The Central Idea
Suppose a vector field is written as
\[ \mathbf{F} = \nabla f. \]Then \(\mathbf{F}\) is generated by the scalar field \(f\).
If a curve \(C\) runs from point \(A\) to point \(B\), then
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = f(B)-f(A) }. \]The integral depends only on the endpoint values of the potential.
What Is a Conservative Vector Field?
A vector field \(\mathbf{F}\) is called conservative on a region if there exists a scalar field \(f\) such that
\[ \boxed{ \mathbf{F}=\nabla f }. \]The scalar field \(f\) is called a potential function for \(\mathbf{F}\).
Potential Functions
In two dimensions,
\[ \mathbf{F} = \langle P,Q\rangle \]is conservative when there is a scalar field \(f(x,y)\) satisfying
\[ f_x=P \]and
\[ f_y=Q. \]In three dimensions,
\[ \mathbf{F} = \langle P,Q,R\rangle \]requires
\[ f_x=P, \qquad f_y=Q, \qquad f_z=R. \]Potentials Are Not Unique
If \(f\) is a potential function, then
\[ f+C \]has the same gradient for any constant \(C\):
\[ \boxed{ \nabla(f+C)=\nabla f }. \]Therefore, potential functions are determined only up to an additive constant on each connected region.
Path Independence
A line integral is path independent when every suitable path with the same initial and terminal points produces the same value.
If
\[ \mathbf{F}=\nabla f, \]then
\[ \int_C \mathbf{F}\cdot d\mathbf{r} \]is path independent.
Why Conservative Fields Are Path Independent
Along a parameterized curve \(\mathbf{r}(t)\),
\[ \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t). \]By the multivariable chain rule,
\[ \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \frac{d}{dt} f(\mathbf{r}(t)). \]Therefore,
\[ \int_a^b \frac{d}{dt} f(\mathbf{r}(t)) \,dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a)). \]The Fundamental Theorem for Line Integrals
If
\[ \mathbf{F}=\nabla f \]and \(C\) runs from \(A\) to \(B\), then
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = f(B)-f(A) }. \]This is the vector-calculus analogue of the Fundamental Theorem of Calculus.
The Route No Longer Matters
For a conservative field, a complicated curve can be replaced by any other convenient path with the same endpoints.
Better still, once a potential is known, no path parameterization is needed at all.
This can turn a long integration problem into two substitutions.
Example: Endpoint Evaluation
Let
\[ \mathbf{F}(x,y) = \langle2x,2y\rangle. \]A potential is
\[ f(x,y)=x^2+y^2. \]Find the line integral from
\[ A=(1,2) \]to
\[ B=(4,3). \]Then
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = f(4,3)-f(1,2). \]Therefore,
\[ \boxed{ 25-5=20 }. \]Closed-Loop Integrals
If a conservative field is integrated around a closed curve, the initial and final points are identical.
Therefore,
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 }. \]This is one of the most important signatures of conservative behavior.
Closed-Loop Test for Path Independence
On a connected region where the relevant curves are allowed, path independence is equivalent to saying that every closed-loop line integral vanishes:
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 }. \]The logic is that two paths between the same endpoints can be combined into one closed loop.
How Two Paths Form a Closed Loop
Suppose \(C_1\) and \(C_2\) both run from \(A\) to \(B\).
Travel along \(C_1\) from \(A\) to \(B\), then return along \(C_2\) in the reverse direction.
The combined curve is closed.
Therefore,
\[ \oint \mathbf{F}\cdot d\mathbf{r} = \int_{C_1} \mathbf{F}\cdot d\mathbf{r} - \int_{C_2} \mathbf{F}\cdot d\mathbf{r}. \]If every closed integral is zero, the two path integrals must be equal.
Finding a Potential in Two Dimensions
Suppose
\[ \mathbf{F} = \langle P(x,y),Q(x,y)\rangle. \]We seek \(f\) satisfying
\[ f_x=P \]and
\[ f_y=Q. \]Integrate one component, then use the other to determine the missing function.
