Conservative Fields and the Fundamental Theorem for Line Integrals, multivariable and vector calculus lesson

Carry a backpack from the bottom of a hill to the top. Gravity does the same work on it whether you take the steep trail or the long switchbacks, because only the change in height matters. Fields with this property are called conservative, and for them a line integral collapses to a subtraction of two numbers. The catch is deciding when a field really has this property, and that turns out to depend on the shape of its domain as well as on its formula.

Key terms

Conservative fields and potentials

Definition Let $D$ be an open set in $\mathbb{R}^2$ or $\mathbb{R}^3$ and $\mathbf{F}$ a vector field that is continuous on $D$ . The field is conservative on $D$ if there is a differentiable function $f$ on $D$ with

\nabla f = \mathbf{F} \quad \text{at every point of } D.

Such an $f$ is a potential function for $\mathbf{F}$ . Since its partial derivatives are the continuous components of $\mathbf{F}$ , a potential is automatically $C^1$ on $D$ . In physics the potential energy is usually taken to be $-f$ .

In plain words A conservative field is the gradient of some function. That function plays the role of height, and wherever the field is nonzero it points in the direction where the height increases fastest.

Example The field $\mathbf{F}(x, y) = \langle 2xy, x^2 \rangle$ is conservative on all of $\mathbb{R}^2$ , since the function $f(x, y) = x^2 y$ has $f_x = 2xy$ and $f_y = x^2$ .

Connected open domains

Definition An open set $D$ is connected if any two points of $D$ can be joined by a piecewise regular curve that lies entirely in $D$ . For open sets this agrees with the general notions of connected and path-connected, and the curve can even be taken to be a chain of line segments. A connected open set is often called a domain. If $D$ is open and connected and $\nabla f = \nabla g$ on $D$ , then $f - g$ is constant on $D$ . So on a connected open set a potential is unique up to an added constant.

In plain words A connected set comes in one piece. Line integrals can only compare points that some path in the set reaches, so this is the natural setting for everything below.

Example The set $D = \{ (x, y) : x \ne 0 \}$ is open but not connected, since no path inside it crosses the $y$ -axis. The function equal to $0$ for $x > 0$ and to $1$ for $x < 0$ has gradient $\mathbf{0}$ on $D$ , yet it is not constant. On a connected set this cannot happen.

Potential difference and the fundamental theorem

Definition Let $D$ be an open set, $f$ a $C^1$ function on $D$ , and $C$ an oriented piecewise regular curve in $D$ from the point $A$ to the point $B$ . Then

\int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A).

The right side is the potential difference between the endpoints. To prove it, take a regular piece with parametrization $\mathbf{r}(t)$ , $a \le t \le b$ . By the chain rule

\frac{d}{dt} f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t),

so the fundamental theorem of calculus gives $f(\mathbf{r}(b)) - f(\mathbf{r}(a))$ on that piece. Adding over the pieces, the values at the joints cancel and only $f(B) - f(A)$ remains.

In plain words For a gradient field the line integral is the final value of the potential minus the starting value. The path in between does not matter.

Example For $\mathbf{F} = \langle 2xy, x^2 \rangle = \nabla(x^2 y)$ , along any piecewise regular curve from $(1, 1)$ to $(2, 3)$ ,

\int_C \mathbf{F} \cdot d\mathbf{r} = 2^2 \cdot 3 - 1^2 \cdot 1 = 11.

As a check, the parabola $\mathbf{r}(t) = \langle 1 + t, 1 + 2t^2 \rangle$ , $0 \le t \le 1$ , joins these points, and the direct computation also gives $11$ .

Path independence

Definition Let $\mathbf{F}$ be continuous on an open set $D$ . The integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ is path independent in $D$ if any two oriented piecewise regular curves in $D$ with the same initial point and the same terminal point give the same value.

If $\mathbf{F}$ is continuous on a connected open set $D$ , then the integral of $\mathbf{F}$ is path independent in $D$ if and only if $\mathbf{F}$ is conservative on $D$ .

One direction is the fundamental theorem. For the other, fix a point $A$ in $D$ and define $f(X)$ as the integral of $\mathbf{F}$ along any path in $D$ from $A$ to $X$ . Connectedness makes such a path exist, and path independence makes the value well defined. Because $D$ is open, a path to $(x, y)$ can end with a short horizontal segment, and differentiating along that segment gives $f_x = P$ . A vertical segment gives $f_y = Q$ , and similarly in space.

