The Fundamental Theorem for Line Integrals

Quick answer

If F = ∇f, then the line integral of F along any curve from A to B equals f(B) − f(A): the path does not matter, and around any closed curve the integral is zero. Such fields are called conservative, and f is a potential function. In the plane, a field (P, Q) on a region without holes is conservative exactly when ∂P/∂y = ∂Q/∂x. A potential is found by integrating one component and matching the other.

What you'll learn

Evaluate line integrals of gradient fields from endpoint values
Explain path independence and zero circulation
Test a vector field for being conservative
Find a potential function by integrating components

Integrals that only see the endpoints

The fundamental theorem of calculus says $\int_a^b F'(x)\,dx = F(b) - F(a)$ : integrating a derivative gives the change in the function. Line integrals have the same shortcut, with the gradient as the derivative.

Fundamental theorem for line integrals. Let $C$ be a smooth curve from $A$ to $B$ , and let $f$ have a continuous gradient on $C$ . Then

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

The proof is the chain rule. Along $\mathbf{r}(t)$ , the multivariable chain rule gives $\tfrac{d}{dt} f\big(\mathbf{r}(t)\big) = \nabla f \cdot \mathbf{r}'(t)$ , so the line integral is $\int_a^b \tfrac{d}{dt} f\big(\mathbf{r}(t)\big)\,dt$ , and the ordinary fundamental theorem finishes it.

A field that is a gradient, $\mathbf{F} = \nabla f$ , is called conservative, and $f$ is a potential function for it.

segment
curve

Two paths through the gradient field of x²y

Why path independence and zero loops go together

Suppose $\mathbf{F}$ is conservative, and $C$ is a closed curve, starting and ending at the same point $A$ . The theorem gives $f(A) - f(A) = 0$ . So the integral of a conservative field around every closed curve is zero.

The converse holds too. Take two paths $C_1$ and $C_2$ from $A$ to $B$ . Going out along $C_1$ and back along $C_2$ reversed makes a closed loop, whose integral is $\int_{C_1} \mathbf{F} \cdot d\mathbf{r} - \int_{C_2} \mathbf{F} \cdot d\mathbf{r}$ . If every loop integral is zero, the two paths agree: the integral is independent of path. And a path-independent field is conservative, with potential $f(P) = \int_A^P \mathbf{F} \cdot d\mathbf{r}$ along any path. The three properties, being a gradient, path independence and zero circulation, are one property seen three ways.

That is why the rotation field $(-y, x)$ of line integrals cannot be a gradient: its circulation around the unit circle is $2\pi$ , not $0$ .

A test for conservative fields

If $\mathbf{F} = (P, Q) = \nabla f$ , then $P = f_x$ and $Q = f_y$ , so $\tfrac{\partial P}{\partial y} = f_{xy}$ and $\tfrac{\partial Q}{\partial x} = f_{yx}$ . These mixed partials are equal. So

Test. If $\mathbf{F} = (P, Q)$ is conservative, then $\displaystyle\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ . Conversely, if the equation holds on an open region with no holes, then $\mathbf{F}$ is conservative there.

The condition “no holes” is essential, as the third mistake below shows. In space, the test is that all three pairs of mixed partials match, which the lesson on curl and divergence packages as $\operatorname{curl}\mathbf{F} = \mathbf{0}$ .

Worked examples

Example 2 — Testing and finding a potential

Show that $\mathbf{F} = (3 + 2xy,\ x^2 - 3y^2)$ is conservative and find a potential.

Test: $\tfrac{\partial P}{\partial y} = 2x$ and $\tfrac{\partial Q}{\partial x} = 2x$ . They match on the whole plane, so $\mathbf{F}$ is conservative.

Integrate $P$ in $x$ , with an unknown function of $y$ in place of a constant: $f = 3x + x^2 y + g(y)$ . Then $f_y = x^2 + g'(y)$ must equal $Q = x^2 - 3y^2$ , so $g'(y) = -3y^2$ and $g(y) = -y^3$ . A potential is

f(x, y) = 3x + x^2 y - y^3

Checking: $f_x = 3 + 2xy$ and $f_y = x^2 - 3y^2$ .

Example 5 — Conservation of energy

A particle of mass $m$ moves along $\mathbf{r}(t)$ under a force $\mathbf{F} = -\nabla U$ , where $U$ is the potential energy. Show that kinetic plus potential energy stays constant.

By Newton’s law, $\mathbf{F} = m\mathbf{r}''$ . The work from time $a$ to time $b$ is $\int_a^b m\mathbf{r}'' \cdot \mathbf{r}'\,dt = \Big[\tfrac{1}{2}m|\mathbf{r}'|^2\Big]_a^b$ , the change in kinetic energy $K$ . By the theorem, the same work is $-\big(U(b) - U(a)\big)$ . So $K(b) - K(a) = U(a) - U(b)$ , which rearranges to $K(a) + U(a) = K(b) + U(b)$ . This is where the name “conservative” comes from.