Example: Recovering a Two-Dimensional Potential
Let
\[ \mathbf{F}(x,y) = \langle 2xy+3, x^2+4y \rangle. \]From
\[ f_x=2xy+3, \]integrate with respect to \(x\):
\[ f(x,y) = x^2y+3x+g(y). \]Differentiate with respect to \(y\):
\[ f_y = x^2+g'(y). \]Match with
\[ x^2+4y. \]Thus,
\[ g'(y)=4y, \]so
\[ g(y)=2y^2+C. \]Therefore,
\[ \boxed{ f(x,y) = x^2y+3x+2y^2+C }. \]Why the Extra Function Is Necessary
When integrating
\[ f_x=P(x,y) \]with respect to \(x\), any function of \(y\) differentiates to zero with respect to \(x\).
Therefore,
\[ \boxed{ f(x,y) = \int P(x,y)\,dx + g(y) }. \]Writing only \(+C\) at this stage can lose essential terms.
Finding a Potential in Three Dimensions
For
\[ \mathbf{F} = \langle P,Q,R\rangle, \]seek \(f\) satisfying
\[ f_x=P, \qquad f_y=Q, \qquad f_z=R. \]Integrate one component and use the other two to determine the unknown functions.
Example: Recovering a Three-Dimensional Potential
Let
\[ \mathbf{F}(x,y,z) = \langle yz+2x, xz+2y, xy+2z \rangle. \]From
\[ f_x=yz+2x, \]integrate with respect to \(x\):
\[ f = xyz+x^2+g(y,z). \]Differentiate with respect to \(y\):
\[ f_y = xz+g_y. \]Match with
\[ xz+2y. \]Therefore,
\[ g_y=2y. \]Integrate with respect to \(y\):
\[ g(y,z)=y^2+h(z). \]Hence,
\[ f = xyz+x^2+y^2+h(z). \]Differentiate with respect to \(z\):
\[ f_z = xy+h'(z). \]Match with
\[ xy+2z. \]Thus,
\[ h'(z)=2z, \]so
\[ h(z)=z^2+C. \]Therefore,
\[ \boxed{ f(x,y,z) = xyz+x^2+y^2+z^2+C }. \]The Mixed-Partial Test in Two Dimensions
If
\[ \mathbf{F} = \langle P,Q\rangle = \nabla f \]and the relevant second partial derivatives are continuous, then
\[ P_y=f_{xy} \]and
\[ Q_x=f_{yx}. \]Therefore,
\[ \boxed{ P_y=Q_x }. \]What the Mixed-Partial Test Can Tell Us
If
\[ P_y\ne Q_x \]at a point, then the field is not conservative on a smooth region containing that point.
If
\[ P_y=Q_x, \]then the field may be conservative, but the geometry of the domain still matters.
Example: Field That Passes the Mixed-Partial Test
Consider
\[ \mathbf{F}(x,y) = \langle 2xy+3, x^2+4y \rangle. \]Then
\[ P_y=2x \]and
\[ Q_x=2x. \]The equality is consistent with the potential found earlier.
Example: Field That Fails the Mixed-Partial Test
Let
\[ \mathbf{F}(x,y) = \langle-y,x\rangle. \]Then
\[ P_y=-1 \]while
\[ Q_x=1. \]Since
\[ P_y\ne Q_x, \]the field is not conservative on any open region where these derivatives apply.
Curl Test in Three Dimensions
If
\[ \mathbf{F}=\nabla f \]and \(f\) is sufficiently smooth, then
\[ \boxed{ \nabla\times\mathbf{F} = \mathbf{0} }. \]Therefore, every smooth conservative field is irrotational.
Why Conservative Fields Have Zero Curl
If
\[ \mathbf{F} = \nabla f, \]then
\[ \nabla\times\mathbf{F} = \nabla\times(\nabla f). \]Equality of mixed partial derivatives gives
\[ \boxed{ \nabla\times(\nabla f) = \mathbf{0} }. \]Zero Curl Is Not Always Enough
A curl-free field is not automatically globally conservative on every possible domain.
