Line integrals and conservative fields

Two kinds of line integral

Let $C$ be traced by a piecewise $C^1$ curve $\mathbf r(t)$, $a\le t\le b$. For a scalar density $f$ along the curve,

$$\int_C f\,ds=\int_a^b f(\mathbf r(t))\|\mathbf r'(t)\|\,dt.$$

For a vector field $\mathbf F$, the oriented work integral is

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

Reversing orientation leaves the scalar arc-length integral unchanged but reverses the work integral. Both formulas include every traversal: a path that goes around twice is counted twice.

Worked example

The same curve, different integrals

On $\mathbf r(t)=(t,t^2)$, $0\le t\le1$, let $f(x,y)=x$ and $\mathbf F=(y,x)$. The arc-length integral is

$$\int_C x\,ds=\int_0^1t\sqrt{1+4t^2}\,dt=\frac{5\sqrt5-1}{12}.$$

The work integral is

$$\int_C\mathbf F\cdot d\mathbf r=\int_0^1(t^2,t)\cdot(1,2t)\,dt=1.$$

The two objects measure different quantities and cannot be substituted for each other.

Potentials and path independence

Definition

Conservative vector field

A vector field $\mathbf F$ on a domain is conservative if $\mathbf F=\nabla\phi$ for a $C^1$ scalar potential $\phi$ on that domain.

Theorem

Fundamental theorem for line integrals

If $C$ is piecewise $C^1$ from $A$ to $B$, $\phi$ is $C^1$ on an open neighbourhood of $C$, and $\mathbf F=\nabla\phi$ there, then

$$\int_C\mathbf F\cdot d\mathbf r=\phi(B)-\phi(A).$$

In particular, every piecewise $C^1$ closed-curve integral in that neighbourhood is zero, and such integrals depend only on endpoints.

Proof

The chain rule along a path

On each smooth segment, $\nabla\phi(\mathbf r(t))\cdot\mathbf r'(t)=\frac{d}{dt}\phi(\mathbf r(t))$. Integrate and sum over segments. Intermediate endpoint contributions cancel.

Conversely, if $\mathbf F$ is continuous on an open, path-connected domain and its integrals along piecewise $C^1$ paths are path independent, integrating from a fixed base point defines a $C^1$ potential. For a $C^1$ planar field $(P,Q)$, a potential implies $P_y=Q_x$. On a simply connected open domain, this equality is also sufficient under the standard hypotheses. Without the domain condition it is only a local test.

Proof

Constructing the potential from path independence

Let $\Omega\subset\mathbb R^m$ be nonempty, open and path connected, and suppose $F:\Omega\to\mathbb R^m$ is continuous. The points reachable from a base point by polygonal paths form a relatively open subset of $\Omega$: a small ball inside $\Omega$ lets us append one line segment. Its complement is relatively open by the same argument; otherwise a reachable point in such a ball would make its center reachable. Connectedness therefore makes every point reachable. Polygonal paths are piecewise $C^1$.

Fix $x_0\in\Omega$ and define $\phi(x)$ as the integral of $F$ along any such path from $x_0$ to $x$. Path independence makes this a single number. For $x\in\Omega$, choose a ball about $x$ contained in $\Omega$. If $e_i$ is the $i$th coordinate vector and $h$ is sufficiently small, append the coordinate segment to obtain

$$\phi(x+he_i)-\phi(x)=\int_0^h F_i(x+te_i)\,dt.$$

After division by $h\ne0$, continuity gives $\partial_i\phi(x)=F_i(x)$ as $h\to0$ from either side. These partial derivatives are continuous because $F$ is, so $\phi$ is $C^1$ and $\nabla\phi=F$. This proves the converse without assuming differentiability of $F$. If $F$ is $C^1$, the potential is $C^2$, and equality of mixed partials gives $P_y=Q_x$ in the plane.

The full simply connected version of the zero-curl criterion is not proved by this argument: it would first require a proof that the field's path integrals are path independent. The vector-field supplement proves the concrete star-shaped-domain sufficient case, with its regularity assumptions explicit.

Worked example

A zero-curl field with a hole

On $\mathbb R^2\setminus\{0\}$, $\mathbf F=(-y/(x^2+y^2),x/(x^2+y^2))$ satisfies $Q_x-P_y=0$. But on the counterclockwise unit circle $\mathbf r(t)=(\cos t,\sin t)$, $0\le t\le2\pi$, the integral is $\int_0^{2\pi}1\,dt=2\pi$. Hence no globally single-valued potential exists on the punctured plane. The missing origin is essential; the field is not defined there.

Method comparison

Use a direct parametrization for arbitrary fields or curves, carefully respecting orientation. Use a potential when available to reduce work to endpoints. A curl test can help find a potential only after checking domain topology and differentiability. For scalar density use $ds$, not $d\mathbf r$.

Practice

1. Evaluate $\int_C(x,y)\cdot d\mathbf r$ from $(0,0)$ to $(2,1)$ along any piecewise $C^1$ path.
2. Compute $\int_C1\,ds$ on the straight segment from $(0,0)$ to $(3,4)$.
3. Is $(2xy,x^2+1)$ conservative on $\mathbb R^2$? Find a potential and its integral from $(0,0)$ to $(1,2)$.
4. Explain how the work integral changes if an oriented path is traversed in reverse.

Full solutions

1. A potential is $\phi=(x^2+y^2)/2$, so the integral is $\phi(2,1)-\phi(0,0)=5/2$.
2. It is the arc length, $5$.
3. Take $\phi=x^2y+y$; its gradient is the field. The integral is $\phi(1,2)-\phi(0,0)=4$.
4. Reparametrize by $\tilde{\mathbf r}(t)=\mathbf r(a+b-t)$. Its derivative is the negative of the original derivative evaluated at the reversed parameter, and change of variables gives the negative of the original integral.

Synthesis

Scalar line integrals accumulate along length; vector line integrals record oriented work. A potential turns a path calculation into an endpoint difference, but local derivative tests alone do not remove global domain obstructions.