The Fundamental Theorem for Line Integrals

Main idea

When a field is the gradient of some function $f$ , the work along a path is just the change in $f$ from start to finish: $\int_C\nabla f\cdot d\vec r = f(\text{end}) - f(\text{start})$ . The route in between doesn't matter, the same way your change in elevation on a hike only depends on where you started and stopped.

\int_C\nabla {\htmlData{sym=0}{f}}\cdot d\vec r = {\htmlData{sym=0}{f}}\big(\vec r(b)\big) - {\htmlData{sym=0}{f}}\big(\vec r(a)\big)

What the symbols mean

$f$
Potential function
Any f with ∇f = F. For gravity, it's (minus) potential energy.

Explanation

This theorem simply says that for the right kind of field, you can skip the line integral entirely. If the field $\vec F$ is the gradient of some function $f$ (called a potential), then the work along any path is $f$ at the end minus $f$ at the start. Think of a hike: however much you wind up and down the trail, your total change in elevation is just (finish height) − (start height). The formula $\int_C\nabla f\cdot d\vec r = f(\vec r(b)) - f(\vec r(a))$ says exactly that, with $f$ playing the role of elevation.

Take a look at the Fundamental Theorem of Calculus from Calculus 1: $\int_a^b F'(x)\,dx = F(b) - F(a)$ . Integrate a derivative and you get the total change. This is the same statement along a curve, with the gradient playing the role of the derivative.

Why it's true. Follow $f$ along the path: $g(t) = f(\vec r(t))$ . By the multivariable chain rule, $g'(t) = f_x x'(t) + f_y y'(t) = \nabla f(\vec r(t))\cdot\vec r\,'(t)$ , which is exactly the integrand of the line integral. So

\int_C\nabla f\cdot d\vec r = \int_a^b g'(t)\,dt = g(b) - g(a),

by the ordinary Fundamental Theorem. That's the whole proof: the chain rule plus one-variable calculus.

Consequences for gradient fields:

Path independence: every path between the same two points gives the same answer.
Closed loops give zero: $\oint_C\nabla f\cdot d\vec r = 0$ , since the start and end are the same point.

That's why lifting a box and putting it back down does zero net work against gravity, whatever route you take. Gravity is the gradient of (minus) potential energy, and you came back to the same height.

A common mistake. It's tempting to use this shortcut on every line integral. It only works when $\vec F$ really is a gradient. The sailboat's wind in the last lesson gave nonzero work around a closed loop, so no potential could exist for it. How to test a field before trusting the shortcut is the next lesson.

See it

The gradient field of f(x, y) = sin x + y/2, drawn over its level curves, with a straight path and a curved path from A to B. Watch the two work readouts: they differ along the way, since the curved path climbs higher first, but each one is just the change in f so far. Both finish at f(B) − f(A) ≈ 2.51.

Worked examples

Example 1

Evaluate $\int_C\nabla f\cdot d\vec r$ for $f = x^2y$ along any path from $(0, 0)$ to $(2, 3)$ .

1
The field is a gradient, so the path doesn't matter: the integral is $f(\text{end}) - f(\text{start})$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Evaluate $\int_C\nabla(xy)\cdot d\vec r$ from $(1, 1)$ to $(2, 3)$ .

Problem 2+5 XP

$\vec F = \langle 2x, 2y\rangle = \nabla(x^2 + y^2)$ . Find the work from $(0, 0)$ to $(3, 4)$ .

Problem 3+5 XP

What is $\oint_C\nabla f\cdot d\vec r$ around any closed curve?

Problem 4+5 XP

$\vec F = \langle yz, xz, xy\rangle = \nabla(xyz)$ . Find the work from $(0, 0, 0)$ to $(1, 2, 3)$ .

Problem 5+5 XP

Evaluate $\int_C\nabla\left(e^x\sin y\right)\cdot d\vec r$ from $(0, 0)$ to $\left(1, \frac\pi2\right)$ .

Problem 6+5 XP

A 2 kg delivery drone flies a winding route from a rooftop at height 30 m down to a balcony at height 10 m. Gravity is $\vec F = \langle 0, 0, -19.6\rangle = \nabla(-19.6z)$ N. How much work does gravity do on the drone, in joules?

Problem 7+5 XP

Let $f(x, y) = xe^y$ . Evaluate $\int_C\nabla f\cdot d\vec r$ along the curve $\vec r(t) = \left\langle t^2, \sin\frac{\pi t}{2}\right\rangle$ , $0 \le t \le 1$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

The gravitational field of a point mass at the origin is $\vec F = -k\frac{\vec r}{\|\vec r\|^3}$ , with $k = 10$ . Find the work $\vec F$ does on an object moving from $(3, 0, 4)$ to $(0, 0, 2)$ along any path.

Where it's used

Energy conservation is this theorem: in a conservative force field (gravity, electrostatics), the work between two points equals the drop in potential energy, so energy can be stored and recovered.

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