The domain must also have suitable topology.
Holes, excluded points, or singularities can prevent one global potential from existing even when curl vanishes everywhere in the domain.
Simply Connected Regions
Informally, a region is simply connected if it has no holes that obstruct continuously shrinking closed loops to points while remaining in the region.
On a simply connected open region, a continuously differentiable curl-free vector field is conservative.
This gives a practical theorem:
\[ \boxed{ \nabla\times\mathbf{F}=\mathbf{0} \quad+\quad \text{suitable domain} \quad\Longrightarrow\quad \mathbf{F}\text{ conservative} }. \]Why Domain Geometry Matters
Vector calculus is not determined by formulas alone.
The domain tells us where the field exists and which paths are allowed.
Removing a point or line can create a hole in the domain and change global path behavior even when local derivative tests remain unchanged elsewhere.
A Classic Punctured-Plane Pattern
Consider
\[ \mathbf{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle. \]The field is undefined at
\[ (0,0). \]Its domain is the punctured plane
\[ \mathbb{R}^2\setminus\{(0,0)\}. \]Away from the origin, the scalar curl is zero.
Yet the field has nonzero circulation around loops enclosing the origin.
Circulation Around the Unit Circle
For the field
\[ \mathbf{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle, \]parameterize the unit circle counterclockwise:
\[ \mathbf{r}(t) = \langle\cos t,\sin t\rangle, \qquad 0\le t\le2\pi. \]On the circle,
\[ \mathbf{F} = \langle-\sin t,\cos t\rangle, \]while
\[ \mathbf{r}'(t) = \langle-\sin t,\cos t\rangle. \]Their dot product is \(1\), so
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} = 2\pi }. \]Therefore the field is not conservative on the punctured plane.
Local Versus Global Information
The condition
\[ \nabla\times\mathbf{F} = \mathbf{0} \]is local derivative information.
Path independence is global information about all allowable paths.
Domain topology is what connects or separates these two levels of information.
Conservative Force Fields
In mechanics, a conservative force field is often written
\[ \boxed{ \mathbf{F} = -\nabla U }. \]Here \(U\) is potential energy.
The negative sign means the force points toward decreasing potential energy.
Work and Potential Energy
If
\[ \mathbf{F} = -\nabla U, \]then
\[ W = \int_C \mathbf{F}\cdot d\mathbf{r}. \]By the Fundamental Theorem for Line Integrals,
\[ W = -[U(B)-U(A)]. \]Therefore,
\[ \boxed{ W = U(A)-U(B) }. \]Potential Difference
If
\[ \mathbf{F}=\nabla f, \]then
\[ f(B)-f(A) \]measures the potential difference associated with the field between the endpoints.
Different applications assign different physical meanings and sign conventions to the potential.
Level Sets of a Potential
A potential function has level sets
\[ f(x,y,z)=c. \]Because
\[ \mathbf{F}=\nabla f, \]the conservative field is normal to regular level sets of the potential.
Thus field lines cross equipotential curves or surfaces orthogonally.
Equipotential Curves
In two dimensions, an equipotential curve satisfies
\[ f(x,y)=c. \]The gradient field
\[ \nabla f \]is perpendicular to the tangent direction of that curve.
Equipotential Surfaces
In three dimensions, a level surface
\[ f(x,y,z)=c \]is an equipotential surface.
The field
\[ \mathbf{F}=\nabla f \]is normal to that surface.
Motion Along an Equipotential
If a particle moves tangent to an equipotential surface, then
\[ d\mathbf{r} \]is perpendicular to
\[ \nabla f. \]Therefore,
\[ \nabla f\cdot d\mathbf{r}=0. \]The potential does not change to first order along that motion.
Testing a Two-Dimensional Field Efficiently
For
\[ \mathbf{F} = \langle P,Q\rangle, \]a practical workflow is:
- check the domain;
- compute \(P_y\);
- compute \(Q_x\);
- compare them;
- if they match and the domain is suitable, recover a potential.
