Multivariable-Calculus · Unit 18 · Video 2 · Interactive Practice

The Fundamental Theorem for Line Integrals: ∫ ∇f · dr

IKey Formulas

Formula	Name	What it says
$\vec F = \nabla f = f_x\,\hat\imath + f_y\,\hat\jmath$	Gradient field, potential $f$	$M = f_x$ , $N = f_y$ (physics writes $\vec F = -\nabla f$ )
$\displaystyle\int_C \nabla f \cdot d\vec r = f(P_1) - f(P_0)$	Fundamental theorem	$C$ runs from $P_0$ to $P_1$ ; valid only when $\vec F$ is a gradient
$\displaystyle\int_C f_x\,dx + f_y\,dy = \int_C df$	Differential form	Adding the small changes $df$ gives the total change in $f$
$f_x \dfrac{dx}{dt} + f_y \dfrac{dy}{dt} = \dfrac{d}{dt}\, f\big(x(t), y(t)\big)$	Chain rule (the proof)	Then the one-variable fundamental theorem finishes it

Key Insight: $\langle y, x\rangle = \nabla(xy)$ , so the three pieces of the sector are just changes in $xy$ : $0 - 0$ , $\tfrac12 - 0$ and $0 - \tfrac12$ .

IIOnly the Endpoints Matter

For $\vec F = \nabla(xy)$ the work along any path equals $f(P_1) - f(P_0)$ .

Path

IIIThe Sector, Read off the Level Curves

Around the sector the potential $xy$ goes $0 \to 0 \to \tfrac12 \to 0$ , and the work so far is always $f(\text{now}) - f(0,0)$ .

position along C C₂

IVQuiz Questions

Problem 1 · Work from a Known Potential

Given: $\vec F = \nabla f$ with $f(x, y) = x^2 y$ , and $C$ any curve from $(1, 1)$ to $(2, 3)$ — find $\int_C \vec F \cdot d\vec r$ .

✅ Correct!

f(2, 3) - f(1, 1) = 12 - 1 = 11

❌ That is only $f(P_1)$ . The work is the change in

f

: subtract

f(1, 1) = 1

❌ The order is reversed. The theorem gives

f(P_1) - f(P_0)

, end minus start.

❌ Not for a gradient field. The theorem gives the same value

f(P_1) - f(P_0)

for every curve from

P_0

P_1

Show solution

Since $\vec F = \nabla f$ , the fundamental theorem applies:

\int_C \nabla f \cdot d\vec r = f(2, 3) - f(1, 1) = 2^2 \cdot 3 - 1^2 \cdot 1 = 12 - 1 = 11.

Problem 2 · Downhill Work

Given: $\vec F = \nabla f$ with $f(x, y) = x + y^2$ , and $C$ a curve from $P_0 = (0, 2)$ to $P_1 = (1, 0)$ — find $\int_C \vec F \cdot d\vec r$ .

✅ Correct!

f(1, 0) - f(0, 2) = 1 - 4 = -3

: the path moves to lower potential, so the work is negative.

❌ That is $f(P_0) - f(P_1)$ . The work is end minus start,

1 - 4

❌ That adds the two values. The work is the difference

f(P_1) - f(P_0)

❌ Check $f(0, 2)$ .

f = x + y^2

, so

f(0, 2) = 0 + 4 = 4

, not

2

Show solution

\int_C \nabla f \cdot d\vec r = f(1, 0) - f(0, 2) = (1 + 0) - (0 + 4) = -3.

A path that ends at a lower value of the potential gets negative work.

Problem 3 · Spot the Potential

Given: $\vec F = 2xy\,\hat\imath + x^2\,\hat\jmath$ and $C$ the parabola $y = x^2$ from $(0, 0)$ to $(2, 4)$ — find a potential $f$ with $\nabla f = \vec F$ , then $\int_C \vec F \cdot d\vec r$ .

Which function is a potential?

What is the work along $C$ ?

✅ Correct!

f = x^2 y

has

f_x = 2xy

f_y = x^2

, so the work is

f(2, 4) - f(0, 0) = 16

❌ Check both partials. A potential needs

f_x = 2xy

and

f_y = x^2

❌ Evaluate the potential at the endpoints.

f(2, 4) = 2^2 \cdot 4

and

f(0, 0) = 0

Show solution

$f = x^2 y$ gives $f_x = 2xy = M$ and $f_y = x^2 = N$ , so $\vec F = \nabla f$ . Then

\int_C \vec F \cdot d\vec r = f(2, 4) - f(0, 0) = 4 \cdot 4 - 0 = 16.

Check by parametrising: $x = t$ , $y = t^2$ , $0 \le t \le 2$ : $2t \cdot t^2\,dt + t^2 \cdot 2t\,dt = 4t^3\,dt$ , and $\int_0^2 4t^3\,dt = 16$ .

Problem 4 · When the Shortcut Fails

Given: a student computes the work of $\vec G = -y\,\hat\imath + x\,\hat\jmath$ along the upper half of the unit circle from $(1, 0)$ to $(-1, 0)$ as $f(-1, 0) - f(1, 0) = 0$ with $f = xy$ — find what is wrong.

✅ Correct! The theorem needs

\vec G = \nabla f

, and

xy

is the potential of

\langle y, x\rangle

, not of

\langle -y, x\rangle

❌ The arithmetic is fine; the premise is not.

\nabla(xy) = \langle y, x\rangle

, which is not

\vec G

. On the circle

\vec G \cdot \hat T = 1

, so the work is the length

\pi

❌ The order end minus start is right. The problem is that

xy

is not a potential for

\vec G

❌ It applies to any curve from $P_0$ to $P_1$ . The condition is on the field: it must be a gradient.

Show solution

The fundamental theorem holds only for $\vec F = \nabla f$ . Here $\nabla(xy) = y\,\hat\imath + x\,\hat\jmath$ , while $\vec G = -y\,\hat\imath + x\,\hat\jmath$ : the first components differ.

Computed directly: on the unit circle $\vec G$ is tangent with length $1$ and points counterclockwise, the direction of travel, so $\vec G \cdot \hat T = 1$ and

\int_C \vec G \cdot d\vec r = \int_C 1\,ds = \pi.

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