Line Integrals from Endpoint Potentials

Why Endpoint Values Can Replace a Path Calculation

A work integral is ordinarily defined by following a path through a vector field. A parametrization records where the path goes and in what direction, and the integral combines the field with the path’s changing position. When the field has a potential function, however, the whole path calculation can be replaced by two evaluations: one at the starting point and one at the ending point.

This shortcut is useful only when its hypotheses are met. In particular, the vector field must be the gradient of a sufficiently smooth potential on an open set containing the path. When that condition holds, any two paths in that set with the same starting and ending points give the same work.

Central rule: If $\mathbf F=\nabla \phi$, where $\phi$ is a continuously differentiable potential on an open neighborhood of a piecewise smooth path $C$ from $A$ to $B$, then $$ \int_C \mathbf F\cdot d\mathbf r=\phi(B)-\phi(A). $$

Work Integrals and Potentials

For a vector field $\mathbf F$ and a smooth parametrized path $\mathbf r(t)$, $a\le t\le b$, the work integral is defined by

\int_C \mathbf F\cdot d\mathbf r = \int_a^b \mathbf F(\mathbf r(t))\cdot \mathbf r'(t)\,dt.

Here, $\mathbf r'(t)$ is the velocity vector along the path. If the path is piecewise smooth, the line integral is the sum of the integrals over its smooth pieces.

A scalar function $\phi$ is a potential for $\mathbf F$ on an open set $U$ if $\phi$ is differentiable there and $\mathbf F=\nabla\phi$ throughout $U$. For the endpoint formula below, we use the stronger, standard regularity condition $\phi\in C^1(U)$: its first partial derivatives exist and are continuous on $U$. This ensures the integrand is continuous on each smooth piece of the path.

Definition: A vector field $\mathbf F$ is conservative on an open set $U$ if there is a potential $\phi\in C^1(U)$ such that $\mathbf F=\nabla\phi$ on $U$.

Why the Endpoint Formula Holds

The key observation is that as a path is traversed, the potential changes at a rate determined by the field’s component in the direction of travel. The chain rule makes that statement precise.

Theorem: Fundamental Theorem for Line Integrals. Let $U$ be open, let $\phi\in C^1(U)$, and suppose $\mathbf F=\nabla\phi$ on $U$. If $C$ is a piecewise smooth path in $U$ from $A$ to $B$, then $$ \int_C \mathbf F\cdot d\mathbf r=\phi(B)-\phi(A). $$

To justify the theorem, first take one smooth piece, parametrized by $\mathbf r(t)$, $a\le t\le b$. The path lies in $U$, so the chain rule applies to the composite function $\phi(\mathbf r(t))$. Because $\phi\in C^1(U)$, this composite has a continuous derivative along the piece.

\frac{d}{dt}\phi(\mathbf r(t)) = \nabla\phi(\mathbf r(t))\cdot\mathbf r'(t) = \mathbf F(\mathbf r(t))\cdot\mathbf r'(t).

Integrating over the parameter interval and applying the one-variable Fundamental Theorem of Calculus gives

\int_a^b \mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt = \int_a^b \frac{d}{dt}\phi(\mathbf r(t))\,dt = \phi(\mathbf r(b))-\phi(\mathbf r(a)).

For a piecewise smooth path, apply this argument on each smooth piece and add the results. Each intermediate endpoint contributes once with a plus sign and once with a minus sign, so those terms cancel. Only the initial and final values remain. This proves the formula.

Worked Example: Two Different Paths with the Same Endpoints

Worked Example: Compare a Straight Path and a Curved Path

Let $\mathbf F(x,y)=(2xy+3,\ x^2+4y)$, and evaluate the work from $A=(0,0)$ to $B=(1,2)$. Compare the straight path $\mathbf r_1(t)=(t,2t)$ and the curved path $\mathbf r_2(t)=(t,2t^2)$, for $0\le t\le1$.

First identify a potential. Set $\phi(x,y)=x^2y+3x+2y^2$. Its partial derivatives are

\frac{\partial\phi}{\partial x}=2xy+3, \qquad \frac{\partial\phi}{\partial y}=x^2+4y.

Thus $\nabla\phi=\mathbf F$. The potential is a polynomial, so it is $C^1$ on all of $\mathbb R^2$. Both paths lie in that domain and have the same endpoints: $\mathbf r_1(0)=\mathbf r_2(0)=(0,0)$ and $\mathbf r_1(1)=\mathbf r_2(1)=(1,2)$.

Evaluate the potential at the endpoints:

\phi(0,0)=0^2(0)+3(0)+2(0)^2=0,

\phi(1,2)=1^2(2)+3(1)+2(2)^2 =2+3+8=13.

Therefore, along either path,

\int_C\mathbf F\cdot d\mathbf r = \phi(1,2)-\phi(0,0) = 13-0 = 13.