Testing a Three-Dimensional Field Efficiently
For
\[ \mathbf{F} = \langle P,Q,R\rangle, \]first compute
\[ \nabla\times\mathbf{F}. \]If the curl is nonzero, the field is not conservative on the region.
If the curl is zero, inspect the domain before concluding global conservativeness.
Example: Testing and Finding a Potential
Let
\[ \mathbf{F}(x,y) = \langle 3x^2+2y, 2x+6y \rangle. \]First,
\[ P_y=2 \]and
\[ Q_x=2. \]The domain is all of \(\mathbb{R}^2\), so the field is conservative.
Integrate
\[ f_x=3x^2+2y: \] \[ f=x^3+2xy+g(y). \]Differentiate:
\[ f_y=2x+g'(y). \]Match with
\[ 2x+6y. \]Therefore,
\[ g'(y)=6y, \]so
\[ g(y)=3y^2+C. \]Thus,
\[ \boxed{ f(x,y) = x^3+2xy+3y^2+C }. \]Example: Use the Potential to Evaluate a Line Integral
Using the field from the previous example, evaluate the integral from
\[ A=(0,0) \]to
\[ B=(2,1). \]Use
\[ f(x,y)=x^3+2xy+3y^2. \]Then
\[ f(2,1) = 8+4+3 = 15 \]and
\[ f(0,0)=0. \]Therefore,
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = 15 }. \]Checking a Potential
Once a candidate potential is found, differentiate it.
If
\[ f(x,y,z) \]is correct, then
\[ \boxed{ \nabla f=\mathbf{F} }. \]This is the fastest reliable verification.
Conservative Fields and Energy Accounting
Conservative fields allow accumulated work to be tracked using a scalar state function.
Instead of remembering the full path history, we need only compare the potential at the beginning and end.
This is why potential methods are so powerful in mechanics and field theory.
Path-Dependent Fields
If a field is not conservative, then different routes between the same endpoints may produce different line-integral values.
In such cases, there is no single global scalar potential that reduces every line integral to an endpoint difference.
Common Error: Assuming Every Curl-Free Field Is Conservative
Zero curl is not sufficient by itself on arbitrary domains.
Always inspect whether the domain has holes, singularities, or excluded regions.
Common Error: Ignoring the Domain
A formula may satisfy
\[ \nabla\times\mathbf{F} = \mathbf{0} \]wherever it is defined, yet fail to be conservative on the entire domain because the domain is not simply connected.
Common Error: Forgetting the Extra Function
If
\[ f_x=P(x,y), \]then
\[ f = \int P\,dx+g(y), \]not merely
\[ \int P\,dx+C. \]Common Error: Matching Components Incorrectly
In
\[ \mathbf{F} = \langle P,Q,R\rangle = \nabla f, \]the matching is
\[ P=f_x, \qquad Q=f_y, \qquad R=f_z. \]Mixing coordinate derivatives produces an incorrect potential.
Common Error: Parameterizing When a Potential Is Already Known
If the field is conservative and a potential is available, use
\[ \boxed{ f(B)-f(A) }. \]Direct parameterization is usually unnecessary unless the problem explicitly asks for it.
Common Error: Forgetting the Sign Convention for Potential Energy
A mathematical potential field may satisfy
\[ \mathbf{F}=\nabla f. \]In mechanics, potential energy often satisfies
\[ \boxed{ \mathbf{F}=-\nabla U }. \]Keep the convention used by the model explicit.
A Reliable Procedure for Testing Conservativeness in Two Dimensions
- Write \(\mathbf{F}=\langle P,Q\rangle\).
- Determine the domain.
- Compute \(P_y\).
- Compute \(Q_x\).
- If they differ, stop: the field is not conservative.
- If they agree, inspect whether the domain is suitable.
- If appropriate, recover a potential function.
- Differentiate the potential to verify.
A Reliable Procedure for Testing Conservativeness in Three Dimensions
- Write \(\mathbf{F}=\langle P,Q,R\rangle\).
- Determine the domain.
- Compute \(\nabla\times\mathbf{F}\).