In plain words Same endpoints, same answer, for every route. That happens exactly when the field has a potential.

Example For $\mathbf{F} = \langle 2xy, x^2 \rangle$ the straight segment $\langle 1 + t, 1 + 2t \rangle$ from $(1, 1)$ to $(2, 3)$ gives

\begin{aligned} &\int_0^1 \big( 2(1 + t)(1 + 2t) \\[0.4em] &\qquad + 2(1 + t)^2 \big) \, dt \\[0.4em] &\quad = \int_0^1 (4 + 10t + 6t^2) \, dt = 11, \end{aligned}

the same as along the parabola. In contrast, the field $\langle -y, x \rangle$ gives work $1$ along the segment from $(1, 0)$ to $(0, 1)$ and $\pi/2$ along the quarter circle, so it is not conservative on any open set containing both paths.

Closed-loop integrals

Definition Let $\mathbf{F}$ be continuous on an open set $D$ . The integral of $\mathbf{F}$ is path independent in $D$ if and only if

\oint_C \mathbf{F} \cdot d\mathbf{r} = 0

for every closed oriented piecewise regular curve $C$ in $D$ . The reason is that two paths $C_1$ and $C_2$ from $A$ to $B$ form the closed curve $C_1$ followed by $-C_2$ , whose integral is the difference of their integrals. Conversely, choosing two points $A$ and $B$ on a loop splits it into two paths from $A$ to $B$ , one of them traversed backwards.

In plain words A conservative field does zero net work around every loop. You end where you started, so the potential difference is zero.

Example For $\mathbf{F} = \langle -y, x \rangle$ the circulation around the unit circle counterclockwise is $2\pi$ . One loop with a nonzero integral is enough to prove that the field is not conservative on any open set containing that loop.

The cross-partial test

Definition Let $\mathbf{F} = \langle P, Q \rangle$ have continuous first partial derivatives on an open set $D$ in $\mathbb{R}^2$ . If $\mathbf{F}$ is conservative on $D$ , then

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \quad \text{on } D.

Indeed a potential $f$ then has continuous second partials, and Clairaut’s theorem gives $f_{xy} = f_{yx}$ . For $\mathbf{F} = \langle P, Q, R \rangle$ with continuous first partials on an open set in $\mathbb{R}^3$ , conservative implies the three equations

R_y = Q_z, \quad P_z = R_x, \quad Q_x = P_y.

These say that the curl $\langle R_y - Q_z, \, P_z - R_x, \, Q_x - P_y \rangle$ is $\mathbf{0}$ , a quantity studied in the next lesson.

In plain words This is a quick local check. If the cross partials differ anywhere in $D$ , the field cannot be conservative on $D$ . Passing the test, though, does not by itself prove the field is conservative.

Example For $\mathbf{F} = \langle y, 2x \rangle$ we get $P_y = 1$ and $Q_x = 2$ . They differ everywhere, so $\mathbf{F}$ is not conservative on any open set.

Curl-free fields on simply connected domains

Definition A connected open set $D$ is simply connected if every closed curve in $D$ can be continuously shrunk to a point within $D$ , that is, deformed through a continuous family of closed curves, all lying in $D$ , that ends at a single point. In the plane this means $D$ has no holes. In space, $\mathbb{R}^3$ with one point removed is still simply connected, while $\mathbb{R}^3$ with the $z$ -axis removed is not, since a loop around the axis cannot be shrunk. The converse of the cross-partial test holds under the following hypotheses.

Let $D$ be an open, simply connected set in $\mathbb{R}^2$ and let $P$ and $Q$ have continuous first partial derivatives on $D$ . If $P_y = Q_x$ everywhere on $D$ , then $\langle P, Q \rangle$ is conservative on $D$ . In $\mathbb{R}^3$ the same holds for $\langle P, Q, R \rangle$ with continuous first partials on an open, simply connected $D$ satisfying all three cross-partial equations, that is with zero curl on $D$ .

In plain words Passing the local test is enough only when the domain has no holes for a loop to wrap around. On a domain with a hole, a field can pass the test at every point and still circulate around the hole.

