Line Integrals & Conservative Fields

A parameterized curve in $\mathbb{R}^n$ is a continuous function $\mathbf{r}: [a, b] \to \mathbb{R}^n$. The curve is smooth if $\mathbf{r}$ is $C^1$ and $\mathbf{r}'(t) \neq \mathbf{0}$ for all $t \in (a, b)$ — the velocity never vanishes, so the particle never stops. The curve is piecewise smooth if $[a, b]$ can be partitioned into finitely many subintervals on each of which $\mathbf{r}$ is smooth.

The arc length of a smooth curve $\mathbf{r}: [a, b] \to \mathbb{R}^n$ is

$L(C) = \int_a^b \|\mathbf{r}'(t)\|\,dt.$

The arc length element is $ds = \|\mathbf{r}'(t)\|\,dt$, representing the infinitesimal distance traveled in an infinitesimal time $dt$.

If $\phi: [\alpha, \beta] \to [a, b]$ is a $C^1$ bijection with $\phi'(\tau) > 0$ (orientation-preserving), then $\tilde{\mathbf{r}}(\tau) = \mathbf{r}(\phi(\tau))$ traces the same curve in the same direction. By the substitution rule (Topic 14, Theorem 1):

$\int_\alpha^\beta \|\tilde{\mathbf{r}}'(\tau)\|\,d\tau = \int_a^b \|\mathbf{r}'(t)\|\,dt.$

Arc length does not depend on how fast we traverse the curve — it is a geometric property of the curve itself.

Let $\mathbf{r}(t) = (R\cos t,\; R\sin t)$ for $t \in [0, 2\pi]$. Then $\mathbf{r}'(t) = (-R\sin t,\; R\cos t)$ and $\|\mathbf{r}'(t)\| = R$. The arc length is

$L = \int_0^{2\pi} R\,dt = 2\pi R.$

The constant speed $R$ means the particle moves uniformly — the arc length is simply speed times time.

A helix $\mathbf{r}(t) = (\cos t, \sin t, t)$ for $t \in [0, 2\pi]$ climbs one full turn. We have $\|\mathbf{r}'(t)\| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$, giving $L = 2\pi\sqrt{2}$. The vertical climb adds length to the horizontal circle.

The parabola $\mathbf{r}(t) = (t, t^2)$ for $t \in [0, 1]$ has $\|\mathbf{r}'(t)\| = \sqrt{1 + 4t^2}$. The arc length integral $\int_0^1 \sqrt{1 + 4t^2}\,dt$ requires the $\sinh^{-1}$ formula or numerical quadrature — not every arc length computation is elementary.

Let $C$ be a smooth curve parameterized by $\mathbf{r}: [a, b] \to \mathbb{R}^n$, and let $f: C \to \mathbb{R}$ be continuous. The scalar line integral of $f$ over $C$ is:

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

This is a Riemann integral (Topic 7) of the composite function $t \mapsto f(\mathbf{r}(t)) \cdot \|\mathbf{r}'(t)\|$ over $[a, b]$.

The scalar line integral $\int_C f\,ds$ is independent of the parameterization of $C$ (including orientation). Any two smooth parameterizations of the same curve give the same value.

Let $\mathbf{r}: [a, b] \to \mathbb{R}^n$ and $\tilde{\mathbf{r}} = \mathbf{r} \circ \phi: [\alpha, \beta] \to \mathbb{R}^n$ with $\phi$ a $C^1$ bijection. By the chain rule, $\tilde{\mathbf{r}}'(\tau) = \mathbf{r}'(\phi(\tau)) \cdot \phi'(\tau)$, so $\|\tilde{\mathbf{r}}'(\tau)\| = \|\mathbf{r}'(\phi(\tau))\| \cdot |\phi'(\tau)|$. Then:

$\int_\alpha^\beta f(\tilde{\mathbf{r}}(\tau))\,\|\tilde{\mathbf{r}}'(\tau)\|\,d\tau = \int_\alpha^\beta f(\mathbf{r}(\phi(\tau)))\,\|\mathbf{r}'(\phi(\tau))\|\,|\phi'(\tau)|\,d\tau.$

By the substitution rule (Topic 14, Theorem 1) with $t = \phi(\tau)$, this equals $\int_a^b f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt$. The absolute value $|\phi'(\tau)|$ ensures the result holds regardless of whether $\phi$ preserves or reverses orientation.

