Math Core

The Fundamental Theorem of Calculus says $\int_a^b F'(x)\,dx = F(b) - F(a)$ : to integrate a derivative, you only need the values at the endpoints. Line integrals have an exact analogue. When a vector field is a gradient, its line integral depends only on where the curve starts and ends, not on the path in between. This lesson proves that, shows how to recognize such fields, and shows how to find the function they are the gradient of.

The theorem

Recall that $\mathbf{F}$ is conservative if $\mathbf{F} = \nabla f$ for some scalar function $f$ , called a potential function for $\mathbf{F}$ .

Fundamental Theorem for Line Integrals

Let $C$ be a smooth curve given by $\mathbf{r}(t)$ , $a \le t \le b$ , and let $f$ have continuous partial derivatives on an open set containing $C$ . Then

\int_C \nabla f\cdot d\mathbf{r} = f\big(\mathbf{r}(b)\big) - f\big(\mathbf{r}(a)\big).

The proof is one line of the chain rule. Along the curve,

\nabla f\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t) = f_x\frac{dx}{dt} + f_y\frac{dy}{dt} = \frac{d}{dt}\Big[f\big(\mathbf{r}(t)\big)\Big],

so the line integral is $\displaystyle\int_a^b \frac{d}{dt}f(\mathbf{r}(t))\,dt$ , and the ordinary Fundamental Theorem finishes the job. The same argument works in three dimensions, and for curves made of several smooth pieces.

Worked example: Endpoints are all you need

Let $f(x, y) = x^2y + y^3$ . Evaluate $\displaystyle\int_C \nabla f\cdot d\mathbf{r}$ , where $C$ is any smooth curve from $(1, 0)$ to $(2, 1)$ .

No parametrization needed:

\int_C \nabla f\cdot d\mathbf{r} = f(2, 1) - f(1, 0) = (4 + 1) - 0 = 5.

Path independence

Two consequences follow immediately for a conservative field $\mathbf{F}$ :

Path independence. If $C_1$ and $C_2$ have the same initial point and the same terminal point, then $\int_{C_1}\mathbf{F}\cdot d\mathbf{r} = \int_{C_2}\mathbf{F}\cdot d\mathbf{r}$ .
Closed curves give zero. If $C$ is closed (it ends where it starts), then $\oint_C \mathbf{F}\cdot d\mathbf{r} = 0$ .

These two statements are equivalent to each other: going out along $C_1$ and back along $-C_2$ makes a closed loop. The converse is also true on an open connected region: if every closed-loop integral of $\mathbf{F}$ is zero, then $\mathbf{F}$ is conservative. Physically, a conservative force (like gravity) does zero net work around any round trip, which is why mechanical energy is conserved under such forces.

Testing whether a field is conservative

If $\mathbf{F} = \langle P, Q\rangle = \nabla f$ , then $P = f_x$ and $Q = f_y$ . By Clairaut's Theorem, $P_y = f_{xy} = f_{yx} = Q_x$ . So equal cross-partials are necessary. They are also sufficient when the domain has no holes.

Test for conservative fields in the plane

Let $\mathbf{F} = \langle P, Q\rangle$ have continuous first partial derivatives on a simply connected open region $D$ (connected, with no holes). Then

\mathbf{F}\ \text{is conservative on } D \quad\Longleftrightarrow\quad \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\ \text{throughout } D.

Finding a potential function

Once the test passes, recover $f$ by integrating one component and matching the other.

Worked example: Test, find f, evaluate

Let $\mathbf{F}(x, y) = \langle 2xy + 3,\ x^2 + 4y\rangle$ . Show that $\mathbf{F}$ is conservative, find a potential, and evaluate $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ from $(0, 0)$ to $(1, 2)$ .

Test. $P_y = 2x$ and $Q_x = 2x$ , and the domain is all of $\mathbb{R}^2$ , so $\mathbf{F}$ is conservative.

Integrate $P$ in $x$ . $f = \displaystyle\int (2xy + 3)\,dx = x^2y + 3x + g(y)$ . The "constant" of integration can depend on $y$ .