Example The field

\mathbf{F}(x, y) = \left\langle \frac{-y}{x^2 + y^2}, \, \frac{x}{x^2 + y^2} \right\rangle

has continuous partials on $D = \mathbb{R}^2 \setminus \{(0, 0)\}$ , and both cross partials equal

\frac{y^2 - x^2}{(x^2 + y^2)^2}.

Still, around the unit circle counterclockwise the integrand is $\sin^2 t + \cos^2 t = 1$ and the circulation is $2\pi$ , so $\mathbf{F}$ is not conservative on $D$ . The theorem does not apply because $D$ has a hole at the origin. On the half plane $x > 0$ , which is simply connected, the same field equals $\nabla \arctan(y/x)$ .

Finding potentials

Definition Suppose $\mathbf{F} = \langle P, Q, R \rangle$ has continuous first partials and passes the cross-partial test on a simply connected open set $D$ such as all of $\mathbb{R}^3$ , an open box or an open ball, so a potential exists. To find it, integrate the first component with respect to $x$ ,

f(x, y, z) = \int P \, dx + g(y, z),

where the constant of integration $g$ may depend on the other variables. Differentiate with respect to $y$ and match with $Q$ to find $g$ up to a function $h(z)$ , then match $f_z$ with $R$ to find $h$ up to a constant. In the plane stop after the second step. Finish by checking that $\nabla f = \mathbf{F}$ .

In plain words Undo one partial derivative at a time, and let each new step pin down the unknown part left by the previous one.

Example Let $\mathbf{F} = \langle yz, \, xz + 2y, \, xy + \cos z \rangle$ on $\mathbb{R}^3$ . The cross partials agree, $R_y = x = Q_z$ , $P_z = y = R_x$ and $Q_x = z = P_y$ , and $\mathbb{R}^3$ is simply connected. Integrating $P$ gives $f = xyz + g(y, z)$ . Then $f_y = xz + g_y = xz + 2y$ , so $g = y^2 + h(z)$ . Finally $f_z = xy + h'(z) = xy + \cos z$ , so $h = \sin z$ and

f(x, y, z) = xyz + y^2 + \sin z + K.

The work along any path from $(0, 0, 0)$ to $(1, 2, \pi/2)$ is

f\left(1, 2, \tfrac{\pi}{2}\right) - f(0, 0, 0) = \pi + 5.

Check yourself

Find a potential for $\mathbf{F}(x, y) = \langle 3x^2 y + 2, \, x^3 - 4y \rangle$ on $\mathbb{R}^2$ and the work it does along any path from $(0, 0)$ to $(1, 2)$ .

Show the answer

Both cross partials equal $3x^2$ and $\mathbb{R}^2$ is simply connected. Integrating $P$ gives $f = x^3 y + 2x + g(y)$ , and $f_y = x^3 + g'(y) = x^3 - 4y$ gives $g = -2y^2$ . So $f = x^3 y + 2x - 2y^2$ , and the work is

f(1, 2) - f(0, 0) = 2 + 2 - 8 = -4.

The field $\mathbf{G}(x, y) = \left\langle \dfrac{x}{x^2 + y^2}, \, \dfrac{y}{x^2 + y^2} \right\rangle$ satisfies the cross-partial test on $\mathbb{R}^2 \setminus \{(0, 0)\}$ , which is not simply connected. Is it conservative there?

Show the answer

Yes. The function $f = \frac{1}{2} \ln(x^2 + y^2)$ is $C^1$ on the punctured plane and $\nabla f = \mathbf{G}$ . A domain with a hole means the test cannot guarantee a potential, not that a potential cannot exist.

Let $\mathbf{F}$ be the field $\left\langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right\rangle$ . Find its integral along any piecewise regular path from $(1, 0)$ to $(1, 1)$ that stays in the half plane $x > 0$ .

Show the answer

On the half plane $x > 0$ the field has the potential $\arctan(y/x)$ , so the integral is

\arctan 1 - \arctan 0 = \frac{\pi}{4}.

A path that loops around the origin leaves the half plane, and then the answer can differ by a multiple of $2\pi$ .

Conservative Fields and the Fundamental Theorem for Line Integrals

Key terms

Conservative fields and potentials

Connected open domains

Potential difference and the fundamental theorem

Path independence

Closed-loop integrals

The cross-partial test

Curl-free fields on simply connected domains

Finding potentials

Check yourself

How this lesson connects