- If the curl is nonzero, the field is not conservative.
- If the curl is zero, inspect the domain for holes or singularities.
- If the domain is suitable, recover a potential.
- Verify that \(\nabla f=\mathbf{F}\).
A Reliable Procedure for Finding a Potential
- Choose one component to integrate.
- Integrate with respect to its matching variable.
- Add an unknown function of the remaining variables.
- Differentiate the candidate using another variable.
- Match the corresponding field component.
- Determine the unknown function.
- Repeat if another variable remains.
- Add the final constant \(C\).
- Differentiate the completed potential to verify.
A Reliable Procedure for Conservative Line Integrals
- Confirm that the field is conservative.
- Find or identify a potential function \(f\).
- Identify the starting point \(A\).
- Identify the ending point \(B\).
- Compute \(f(B)-f(A)\).
- Do not integrate along the path unless required.
Example: Full Two-Dimensional Workflow
Determine whether
\[ \mathbf{F}(x,y) = \langle 2xy+y^2, x^2+2xy \rangle \]is conservative on \(\mathbb{R}^2\), find a potential, and evaluate the line integral from \((0,0)\) to \((1,2)\).
First,
\[ P_y=2x+2y \]and
\[ Q_x=2x+2y. \]The derivatives match, and the domain is all of \(\mathbb{R}^2\), so the field is conservative.
Integrate
\[ f_x=2xy+y^2: \] \[ f=x^2y+xy^2+g(y). \]Differentiate:
\[ f_y=x^2+2xy+g'(y). \]Match with
\[ x^2+2xy. \]Thus,
\[ g'(y)=0. \]A potential is
\[ f=x^2y+xy^2. \]Evaluate:
\[ f(1,2)=2+4=6, \] \[ f(0,0)=0. \]Therefore,
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = 6 }. \]Example: Full Three-Dimensional Workflow
Let
\[ \mathbf{F}(x,y,z) = \langle 2x+y, x+2y+z, y+2z \rangle. \]Compute the curl.
\[ R_y-Q_z = 1-1 = 0, \] \[ P_z-R_x = 0-0 = 0, \] \[ Q_x-P_y = 1-1 = 0. \]Therefore,
\[ \nabla\times\mathbf{F} = \mathbf{0}. \]The domain is all of \(\mathbb{R}^3\), so the field is conservative.
Integrate
\[ f_x=2x+y: \] \[ f=x^2+xy+g(y,z). \]Differentiate with respect to \(y\):
\[ f_y=x+g_y. \]Match with
\[ x+2y+z. \]Thus,
\[ g_y=2y+z. \]Integrate:
\[ g=y^2+yz+h(z). \]Then
\[ f=x^2+xy+y^2+yz+h(z). \]Differentiate with respect to \(z\):
\[ f_z=y+h'(z). \]Match with
\[ y+2z. \]Hence,
\[ h(z)=z^2+C. \]Therefore,
\[ \boxed{ f(x,y,z) = x^2+xy+y^2+yz+z^2+C }. \]Practice 1: Mixed-Partial Test
Determine whether
\[ \mathbf{F}(x,y) = \langle 2x+y, x+4y \rangle \]satisfies the mixed-partial condition.
Practice 2: Find a Potential
Find a potential function for the field in Practice 1.
Practice 3: Endpoint Evaluation
Using the field from Practice 1, evaluate the line integral from
\[ (0,0) \]to
\[ (2,1). \]Practice 4: Closed Loop
If
\[ \mathbf{F}=\nabla f, \]what is
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} \]for any closed curve \(C\) lying in the domain?
Practice 5: Nonconservative Field
Determine whether
\[ \mathbf{F}(x,y) = \langle y, -x \rangle \]can be conservative on \(\mathbb{R}^2\).
Practice 6: Three-Dimensional Curl Test
Determine whether
\[ \mathbf{F}(x,y,z) = \langle y+z, x+z, x+y \rangle \]is curl-free.
Practice 7: Find a Three-Dimensional Potential
Find a potential for the field in Practice 6.
