}2\right)=\langle0,0\rangle={\bf{r}}\left(\frac{3\pi}2\right)$, and therefore the curve crosses itself at the origin (Figure 2).

Figure 2. A curve that is closed but not simple.

try it

Is the curve given by parameterization ${\bf{r}}(t)=\langle2\cos{t},3\sin{t}\rangle$, $0\leq{t}\leq6\pi$ a simple closed curve?

Show Solution

Yes.

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

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A region $D$ is a connected region if, for any two points $P_1$ and $P_2$, there is a path from $P_1$ to $P_2$ with a trace contained entirely inside $D$. A region $D$ is a simply connected region if D is connected for any simple closed curve $C$ that lies inside $D$, and curve $C$ can be shrunk continuously to a point while staying entirely inside $D$. 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.

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 $f$ has an antiderivative F, then the integral of $f$ from a to b depends only on the values of F at a and at b—that is,

$\large{\displaystyle\int_a^bf(x) \ dx =F(b)-F(a)}$.

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 ${\bf{F}}(x,y)=\langle2x,4y\rangle$. Calculate $\displaystyle\int_C{\bf{F}}. \ d{\bf{r}}$, where $C$ is the line segment from $(0, 0)$ to $(2, 2)$ (Figure 4).

Show Solution

We use $\displaystyle\int_{C} {\bf{F}}\cdot{ds}=\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt$ to calculate $\displaystyle\int_C{\bf{F}}. \ d{\bf{r}}$. Curve $C$ can be parameterized by ${\bf{r}}(t)=\langle2t,2t\rangle$, $0\leq{t}\leq1$. Then, ${\bf{F}}({\bf{r}}(t))=\langle4t,8t\rangle$ and ${\bf{r}}^\prime(t)=\langle2,2\rangle$, which implies that

[latex]\begin{aligned} \displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_0^1\langle4t,8t\rangle\cdot\langle2,2\rangle \ dt \\

&=\displaystyle\int_0^1(8t+16t)dt=\displaystyle\int_0^124t \ dt \\ &=[12t^2]_0^1=12. \end{aligned}[/latex]

Figure 4. The value of line integral $\displaystyle\int_C{\bf{F}}. \ d{\bf{r}}$ depends only on the value of the potential function of $\bf{F}$ at the endpoints of the curve.

Notice that $F=\nabla{f}$, where $f(x, y)=x^{2}+2y^{2}$. If we think of the gradient as a derivative, then $f$ is an “antiderivative” of $\bf{F}$. In the case of single-variable integrals, the integral of derivative $g'(x)$ is $g(b)-g(a)$, where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral $\bf{F}$ to be $f(P_1)-f(P_0)$, where $P_1$ is the endpoint of the curve of integration and $P_0$ is the start point. Notice that this is the case for this example:

$\large{\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_C\nabla{f}. \ d{\bf{r}}=12}$

and

$\large{f(2,2)-f(0,0)=4+8-0=12}$.

In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.

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.

Proof

By $\displaystyle\int_{C} {\bf{F}}\cdot{ds}=\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt$,

$\displaystyle\int_C\nabla{f}\cdot \ d{\bf{r}}=\displaystyle\int_a^b\nabla{f}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t) \ dt$.

By the chain rule,

$\frac{d}{dt}(f({\bf{r}}(t))=\nabla{f}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t)$.

Therefore, by the Fundamental Theorem of Calculus,

$\begin{aligned} \displaystyle\int_C\nabla{f}\cdot \ d{\bf{r}}&=\displaystyle\int_a^b\nabla{f}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t) \ dt \\ &=\displaystyle\int_a^b\frac{d}{dt}(f({\bf{r}}(t)) \ dt \\ &=[{f}({\bf{r}}(t))]_{t=a}^{t=b} \\ &={f}({\bf{r}}(b))-{f}({\bf{r}}(a)) \end{aligned}$.

$_\blacksquare$

We know that if $\bf{F}$ is a conservative vector field, there are potential functions $f$ such that $\nabla{f}={\bf{F}}$. Therefore $\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_C\nabla{f}\cdot{d}{\bf{r}}={f}({\bf{r}}(b))-{f}({\bf{r}}(a))$. In other words, just as with the Fundamental Theorem of Calculus, computing the line integral $\displaystyle\int_c{\bf{F}}\cdot{d}{\bf{r}}$, where $\bf{F}$ is conservative, is a two-step process: (1) find a potential function (“antiderivative”) $f$ for $\bf{F}$ and (2) compute the value of $f$ at the endpoints of $C$ and calculate their difference ${f}({\bf{r}}(b))-{f}({\bf{r}}(a))$. 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” 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.

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 ${\bf{F}}$ is conservative and $C$ is a closed curve, then the circulation of ${\bf{F}}$ along $C$ is zero—that is, $\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=0$. To see why this is true, let $f$ be a potential function for ${\bf{F}}$. Since $C$ is a closed curve, the terminal point ${\bf{r}}(b)$ of $C$ is the same as the initial point ${\bf{r}}(a)$ of $C$—that is, ${\bf{r}}(a)={\bf{r}}(b)$. Therefore, by the Fundamental Theorem for Line Integrals,

$\begin{aligned} \displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\oint_C\nabla{f}\cdot{d}{\bf{r}} \\ &=f({\bf{r}}(b))-f({\bf{r}}(a)) \\ &=f({\bf{r}}(b))-f({\bf{r}}(b)) \\ &=0 \end{aligned}$.