Match $Q$ . $f_y = x^2 + g'(y)$ must equal $x^2 + 4y$ , so $g'(y) = 4y$ and $g(y) = 2y^2$ . Therefore $f(x, y) = x^2y + 3x + 2y^2$ .

Evaluate. $f(1, 2) - f(0, 0) = (2 + 3 + 8) - 0 = 13$ .

In three dimensions the process is the same with one more matching step. (The three-dimensional test uses the curl, which appears two lessons from now.)

Worked example: A potential in space

Evaluate $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle yz,\ xz,\ xy + 2z\rangle$ along any path from $(0, 0, 0)$ to $(1, 2, 3)$ .

Integrate $P$ in $x$ : $f = xyz + g(y, z)$ . Then $f_y = xz + g_y = xz$ , so $g$ depends on $z$ only. Next $f_z = xy + g'(z) = xy + 2z$ , so $g(z) = z^2$ . Hence $f = xyz + z^2$ , and

\int_C \mathbf{F}\cdot d\mathbf{r} = f(1, 2, 3) - f(0, 0, 0) = 6 + 9 = 15.

When the domain has a hole

Consider $\mathbf{F} = \left\langle \dfrac{-y}{x^2 + y^2},\ \dfrac{x}{x^2 + y^2}\right\rangle$ , defined everywhere except the origin. A quick calculation shows $P_y = Q_x = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}$ . Yet around the unit circle, $\mathbf{r}(t) = \langle\cos t, \sin t\rangle$ gives $\mathbf{F}\cdot\mathbf{r}' = \sin^2 t + \cos^2 t = 1$ , so

\oint_C \mathbf{F}\cdot d\mathbf{r} = 2\pi \ne 0.

The field is not conservative on its domain. There is no contradiction: the punctured plane is not simply connected, so the test does not apply.

Common mistake

Equal cross-partials $P_y = Q_x$ prove a field is conservative only on a simply connected region. If the field blows up at a point inside the loop you care about, check the loop integral directly.

Tip

After finding $f$ , check it by computing $\nabla f$ and comparing with $\mathbf{F}$ . It takes seconds and catches a forgotten $g(y)$ .

Practice

Practice 1

Let $f(x, y) = xe^y$ . Evaluate $\displaystyle\int_C \nabla f\cdot d\mathbf{r}$ , where $C$ is any curve from $(1, 0)$ to $(3, \ln 2)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Which vector field is conservative on $\mathbb{R}^2$ ?

Practice 3

Find the potential function $f$ for $\mathbf{F}(x, y) = \langle 2x + y,\ x + 2y\rangle$ that satisfies $f(0, 0) = 0$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Evaluate $\displaystyle\int_C (2x + y)\,dx + (x + 2y)\,dy$ , where $C$ is any path from $(0, 0)$ to $(2, -1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Let $\mathbf{F}(x, y) = \langle y\cos x,\ \sin x + e^y\rangle$ . Evaluate $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ along any path from $(0, 0)$ to $\left(\tfrac{\pi}{2}, 1\right)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Let $\mathbf{F}(x, y, z) = \langle y^2,\ 2xy + z,\ y\rangle$ . Evaluate $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ along any path from $(0, 0, 0)$ to $(2, 1, 3)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

$\mathbf{F}(x, y) = \langle 2xy,\ x^2 - 3y^2\rangle$ . Evaluate $\displaystyle\oint_C \mathbf{F}\cdot d\mathbf{r}$ , where $C$ is the ellipse $x^2 + 4y^2 = 4$ traversed once counterclockwise.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

For $\mathbf{F} = \left\langle \dfrac{-y}{x^2 + y^2},\ \dfrac{x}{x^2 + y^2}\right\rangle$ , $P_y = Q_x$ wherever $\mathbf{F}$ is defined, yet $\oint_C \mathbf{F}\cdot d\mathbf{r} = 2\pi$ around the unit circle. Why doesn't this contradict the test for conservative fields?