A wire follows $C: \mathbf{r}(t) = (\cos t, \sin t)$ for $t \in [0, \pi]$ with density $f(x, y) = y$. Then:

$\int_C f\,ds = \int_0^\pi \sin t \cdot 1\,dt = [-\cos t]_0^\pi = 2.$

The wire is heaviest at the top ($y = 1$) and weightless at the endpoints ($y = 0$). The total mass is 2.

The average value of $f$ over $C$ is $\bar{f} = \frac{1}{L(C)} \int_C f\,ds$, directly analogous to $\bar{f} = \frac{1}{b-a} \int_a^b f(x)\,dx$ from single-variable calculus (Topic 7).

Let $C$ be a smooth curve parameterized by $\mathbf{r}: [a, b] \to \mathbb{R}^n$ and $\mathbf{F}: \mathbb{R}^n \to \mathbb{R}^n$ a continuous vector field. The vector line integral (or work integral) of $\mathbf{F}$ along $C$ is:

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

In $\mathbb{R}^2$, writing $\mathbf{F} = (P, Q)$ and $d\mathbf{r} = (dx, dy)$, this becomes $\int_C P\,dx + Q\,dy = \int_a^b [P(\mathbf{r}(t))\,x'(t) + Q(\mathbf{r}(t))\,y'(t)]\,dt$.

Reversing the curve $C$ — traversing from $\mathbf{r}(b)$ to $\mathbf{r}(a)$ — negates the integral: $\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}$. This is because $\mathbf{r}'(t)$ reverses sign under orientation reversal, and the dot product is linear. By contrast, the scalar line integral $\int_C f\,ds$ is orientation-independent because $\|\mathbf{r}'(t)\|$ is always positive.

The vector line integral is independent of the orientation-preserving parameterization. Any two parameterizations that traverse $C$ in the same direction yield the same value. The proof is the same substitution argument as Proposition 1, but without the absolute value — the sign of $\phi'(\tau)$ cancels the reversed limits, preserving the integral’s value.

Let $C$, $C_1$, $C_2$ be piecewise-smooth curves, $\mathbf{F}$, $\mathbf{G}$ continuous vector fields, and $\alpha, \beta \in \mathbb{R}$.

1. Linearity: $\int_C (\alpha\mathbf{F} + \beta\mathbf{G}) \cdot d\mathbf{r} = \alpha\int_C \mathbf{F} \cdot d\mathbf{r} + \beta\int_C \mathbf{G} \cdot d\mathbf{r}$.

2. Additivity over path concatenation: If $C = C_1 + C_2$ (the endpoint of $C_1$ is the start of $C_2$), then $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{C_1} \mathbf{F} \cdot d\mathbf{r} + \int_{C_2} \mathbf{F} \cdot d\mathbf{r}$.

3. Orientation reversal: $\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}$.

Let $\mathbf{F} = (3, 4)$ and $C$ be the line segment from $(0, 0)$ to $(2, 1)$: $\mathbf{r}(t) = (2t, t)$ for $t \in [0, 1]$. Then $\mathbf{r}'(t) = (2, 1)$ and:

$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (3 \cdot 2 + 4 \cdot 1)\,dt = 10.$

For a constant field, the work equals $\mathbf{F} \cdot \Delta\mathbf{r} = (3, 4) \cdot (2, 1) = 10$ — the integral is just a dot product.

Let $\mathbf{F}(x, y) = (x, y)$ and $C$ be the upper semicircle from $(1, 0)$ to $(-1, 0)$: $\mathbf{r}(t) = (\cos t, \sin t)$ for $t \in [0, \pi]$. Then $\mathbf{r}'(t) = (-\sin t, \cos t)$:

$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^\pi [(\cos t)(-\sin t) + (\sin t)(\cos t)]\,dt = \int_0^\pi 0\,dt = 0.$

The radial field is everywhere perpendicular to the circle — it does zero work along any circular arc.

Let $\mathbf{F}(x, y) = (-y, x)$. Compute the work along two different paths from $(1, 0)$ to $(0, 1)$:

Path $C_1$: Line segment $\mathbf{r}(t) = (1 - t, t)$ for $t \in [0, 1]$. Then $\mathbf{r}'(t) = (-1, 1)$ and $\mathbf{F}(\mathbf{r}(t)) = (-t, 1 - t)$:

$\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 [(-t)(-1) + (1-t)(1)]\,dt = \int_0^1 1\,dt = 1.$

Path $C_2$: Quarter-circle $\mathbf{r}(t) = (\cos t, \sin t)$ for $t \in [0, \pi/2]$. Then $\mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t) = \mathbf{r}'(t)$:

$\int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_0^{\pi/2} (\sin^2 t + \cos^2 t)\,dt = \frac{\pi}{2}.$

Different paths, different integrals ($1 \neq \pi/2$). This field is not conservative.