Practice 8: Domain Warning
Consider
\[ \mathbf{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle. \]Why is checking \(P_y=Q_x\) alone not enough to conclude that the field is conservative on its full domain?
Practice 9: Potential Energy
Suppose
\[ \mathbf{F} = -\nabla U. \]If
\[ U(A)=10 \]and
\[ U(B)=4, \]how much work does the force field do moving from \(A\) to \(B\)?
Practice 10: Equipotential Direction
If
\[ \mathbf{F}=\nabla f, \]what is the relationship between \(\mathbf{F}\) and a regular level curve or level surface
\[ f=c? \]Practice Answers
Practice 1
Here
\[ P=2x+y, \qquad Q=x+4y. \]Therefore,
\[ P_y=1 \]and
\[ Q_x=1. \]Thus,
\[ \boxed{ P_y=Q_x }. \]Practice 2
From
\[ f_x=2x+y, \]integrate:
\[ f=x^2+xy+g(y). \]Then
\[ f_y=x+g'(y). \]Match with
\[ x+4y. \]Therefore,
\[ g'(y)=4y, \]so
\[ g(y)=2y^2+C. \]Hence,
\[ \boxed{ f(x,y) = x^2+xy+2y^2+C }. \]Practice 3
Use
\[ f(x,y) = x^2+xy+2y^2. \]Then
\[ f(2,1) = 4+2+2 = 8 \]and
\[ f(0,0)=0. \]Therefore,
\[ \boxed{8}. \]Practice 4
\[ \boxed{0}. \]Practice 5
Here
\[ P_y=1 \]while
\[ Q_x=-1. \]Since they differ, the field is not conservative on \(\mathbb{R}^2\).
Practice 6
With
\[ P=y+z, \qquad Q=x+z, \qquad R=x+y, \]we obtain
\[ R_y-Q_z=1-1=0, \] \[ P_z-R_x=1-1=0, \] \[ Q_x-P_y=1-1=0. \]Therefore,
\[ \boxed{ \nabla\times\mathbf{F} = \mathbf{0} }. \]Practice 7
Integrate
\[ f_x=y+z: \] \[ f=xy+xz+g(y,z). \]Then
\[ f_y=x+g_y. \]Match with
\[ x+z, \]giving
\[ g_y=z. \]Therefore,
\[ g=yz+h(z). \]Then
\[ f_z=x+y+h'(z). \]Match with
\[ x+y, \]so
\[ h'(z)=0. \]Thus,
\[ \boxed{ f(x,y,z) = xy+xz+yz+C }. \]Practice 8
The field is undefined at the origin, so its domain has a hole.
The mixed-partial equality is local information and does not guarantee a global potential on the punctured plane.
Practice 9
\[ W = U(A)-U(B) = 10-4 = \boxed{6}. \]Practice 10
Since
\[ \mathbf{F}=\nabla f, \]the field is normal to regular level curves and level surfaces
\[ f=c. \]What to Remember
A conservative field has the form
\[ \boxed{ \mathbf{F}=\nabla f }. \]Its line integrals satisfy
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = f(B)-f(A) }. \]- Conservative fields are gradient fields.
- Potential functions are determined up to an additive constant.
- Conservative line integrals are path independent.
- Closed-loop integrals vanish in conservative fields.
- In two dimensions, \(P_y=Q_x\) is a necessary local test.
- In three dimensions, conservative fields satisfy \(\nabla\times\mathbf{F}=\mathbf{0}\).
- Zero curl plus suitable domain conditions gives conservativeness.
- Domain holes and singularities can destroy global path independence.
- Equipotential curves and surfaces are normal to the gradient field.
- For conservative forces, \(\mathbf{F}=-\nabla U\) is a common physical convention.
Where We Go Next
Conservative fields show how local derivative structure can control global line-integral behavior. The next major theorem applies a similar local-to- global principle in the plane.
The next lesson, Green's Theorem, connects circulation around a closed planar curve with curl accumulated across the enclosed region, and also develops a flux form involving divergence.