WVWC-AEM Path Independence and the Fundamental Theorem of Line Integrals

Section 4.2 Path Independence and the Fundamental Theorem of Line Integrals

Given a continuous vector field \(\FF\) defined over a smooth curve \(C:\rr(t), a\leq t\leq b\text{,}\) we can find the corresponding vector line integral using the formula

\begin{equation*} \int_C\FF\cdot\TT\dd{s} = \int_a^b\FF\cdot\dd{\rr}\text{.} \end{equation*}

If \(F(t)\) is an antiderivative of \(\FF\cdot\rrp\text{,}\) then the Fundamental Theorem of Calculus implies that

\begin{equation*} \int_C\FF\cdot\TT\dd{s} = F(b) - F(a)\text{,} \end{equation*}

which suggests that the line integral only depends on the endpoints of the curve. Unfortunately this is not guaranteed as line integrals of vector fields are generally path dependent. Our primary goal in this section is to determine the vector fields \(\FF\) for which our intuition does hold.

Subsection Path Independence

Consider the vector field \(\FF = \mqty[y \amp -x]\) and the curves

\begin{align*} C_1: \amp \rr_1(t) = \mqty[1-t \amp t], 0\leq t\leq 1 \\ C_2: \amp \rr_2(t) = \mqty[\cos(t) \amp \sin(t)], 0\leq t\leq\frac{\pi}{2} \end{align*}

Note that both curves have the same initial and terminal points, however the corresponding line integrals are different:

\begin{align*} \int_{C_1}\FF\cdot\TT\dd{s} \amp = -1 \\ \int_{C_2}\FF\cdot\TT\dd{s} \amp = -\frac{\pi}{2} \end{align*}

This is an example of path dependence.

Definition 4.2.1. Path Dependence and Path Independence.

Let \(\FF\) be a continuous vector field defined on an open set \(U\text{.}\) We say that the line integral \(\int_C\FF\cdot\TT\dd{s}\) is path independent on \(U\) if for any pair of points \(A,B\) in \(U\) and any pair of piecewise smooth curves \(C_1, C_2\) from \(A\) to \(B\) contained entirely within \(U\) we have

\begin{equation*} \int_{C_1}\FF\cdot\TT\dd{s} = \int_{C_2}\FF\cdot\TT\dd{s}\text{.} \end{equation*}

In other words, any line integral of \(\FF\) along any piecewise smooth path contained within \(U\) depends only on the endpoints of the curve \(C\text{.}\) Otherwise, we say that \(\int_C\FF\cdot\TT\dd{s}\) is path dependent.

Graphically, we can see very quickly that line integrals of \(\FF = \mqty[y \amp -x]\) above should be path dependent. For instance, consider the following plot of \(\FF\) over the unit square \([0,1]\times[0,1]\text{:}\)

A plot of the vector field [y, -x]. — Figure 4.2.2. \(\FF = \mqty[y \amp -x]\text{.}\)

Let \(C_1\) denote the circular arc from \((0,1)\) to \((1,0)\) and let \(C_2\) denote the path piecewise linear path from \((0,1)\) to \((1,0)\) determined by following the axes. Then \(\int_{C_1}\FF\cdot\dd{\rr} > 0\) and \(\int_{C_2}\FF\cdot\dd{\rr} = 0\text{.}\) Therefore \(\FF\) must be path-dependent.

Subsection Fundamental Theorem of Line Integrals

The argument above also gives us some intuition about precisely which vector fields should be path independent. In particular, it took advantage of the fact that \(\FF\) rotates (\(\FF\) always rotates clockwise, but what matters is that it rotates at all) to come up with a line integral that is guaranteed to be positive and another line integral that isn't. This suggests that the path independent vector fields should be related to irrotational fields, i.e., fields with zero curl. Such vector fields are also conservative vector fields if they are defined over a “simply connection domain”. This statement is made precise using the Fundamental Theorem of Line Integrals.

Theorem 4.2.3. Fundamental Theorem of Line Integrals.

Let \(\FF\) be a conservative vector field that is continuous over a piecewise smooth curve \(C:\rr(t),a\leq t\leq b\text{.}\) Then

\begin{equation*} \int_{C}\FF\cdot\dd{\rr} = f(\rr(b)) - f(\rr(a)) \end{equation*}

where \(f\) denotes a potential function of \(\FF\text{.}\)

Proof.

The proof is essentially just applying the Fundamental Theorem of Calculus to \(\int_C\FF\cdot\dd{\rr}\text{:}\)

\begin{align*} \int_C\FF\cdot\dd{\rr} \amp = \int_a^b \FF(\rr(t))\cdot\rrp(t)\dd{t} \\ \amp = \int_a^b \del f(\rr(t))\cdot\rrp(t)\dd{t} \\ \amp = \int_a^b \dv{t} [f(\rr(t))]\dd{t} \\ \amp = f(\rr(b)) - f(\rr(a)) \end{align*}

The Fundamental Theorem of Line Integrals shows that conservative vector fields are always path-independent. The converse is also (almost) true: if \(\FF\) is a path-independent vector field over some simply connected domain, then \(\FF\) is conservative within that domain.

Example 4.2.4. Using the Fundamental Theorem of Line Integrals.

Let \(\FF = \mqty[-y \amp -x]\) and let \(C:\rr(t), 0\leq t\leq 2\pi\) where

\begin{equation*} \rr(t) = \mqty[\sin(t) \amp \cos(t/2)]\text{.} \end{equation*}

Find \(\int_C\FF\cdot\dd{\rr}\text{.}\)

Solution.

We could use the computational formula given in Equation (2), but we'll instead make use of Theorem 4.2.3. To do so, we need to prove that \(\FF\) is conservative and then find a corresponding potential function.

To prove that \(\FF\) is conservative, note that its curl is

\begin{equation*} \pdv{Q}{x} - \pdv{P}{y} = 0\text{.} \end{equation*}

Since \(\FF\) is also continuous everywhere, this is enough to guarantee that it's conservative. Now we know that \(\FF\) must have a potential function \(f\) such that \(\del f = \FF\text{.}\) One such possibility is \(f(x, y) = -xy\text{.}\) Therefore

\begin{equation*} \int_C\FF\cdot\dd{\rr} = f(0, -1) - f(0, 1) = 0\text{.} \end{equation*}

The Fundamental Theorem of Line Integrals has another important consequence. If \(C\) is a piecewise smooth closed curve (so its initial and terminal points are the same) and \(\FF\) is conservative in a region containing \(C\text{,}\) then \(\int_C\FF\cdot\dd{\rr} = 0\text{.}\) This also has a nice physical interpretation. Recall that circulation integrals can also be viewed as work integrals, and so the result that line integrals of conservative fields over closed curves are always \(0\) corresponds to the physical law that the work done by conservative forces along closed curves is \(0\) as well.

Advanced Engineering Mathematics Lecture Notes: West Virginia Wesleyan College

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