A vector field $\mathbf{F}: D \to \mathbb{R}^n$ (where $D \subseteq \mathbb{R}^n$ is open and connected) is conservative if there exists a $C^1$ function $f: D \to \mathbb{R}$ such that $\mathbf{F} = \nabla f$ on $D$. The function $f$ is called a potential function (or scalar potential) for $\mathbf{F}$.

If $f$ and $g$ are both potential functions for $\mathbf{F}$ on a connected domain $D$, then $\nabla(f - g) = \mathbf{0}$ on $D$, so $f - g$ is constant. This follows from the fact that a function with zero gradient on a connected domain must be constant — a consequence of the Mean Value Theorem (Topic 6).

Let $f: D \to \mathbb{R}$ be a $C^1$ function on an open set $D \subseteq \mathbb{R}^n$, and let $C$ be a piecewise-smooth curve in $D$ from $\mathbf{a}$ to $\mathbf{b}$. Then:

$\int_C \nabla f \cdot d\mathbf{r} = f(\mathbf{b}) - f(\mathbf{a}).$

Define $g(t) = f(\mathbf{r}(t))$ for $t \in [a, b]$. By the chain rule (Topic 5 for scalar functions, Topic 10 for the multivariable version):

$g'(t) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t).$

This is the key identity: the integrand of the line integral is exactly $g'(t)$. By the Fundamental Theorem of Calculus (Topic 7, Theorem 2):

$\int_C \nabla f \cdot d\mathbf{r} = \int_a^b g'(t)\,dt = g(b) - g(a) = f(\mathbf{r}(b)) - f(\mathbf{r}(a)). \qquad \square$

Let $\mathbf{F}(x, y) = (2x, 2y)$ with potential $f(x, y) = x^2 + y^2$. For any curve $C$ from $(1, 0)$ to $(0, 3)$:

$\int_C \mathbf{F} \cdot d\mathbf{r} = f(0, 3) - f(1, 0) = 9 - 1 = 8.$

No parameterization needed — just endpoint evaluation.

The radial field $\mathbf{F}(x, y) = (x, y) = \nabla\!\left(\frac{x^2 + y^2}{2}\right)$. The curve from $(1, 0)$ to $(-1, 0)$ gives:

$f(-1, 0) - f(1, 0) = \frac{1}{2} - \frac{1}{2} = 0.$

The Gradient Theorem reproduces Example 7’s result — zero work — without any integration.

A vector field $\mathbf{F}: D \to \mathbb{R}^n$ has path-independent line integrals if $\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_{C_2} \mathbf{F} \cdot d\mathbf{r}$ for every pair of piecewise-smooth curves $C_1, C_2$ in $D$ that share the same endpoints.

A curve $C$ parameterized by $\mathbf{r}: [a, b] \to \mathbb{R}^n$ is closed if $\mathbf{r}(a) = \mathbf{r}(b)$. We write $\oint_C$ for integrals over closed curves.

Let $\mathbf{F}: D \to \mathbb{R}^n$ be a continuous vector field on an open connected domain $D$. The following are equivalent:

1. $\mathbf{F}$ is conservative ($\mathbf{F} = \nabla f$ for some $C^1$ function $f$).
2. $\int_C \mathbf{F} \cdot d\mathbf{r}$ is path-independent in $D$.
3. $\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ for every piecewise-smooth closed curve $C$ in $D$.

(1) $\Rightarrow$ (2): Immediate from the Gradient Theorem — the integral equals $f(\mathbf{b}) - f(\mathbf{a})$, which depends only on the endpoints.

(2) $\Rightarrow$ (3): If $C$ is closed, its start and end points coincide: $\mathbf{a} = \mathbf{b}$. Split $C$ at any interior point $\mathbf{p}$ into two curves $C_1$ (from $\mathbf{a}$ to $\mathbf{p}$) and $C_2$ (from $\mathbf{p}$ to $\mathbf{a}$). By path independence, $\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_{-C_2} \mathbf{F} \cdot d\mathbf{r} = -\int_{C_2} \mathbf{F} \cdot d\mathbf{r}$, so $\oint_C = \int_{C_1} + \int_{C_2} = 0$.

(3) $\Rightarrow$ (1): Fix a base point $\mathbf{a} \in D$ and define $f(\mathbf{x}) = \int_C \mathbf{F} \cdot d\mathbf{r}$ where $C$ is any path from $\mathbf{a}$ to $\mathbf{x}$. The zero-circulation condition ensures this is well-defined (different paths give the same value).

