Multivariable-Calculus · Unit 19 · Video 2 · Interactive Practice

Finding a Potential Function by Line Integrals

IKey Formulas

Formula	Name	What it says
$f(x_1, y_1) = \displaystyle\int_C \vec F\cdot d\vec r + f(0, 0)$	Fundamental theorem, backwards	$C$ runs from the origin to $(x_1, y_1)$ ; any such path
$C_1$ : $y = 0$ , $dy = 0$ $\;\Rightarrow\;$ $\displaystyle\int_0^{x_1} M(x, 0)\,dx$	Along the $x$ -axis	Only $M\,dx$ survives
$C_2$ : $x = x_1$ , $dx = 0$ $\;\Rightarrow\;$ $\displaystyle\int_0^{y_1} N(x_1, y)\,dy$	Vertical piece	Only $N\,dy$ survives
$f = \tfrac43 x^3 + y^3 + 4x^2 y + c$	Potential of $\langle 4x^2 + 8xy,\ 3y^2 + 4x^2\rangle$	Check: $f_x = M$ , $f_y = N$

Key Insight: $f(0, 0)$ is a free constant: $\nabla(f + c) = \nabla f$ , so the line integral determines the potential up to $+\,c$ .

IIAny Path from the Origin Gives the Same Value

Each route to $(x_1, y_1)$ splits the work differently, but the total is always $f(x_1, y_1)$ .

Path

IIIA Different Starting Point Shifts Every Value Equally

Starting from a base point $B$ instead of the origin changes $f$ by the constant $-f(B)$ , the same at every point.

IVQuiz Questions

Problem 1 · Two Straight Pieces

Given: $\vec F = \langle 2x + y,\ x + 2y\rangle$ (it passes the test: $M_y = 1 = N_x$ ). Integrate along the $x$ -axis to $(x_1, 0)$ , then vertically to $(x_1, y_1)$ , with $f(0,0) = 0$ . Find $f(x, y)$ .

✅ Correct!

x_1^2

from

C_1

plus

x_1 y_1 + y_1^2

from

C_2

❌ Counted $xy$ twice. On

C_1

y = 0

, so the

y

M = 2x + y

contributes nothing there.

❌ Lost a term. On

C_2

x

is the fixed number

x_1

, not

0

N(x_1, y) = x_1 + 2y

❌ Not quite. Do each piece separately; check by taking the gradient.

Show solution

On $C_1$ : $y = 0$ , $dy = 0$ , so the integral is $\displaystyle\int_0^{x_1} 2x\,dx = x_1^2$ .

On $C_2$ : $x = x_1$ , $dx = 0$ , so the integral is $\displaystyle\int_0^{y_1} (x_1 + 2y)\,dy = x_1 y_1 + y_1^2$ .

Adding: $f(x_1, y_1) = x_1^2 + x_1 y_1 + y_1^2$ , so $f = x^2 + xy + y^2$ . Check: $f_x = 2x + y = M$ , $f_y = x + 2y = N$ .

Problem 2 · What Survives on the x-Axis

Given: $\vec F = \langle 3x^2 y + 2,\ x^3 + 4y\rangle$ . On the piece $C_1$ along the $x$ -axis from $(0,0)$ to $(x_1, 0)$ , find $\displaystyle\int_{C_1} M\,dx + N\,dy$ .

✅ Correct! With

y = 0

M(x, 0) = 2

and

dy = 0

, so the integral is

\int_0^{x_1} 2\,dx = 2x_1

❌ On $C_1$ , $y$ is $0$ . The end point's

y_1

does not appear until the vertical piece.

❌ The $N\,dy$ term vanishes on $C_1$ , because

dy = 0

along the

x

-axis.

❌ Not quite. Set

y = 0

and

dy = 0

before integrating.

Show solution

On $C_1$ , $y = 0$ and $dy = 0$ . The $N\,dy$ term disappears, and $M(x, 0) = 3x^2\cdot 0 + 2 = 2$ :

\int_{C_1} M\,dx + N\,dy = \int_0^{x_1} 2\,dx = 2x_1.

(The vertical piece then gives $\int_0^{y_1}(x_1^3 + 4y)\,dy = x_1^3 y_1 + 2y_1^2$ , so $f = x^3 y + 2x + 2y^2$ .)

Problem 3 · A Potential with an Exponential

Given: $\vec F = \langle y\,e^{xy},\ x\,e^{xy} + 2y\rangle$ , which passes the test. Using the $x$ -axis then the vertical piece, with $f(0,0) = 0$ , find $f(x_1, y_1)$ .

✅ Correct!

C_1

gives

0

;

C_2

gives

\big[e^{x_1 y} + y^2\big]_0^{y_1} = e^{x_1 y_1} + y_1^2 - 1

❌ Check the lower limit. At

y = 0

e^{x_1\cdot 0} = 1

, which must be subtracted when

f(0,0) = 0

❌ That is the integrand, not its antiderivative.

\partial_y\, e^{x_1 y} = x_1 e^{x_1 y}

❌ Not quite. On

C_1

M(x, 0) = 0

; integrate

N(x_1, y)

y

Show solution

On $C_1$ : $y = 0$ , so $M(x, 0) = 0\cdot e^{0} = 0$ and the integral is $0$ .

On $C_2$ : $x = x_1$ , so

\int_0^{y_1}\big(x_1 e^{x_1 y} + 2y\big)\,dy = \Big[e^{x_1 y} + y^2\Big]_0^{y_1} = e^{x_1 y_1} + y_1^2 - 1.

Check: $\nabla(e^{xy} + y^2 - 1) = \langle y e^{xy},\ x e^{xy} + 2y\rangle$ . (Dropping the $-1$ gives another valid potential, but not the one with $f(0,0) = 0$ .)

Problem 4 · Choosing the Constant

Given: $\vec F = \langle 2x + y,\ x + 2y\rangle$ from Problem 1. Find the potential $f$ with $f(1, 1) = 5$ .

✅ Correct!

f(1,1) = 3 + c = 5

gives

c = 2

❌ Not quite. The constant is

f(1,1)

minus the value of

x^2 + xy + y^2

there, which is

3

❌ Check the sign.

3 + c = 5

means

c = +2

❌ Not quite. Potentials differ only by an added constant; scaling

f

changes its gradient.

Show solution

Every potential of $\vec F$ is $x^2 + xy + y^2 + c$ . At $(1,1)$ this is $1 + 1 + 1 + c = 3 + c$ . Setting $3 + c = 5$ gives $c = 2$ :

f = x^2 + xy + y^2 + 2.

Multiplying by a constant is not allowed: $\nabla\big(\tfrac53(x^2+xy+y^2)\big) = \tfrac53\vec F \ne \vec F$ .

Solved: 0 / 4