Recall that the reason a conservative vector field ${\bf{F}}$ 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 ${\bf{F}}$ in integral $\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}$ as a gravitational field, then the equation $\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}=0$ 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

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Let ${\bf{F}}$ be a vector field with domain $D$. The vector field ${\bf{F}}$ is independent of path (or path independent) if $\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}$ for any paths $C_1$ and $C_2$ in $D$ with the same initial and terminal points.

The second consequence is stated formally in the following theorem.

theorem: path independence of conservative fields

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If ${\bf{F}}$ is a conservative vector field, then ${\bf{F}}$ is independent of path.

Proof

Let $D$ denote the domain of ${\bf{F}}$ and let $C_1$ and $C_2$ be two paths in $D$ with the same initial and terminal points (Figure 5). Call the initial point $P_1$ and the terminal point $P_2$. Since ${\bf{F}}$ is conservative, there is a potential function $f$ for ${\bf{F}}$. By the Fundamental Theorem for Line Integrals,

$\large{\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=f({\bf{P}}_2)-f({\bf{P}}_1)=\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}}$.

Therefore, $\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}$ and ${\bf{F}}$ is independent of path.

$_\blacksquare$

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 ${\bf{F}}$ is conservative, then ${\bf{F}}$ is independent of path. It turns out that if the domain of ${\bf{F}}$ is open and connected, then the converse is also true. That is, if ${\bf{F}}$ is independent of path and the domain of ${\bf{F}}$ is open and connected, then ${\bf{F}}$ 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

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If ${\bf{F}}$ is a continuous vector field that is independent of path and the domain $D$ of ${\bf{F}}$ is open and connected, then ${\bf{F}}$ is conservative.

Proof

We prove the theorem for vector fields in $\mathbb{R}^2$. The proof for vector fields in $\mathbb{R}^3$ is similar. To show that ${\bf{F}}=\langle{P},Q\rangle$ is conservative, we must find a potential function $f$ for ${\bf{F}}$. To that end, let $X$ be a fixed point in $D$. For any point $(x, y)$ in $D$, let $C$ be a path from $X$ to $(x, y)$. Define $f(x, y)$ by $f(x,y)=\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}$. (Note that this definition of $f$ makes sense only because ${\bf{F}}$ is independent of path. If ${\bf{F}}$ was not independent of path, then it might be possible to find another path $C'$ from $X$ to $(x, y)$ such that $\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}\ne\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}$, and in such a case $f(x, y)$ would not be a function.) We want to show that $f$ has the property $\nabla{f}={\bf{F}}$.

Since domain $D$ is open, it is possible to find a disk centered at $(x, y)$ such that the disk is contained entirely inside $D$. Let $(a, y)$ with [latex]a

$\large{f(x,y)=\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}}$.

The first integral does not depend on $x$, so

$\large{f_x=\frac{\partial}{\partial{x}}\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}}$.

If we parameterize $C_2$ by ${\bf{r}}(t)=\langle{t},y\rangle$, $a\leq{t}\leq{x}$, then

$\begin{aligned} f_x&=\frac\partial{\partial{x}}\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^x{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t) \ dt \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^x{\bf{F}}({\bf{r}}(t))\cdot\frac{d}{dt}(\langle{t},y\rangle \ dt \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^x{\bf{F}}({\bf{r}}(t))\cdot\langle1,0\rangle \ dt \\ &=\frac\partial{\partial{x}}\displaystyle\int_a^xP(t,y) \ dt \end{aligned}$.

By the Fundamental Theorem of Calculus (part 1),

$\large{f_x=\frac\partial{\partial{x}}\displaystyle\int_a^xP(t,y) \ dt=P(x,y)}$.

Figure 6. Here, $C_1$ is any path from $X$; to $(a, y)$ that stays inside $D$, and $C_2$ is the horizontal line segment from $(a, y)$ to $(x, y)$.

Example: showing that a vector field is not conservative

Use path independence to show that vector field ${\bf{F}}(x,y)=\langle{x^2}y,y+5\rangle$ is not conservative.

Show Solution

We can indicate that ${\bf{F}}$ is not conservative by showing that ${\bf{F}}$ is not path independent. We do so by giving two different paths, $C_1$ and $C_2$, that both start at $(0, 0)$ and end at $(1, 1)$, and yet $\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}\ne\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}$.

Let $C_1$ be the curve with parameterization $r_1(t)=\langle{t},t\rangle$, $0\leq{t}\leq1$ and let $C_2$ be the curve with parameterization $r_2(t)=\langle{t},t\rangle$, $0\leq{t}\leq1$ (Figure 7). Then

$\begin{aligned} \displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_0^1{\bf{F}}(r_1(t))\cdot{r}_1^\prime(t) \ dt \\ &=\displaystyle\int_0^1\langle{t^3},t+5\rangle\cdot\langle1,1\rangle \ dt =\displaystyle\int_0^1(t^3+t+5) \ dt \\ &=\left[\frac{t^4}4+\frac{t^2}2+5t\right]_0^1=\frac{23}4 \end{aligned}$

and

$\begin{aligned} \displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_0^1{\bf{F}}(r_2(t))\cdot{r}_2^\prime(t) \ dt \\ &=\displaystyle\int_0^1\langle{t^4},t^2+5\rangle\cdot\langle1,2t\rangle \ dt =\displaystyle\int_0^1(t^4+2t^2+10t) \ dt \\ &=\left[\frac{t^5}5+\frac{t^4}2+5t^2\right]_0^1=\frac{57}{10} \end{aligned}$

Since $\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}\ne\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}$, the value of a line integral of ${\bf{F}}$ depends on the path between two given points. Therefore, ${\bf{F}}$ is not independent of path, and ${\bf{F}}$ is not conservative.

Figure 7. Curves $C_1$ and $C_2$ are both oriented from left to right.