Common mistakes

Practice problems

Find a potential for $\mathbf{F} = (2x,\ 2y)$ .

Answer

$f = x^2 + y^2$

Full solution

$f_x = 2x$ and $f_y = 2y$ . Any constant may be added.
Decide whether $\mathbf{F} = (x + y,\ x - y)$ is conservative. If so, find a potential.

Answer

Yes; $f = \tfrac{x^2}{2} + xy - \tfrac{y^2}{2}$

Full solution

$\tfrac{\partial P}{\partial y} = 1 = \tfrac{\partial Q}{\partial x}$ . Integrating $P$ gives $\tfrac{x^2}{2} + xy + g(y)$ , and matching $Q$ gives $g' = -y$ .
Decide whether $\mathbf{F} = (y,\ -x)$ is conservative.

Answer

No

Full solution

$\tfrac{\partial P}{\partial y} = 1$ and $\tfrac{\partial Q}{\partial x} = -1$ . Its circulation around the unit circle is $-2\pi$ , confirming it.
Evaluate $\int_C (x + y)\,dx + (x - y)\,dy$ along $\mathbf{r}(t) = (t,\ t^3)$ , $0 \le t \le 2$ .

Answer

$-14$

Full solution

With the potential of Exercise 2, the endpoints $(0, 0)$ and $(2, 8)$ give $f(2, 8) - f(0, 0) = 2 + 16 - 32 = -14$ .
Evaluate $\int_C yz\,dx + xz\,dy + xy\,dz$ along any curve from $(0, 0, 0)$ to $(1, 2, 3)$ .

Answer

$6$

Full solution

The field is $\nabla(xyz)$ , so the integral is $1 \cdot 2 \cdot 3 - 0$ .
Find a potential for $\mathbf{F} = \big(e^y,\ x e^y + 1\big)$ and evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ from $(0, 0)$ to $(2, \ln 3)$ .

Answer

$f = x e^y + y$ ; the integral is $6 + \ln 3$

Full solution

Integrating $P$ gives $x e^y + g(y)$ , and matching $Q$ gives $g' = 1$ . Then $f(2, \ln 3) = 2 \cdot 3 + \ln 3$ and $f(0, 0) = 0$ .
Find $\oint_C 2x\,dx + 2y\,dy$ around the ellipse $x^2 + 4y^2 = 4$ .

Answer

$0$

Full solution

The field is $\nabla(x^2 + y^2)$ , defined on the whole plane. Every closed curve gives $0$ .
For Example 4, evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ along the segment from $(0, 0, 0)$ to $\left(1, 2, \tfrac{1}{2}\right)$ .

Answer

$4 + 2e^{3/2}$

Full solution

$f\left(1, 2, \tfrac{1}{2}\right) = 1 \cdot 4 + 2e^{3/2}$ and $f(0, 0, 0) = 0$ .
A student checks $\mathbf{F} = \left(\tfrac{-y}{x^2 + y^2},\ \tfrac{x}{x^2 + y^2}\right)$ , finds $\tfrac{\partial P}{\partial y} = \tfrac{\partial Q}{\partial x}$ , and concludes that $\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ around the unit circle. What went wrong?

Answer

The field is undefined at the origin, a hole inside the circle; the circulation is $2\pi$ .

Full solution

On the unit circle the field equals $(-\sin t, \cos t)$ , the same as the rotation field, so the circulation is $2\pi$ . The test guarantees a potential only on a region without holes, and every region containing the circle but avoiding the origin has one.

Frequently asked questions

What does the fundamental theorem for line integrals say?

If C runs from point A to point B and f has continuous gradient, then the integral of ∇f · dr along C equals f(B) − f(A).

What is a conservative vector field?

A field that is the gradient of some function f, called a potential function. Its line integrals depend only on the endpoints, and its integral around every closed curve is zero.

How do you test whether a plane field is conservative?

Check whether ∂P/∂y = ∂Q/∂x. If the field is defined on an open region with no holes, equality means the field is conservative; inequality always means it is not.

How do you find a potential function?

Integrate P with respect to x, adding an unknown function g(y) instead of a constant. Differentiate the result with respect to y, set it equal to Q, and solve for g.

Why is it called conservative?

For a conservative force F = −∇U, kinetic energy plus potential energy U stays constant along any motion: energy is conserved.

What to learn next

Green's TheoremGreen's theorem turns a line integral around a closed curve into a double integral over the region inside. Use it to compute circulation and area.

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