The straight and curved paths need not have the same shape or speed. Since the same potential applies on a domain containing both, their work integrals are equal. As a check, the endpoint difference is $13$ for each path, exactly as the theorem requires.

Worked Example: A Three-Dimensional Field

Worked Example: Evaluate Work in Three Dimensions

Let $\mathbf F(x,y,z)=(2x,3y^2,-2)$. Find the work from $A=(0,1,2)$ to $B=(2,-1,0)$ along any piecewise smooth path in $\mathbb R^3$.

Define $\phi(x,y,z)=x^2+y^3-2z$. Its gradient is

\nabla\phi(x,y,z) = \left( \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z} \right) = (2x,3y^2,-2) = \mathbf F(x,y,z).

Because $\phi$ is a polynomial, it is $C^1$ throughout $\mathbb R^3$. The endpoint values are

\phi(0,1,2)=0^2+1^3-2(2)=1-4=-3,

\phi(2,-1,0)=2^2+(-1)^3-2(0)=4-1=3.

Hence the work is

\int_C\mathbf F\cdot d\mathbf r = \phi(2,-1,0)-\phi(0,1,2) = 3-(-3) = 6.

The answer is $6$ for every permitted path between these points. The calculation uses the values of the potential at the endpoints, not the individual coordinates of an intermediate route. Substitution back into the endpoint formula gives $3-(-3)=6$, confirming the result.

Worked Example: Nonzero Field but Zero Work

Worked Example: Equal Potential Values at Distinct Points

Consider the field $\mathbf F(x,y)=(e^x\cos y,-e^x\sin y)$. Evaluate the work from $A=(0,0)$ to $B=(\ln 2,\pi/3)$ along any piecewise smooth path in $\mathbb R^2$.

A potential is $\phi(x,y)=e^x\cos y$, since

\nabla\phi(x,y) = (e^x\cos y,-e^x\sin y) = \mathbf F(x,y).

The potential is continuously differentiable on the whole plane. At the starting point,

\phi(0,0)=e^0\cos 0=1\cdot1=1.

At the ending point, using $e^{\ln 2}=2$ and $\cos(\pi/3)=1/2$,

\phi(\ln 2,\pi/3) = e^{\ln 2}\cos(\pi/3) = 2\left(\frac12\right) = 1.

Consequently,

\int_C\mathbf F\cdot d\mathbf r = 1-1 = 0.

The field is not identically zero along the path; the zero result comes from the equality of the potential values at the two endpoints. Checking the endpoint values confirms that the net change in potential, and therefore the work, is zero.

Conditions and Practical Use

To use endpoint values safely, check that the field has a potential on an open set containing the path and that the potential is $C^1$ there. If comparing two paths, the potential must be defined on a region containing both paths, with the same endpoint values and field relation on that region. It is not enough to find a potential near just one endpoint.

Once those conditions hold, the procedure is short:

Identify a potential.
Use a function $\phi$ satisfying $\nabla\phi=\mathbf F$ on a suitable open set.

Check the endpoints.
Read the starting point $A$ and ending point $B$ from the direction in which the path is traversed.

Subtract in endpoint order.
Compute $\phi(B)-\phi(A)$, then simplify and check the substitutions.

A potential is not unique: adding a constant $c$ gives another potential, because $\nabla(\phi+c)=\nabla\phi$. This does not change the work, since

(\phi(B)+c)-(\phi(A)+c)=\phi(B)-\phi(A).

Common Mistake: Using the Field Value at the Start

A frequent error is to treat the work as the field at the starting point dotted with the displacement. That expression generally does not give the line integral, because the field can change along the path.

Worked Example: Correct a False Shortcut

Use the field and endpoints from the first example. The displacement is $B-A=(1,2)$, and $\mathbf F(0,0)=(3,0)$. The tempting calculation gives

\mathbf F(A)\cdot(B-A) = (3,0)\cdot(1,2) = 3.

This is not the work: the correct endpoint calculation gives $\phi(B)-\phi(A)=13$. The value $3$ only uses the field at one point and misses how the field acts elsewhere along the path. The correction is to use a potential and evaluate it at both endpoints. Indeed, $\phi(1,2)-\phi(0,0)=13-0=13$, as required.

The endpoint method does not apply merely because a field is written down or because one can compute its values at the endpoints. It requires a potential satisfying the stated conditions on a region containing the path. If that requirement is unavailable, the line integral must be evaluated by another valid method.

Check Your Understanding

Use the endpoint formula where its hypotheses hold.

For $\mathbf F=\nabla\phi$ on an open set containing a path from $A$ to $B$, which potential value is subtracted from which?
Why does the $C^1$ condition on the potential allow the chain-rule and Fundamental Theorem of Calculus argument?
Two different paths in the same suitable domain join the same endpoints. What does the endpoint formula imply about their work integrals?
If $\phi(A)=5$ and $\phi(B)=-2$, what is the work from $A$ to $B$?
Why is $\mathbf F(A)\cdot(B-A)$ generally not a valid replacement for the work integral?