To show $\nabla f = \mathbf{F}$: compute $\frac{\partial f}{\partial x_i}(\mathbf{x})$ by choosing the path to $\mathbf{x} + h\mathbf{e}_i$ as any path from $\mathbf{a}$ to $\mathbf{x}$, then a straight segment from $\mathbf{x}$ to $\mathbf{x} + h\mathbf{e}_i$. The difference $f(\mathbf{x} + h\mathbf{e}_i) - f(\mathbf{x}) = \int_0^h F_i(\mathbf{x} + s\mathbf{e}_i)\,ds$. By the FTC (Topic 7), dividing by $h$ and taking $h \to 0$ gives $\frac{\partial f}{\partial x_i}(\mathbf{x}) = F_i(\mathbf{x})$. $\square$

An open connected domain $D \subseteq \mathbb{R}^2$ is simply connected if every closed curve in $D$ can be continuously shrunk to a point without leaving $D$. Informally: $D$ has no holes. The full plane $\mathbb{R}^2$ is simply connected; the punctured plane $\mathbb{R}^2 \setminus \{(0,0)\}$ is not.

Let $\mathbf{F} = (P, Q): D \to \mathbb{R}^2$ be a $C^1$ vector field on an open, simply connected domain $D \subseteq \mathbb{R}^2$. Then $\mathbf{F}$ is conservative if and only if:

$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \quad \text{on } D.$

The condition $\partial P / \partial y = \partial Q / \partial x$ says $\mathbf{F}$ is closed — its 1-form $P\,dx + Q\,dy$ is closed. On simply connected domains, closed = exact (= conservative). On domains with holes, closed $\neq$ exact. The gap is topological, not analytical — it is measured by de Rham cohomology $H^1_{\text{dR}}$ (→ Smooth Manifolds on formalML).

Let $\mathbf{F}(x, y) = (2xy + y^2,\; x^2 + 2xy)$. Check: $\frac{\partial P}{\partial y} = 2x + 2y = \frac{\partial Q}{\partial x}$. Conservative.

Find $f$: from $f_x = 2xy + y^2$ we get $f(x, y) = x^2 y + xy^2 + g(y)$. Then $f_y = x^2 + 2xy + g'(y) = x^2 + 2xy$ forces $g'(y) = 0$, so $f(x, y) = x^2 y + xy^2 + C$.

The vortex field $\mathbf{F}(x, y) = \left(\frac{-y}{x^2+y^2},\; \frac{x}{x^2+y^2}\right)$ on $D = \mathbb{R}^2 \setminus \{(0,0)\}$.

Check the exactness condition: $\frac{\partial P}{\partial y} = \frac{y^2 - x^2}{(x^2+y^2)^2} = \frac{\partial Q}{\partial x}$. The condition holds, yet the circulation around the unit circle is:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = 2\pi \neq 0.$

The catch: $D$ is not simply connected — it has a hole at the origin. The “potential function” $f(x,y) = \arctan(y/x)$ is multi-valued; it gains $2\pi$ each time we circle the origin. The vortex field is the canonical example showing that topology matters.

Let $D \subseteq \mathbb{R}^2$ be a bounded region with piecewise-smooth boundary $\partial D$ oriented counterclockwise. Let $\mathbf{F} = (P, Q): \bar{D} \to \mathbb{R}^2$ be a $C^1$ vector field. Then:

$\oint_{\partial D} P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dA.$

We show $\oint_{\partial D} P\,dx = -\iint_D \frac{\partial P}{\partial y}\,dA$ and $\oint_{\partial D} Q\,dy = \iint_D \frac{\partial Q}{\partial x}\,dA$ separately, then add.

Proof that $\oint P\,dx = -\iint \frac{\partial P}{\partial y}\,dA$: Let $D$ be a Type I region: $a \le x \le b$, $g_1(x) \le y \le g_2(x)$. The right side is:

$-\iint_D \frac{\partial P}{\partial y}\,dA = -\int_a^b \int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y}(x, y)\,dy\,dx = -\int_a^b \bigl[P(x, g_2(x)) - P(x, g_1(x))\bigr]\,dx.$

The boundary $\partial D$ traversed counterclockwise consists of: the bottom curve $C_1: y = g_1(x)$ from $x = a$ to $x = b$, the right side, the top curve $C_3: y = g_2(x)$ from $x = b$ to $x = a$, and the left side. On $C_1$: $\int_{C_1} P\,dx = \int_a^b P(x, g_1(x))\,dx$. On $C_3$ (reversed): $\int_{C_3} P\,dx = -\int_a^b P(x, g_2(x))\,dx$. On the vertical sides, $dx = 0$, so their contributions vanish. Adding:

$\oint_{\partial D} P\,dx = \int_a^b P(x, g_1(x))\,dx - \int_a^b P(x, g_2(x))\,dx = -\int_a^b [P(x, g_2(x)) - P(x, g_1(x))]\,dx.$

The proof for $Q\,dy$ is analogous using a Type II description of $D$. For general regions, decompose into Type I and Type II pieces; interior boundary contributions cancel in pairs. $\square$

Direct computation: $\oint_C (-y\,dx + x\,dy)$ with $\mathbf{r}(t) = (\cos t, \sin t)$ gives:

$\int_0^{2\pi} [\sin^2 t + \cos^2 t]\,dt = 2\pi.$

Via Green’s theorem: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$, so:

$\iint_D 2\,dA = 2 \cdot \pi = 2\pi.$

Both give $2\pi$. The rotation field $(-y, x)$ has constant curl 2 — every point in the disk contributes equally to the circulation.

Setting $P = -y/2$, $Q = x/2$ gives $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{2} + \frac{1}{2} = 1$, so:

$A(D) = \iint_D dA = \frac{1}{2}\oint_{\partial D} (x\,dy - y\,dx).$

This is the Shoelace formula for polygonal areas (a special case when $\partial D$ is a polygon) and the formula used by mechanical planimeters.

Green’s theorem says that the “total rotation inside $D$” equals the “total circulation around $\partial D$.” The interior quantity (curl) and the boundary quantity (circulation) are related by an exact balance. This is the prototype of all conservation laws in physics — and the 2D instance of Stokes’ theorem, which will be generalized to surfaces and volumes in Surface Integrals & the Divergence Theorem.

For $\mathbf{F} = (P, Q): D \to \mathbb{R}^2$ of class $C^1$, the 2D curl (or scalar curl) is:

$\operatorname{curl}\mathbf{F}(x, y) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}.$

This is the $\hat{\mathbf{k}}$-component of the 3D curl $\nabla \times \mathbf{F}$, with $\mathbf{F}$ viewed as the 3D field $(P, Q, 0)$.

Let $\mathbf{F}$ be $C^1$ at $\mathbf{p}$, and let $C_r$ be the circle of radius $r$ centered at $\mathbf{p}$, oriented counterclockwise. Then:

$\operatorname{curl}\mathbf{F}(\mathbf{p}) = \lim_{r \to 0} \frac{1}{\pi r^2} \oint_{C_r} \mathbf{F} \cdot d\mathbf{r}.$

The curl is the circulation per unit area in the limit of infinitesimally small loops.

By Green’s theorem, $\oint_{C_r} \mathbf{F} \cdot d\mathbf{r} = \iint_{D_r} \operatorname{curl}\mathbf{F}\,dA$. By the Mean Value Theorem for double integrals (Topic 13), $\iint_{D_r} \operatorname{curl}\mathbf{F}\,dA = \operatorname{curl}\mathbf{F}(\mathbf{p}_r) \cdot \pi r^2$ for some $\mathbf{p}_r \in D_r$. As $r \to 0$, $\mathbf{p}_r \to \mathbf{p}$ and continuity of $\operatorname{curl}\mathbf{F}$ gives the limit. $\square$

Theorem 4 restated: on a simply connected domain, $\mathbf{F}$ is conservative if and only if $\operatorname{curl}\mathbf{F} = 0$ everywhere. Green’s theorem explains why: if $\operatorname{curl}\mathbf{F} = 0$ on $D$, then $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \operatorname{curl}\mathbf{F}\,dA = 0$ for every closed curve $C$ bounding a region in $D$. On simply connected domains, every closed curve bounds a region in $D$, so the zero-circulation condition (Theorem 3) is satisfied.

Three vector fields, three curl values:

• Rotation field $\mathbf{F} = (-y, x)$: $\operatorname{curl}\mathbf{F} = 1 - (-1) = 2$. Constant positive curl — rigid counterclockwise rotation.
• Shear field $\mathbf{F} = (y, 0)$: $\operatorname{curl}\mathbf{F} = 0 - 1 = -1$. Constant negative curl — clockwise shearing.
• Expansion field $\mathbf{F} = (x, y)$: $\operatorname{curl}\mathbf{F} = 0 - 0 = 0$. Curl-free — pure expansion, no rotation. This field is conservative (it’s a gradient field).