Conservative Vector Fields
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.3: “Conservative Vector Fields” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-3-conservative-vector-fields |
| Textbook used in class | Stewart, Calculus, Section 16.3: “The Fundamental Theorem for Line Integrals” (Theorem, Examples 1, 2) |
Quick Reference
$\mathbf{F}$ is conservative (path-independent) on a simply-connected domain $D$ if and only if:
- $\mathbf{F} = \nabla f$ for some potential function $f$, OR equivalently
- $\int_C\mathbf{F}\cdot d\mathbf{r} = 0$ for every closed curve $C$ in $D$, OR equivalently (for $\mathbf{F} = \langle P, Q\rangle$ in 2D)
- $\partial P/\partial y = \partial Q/\partial x$ throughout $D$.
Motivation
A force field is conservative if the work it does is independent of the path: it depends only on the starting and ending points. This is why potential energy is well-defined in physics -- if work depended on the path, you could gain or lose energy indefinitely by cycling on a closed loop, violating conservation of energy.
The three equivalent characterizations of conservative fields make it practical to check conservatism and to exploit it: condition 3 is easy to verify, and when satisfied, you can use the FTLI (next skill) to evaluate any line integral instantly.
Key Concepts
1. Path Independence
$\mathbf{F}$ is path-independent on $D$ if $\int_{C_1}\mathbf{F}\cdot d\mathbf{r} = \int_{C_2}\mathbf{F}\cdot d\mathbf{r}$ for any two curves $C_1$, $C_2$ in $D$ with the same endpoints.
Equivalently, $\oint_C\mathbf{F}\cdot d\mathbf{r} = 0$ for all closed curves $C$.
2. Checking Conservatism
For $\mathbf{F} = \langle P, Q\rangle$ on a simply-connected region: $\mathbf{F}$ is conservative if and only if $P_y = Q_x$.
For $\mathbf{F} = \langle P, Q, R\rangle$ in 3D: $\mathbf{F}$ is conservative if and only if $\text{curl}\,\mathbf{F} = \mathbf{0}$ (all mixed partial conditions: $P_y = Q_x$, $P_z = R_x$, $Q_z = R_y$).
Worked Example
Determine if $\mathbf{F}(x,y) = \langle x^2 y, xy^2\rangle$ is conservative. (Adapted from Stewart 16.3.)
$P = x^2 y$, $Q = xy^2$.
$P_y = x^2$, $Q_x = y^2$.
$x^2 \neq y^2$ in general, so $\mathbf{F}$ is NOT conservative. There is no potential function, and the work done by $\mathbf{F}$ depends on the path.
a field is conservative if and only if its components are “nice” functions. Conservatism is a structural property about partial derivatives, not the complexity of the formulas. The field $\mathbf{F} = \langle e^x\sin y, e^x\cos y\rangle$ looks complex but is conservative ($P_y = e^x\cos y = Q_x$). The field $\mathbf{F} = \langle y, x^2\rangle$ looks simple but is NOT conservative ($P_y = 1 \neq 2x = Q_x$).
Common Misconceptions
if $\partial P/\partial y = \partial Q/\partial x$ everywhere, then $\mathbf{F}$ is conservative on any domain.
This is the concept-image-conflicts-definition error. The mixed-partial condition guarantees conservatism only on a simply-connected domain. On a domain with holes, the condition can hold at every point and yet the field may fail to be conservative. The field $\mathbf{F} = \langle -y/(x^2+y^2), x/(x^2+y^2)\rangle$ satisfies $P_y = Q_x$ on $\mathbb{R}^2\setminus\{(0,0)\}$ but has a nonzero circulation around the origin, so no potential function exists on that punctured domain.
Leveled Practice
Problem 1. Determine if $\mathbf{F} = \langle 2xy+1, x^2+e^y\rangle$ is conservative.
Show answer
$P_y = 2x$, $Q_x = 2x$. Equal. So $\mathbf{F}$ is conservative.
Potential function: $f_x = 2xy+1 \implies f = x^2 y + x + g(y)$.
$f_y = x^2 + g'(y) = x^2 + e^y \implies g'(y) = e^y \implies g(y) = e^y$.
$f(x,y) = x^2 y + x + e^y$.