Vector Fields and Line Integrals

A vector field attaches a vector to every point of space: fluid velocity, gravitational pull, and electric force all have this form. The quantities that matter about a field — the work done moving through it, the flow across a boundary — are integrals along curves and across surfaces rather than over regions. The line integral accumulates a field's tangential component along a path. For the special class of conservative fields it collapses to a difference of endpoint values, the exact analog of the Fundamental Theorem of Calculus.

Vector fields

The picture is an arrow drawn at each point, its tail at the point and its direction and length given by $\vec{F}$ there. A velocity field of a flowing fluid, the gravitational field of a mass, and the electric field of a charge all have this form.

A special and central class of vector fields comes from scalar functions.

The gradient points in the direction of steepest increase of $f$ and is perpendicular to the level curves $f=\text{const}$. Gravitational and electrostatic fields are conservative; a potential exists because the force derives from a potential-energy function. Not every field is conservative; deciding which are comes down to a test on the component functions, developed below.

Line integrals of scalar functions

Before integrating a field, integrate a scalar function along a curve. Let $C$ be a smooth curve parametrized by $\vec{r}(t)=⟨x(t),y(t)⟩$, $a\le t\le b$. Cut $C$ into short arcs of length $\Delta s_i$, sample $f$ on each, and sum $f\Delta s_i$. The limit is the line integral with respect to arc length.

Because arc length satisfies $\mathbf{d}s=\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}\mathbf{d}t=|\vec{r}{}^{\prime }(t)|\mathbf{d}t$, the integral reduces to an ordinary integral in $t$:

The value does not depend on the parametrization, as long as $C$ is traced once as $t$ goes from $a$ to $b$. If the density of a wire bent along $C$ is $\rho (x,y)$, then ${\int }_C\rho \mathbf{d}s$ is its mass; with $f=1$ it is the length of $C$. Scalar line integrals are the base case; the vector version measures work.

Line integrals of vector fields: work

A force $\vec{F}$ pushing an object along a path does work equal to the distance times the component of force along the direction of motion. Along a curve, the component of $\vec{F}$ in the direction of travel is $\vec{F}\cdot \hat{T}$, where $\hat{T}$ is the unit tangent. Summing over the curve gives the work.

The middle expression is the working formula: substitute the parametrization into $\vec{F}$, dot with the velocity $\vec{r}{}^{\prime }(t)$, and integrate over $t$. In components with $\vec{F}=P\hat{ı}+Q\hat{ȷ}$ and $\mathbf{d}\vec{r}=⟨\mathbf{d}x,\mathbf{d}y⟩$, the same integral is written

The two terms cancel identically; the field does no net work along this arc. A different field would give a nonzero value, and reversing the orientation of $C$ flips the sign of a vector line integral (though not of a scalar one, since arc length is unsigned).

The Fundamental Theorem for Line Integrals

For a conservative field the line integral has a shortcut. If $\vec{F}=\nabla f$, then along a curve $\vec{F}(\vec{r}(t))\cdot \vec{r}{}^{\prime }(t)=\nabla f(\vec{r}(t))\cdot \vec{r}{}^{\prime }(t)$, which by the chain rule is exactly $\frac{d}{dt}f(\vec{r}(t))$. Integrating a derivative recovers the endpoint values.

This is the multivariable version of the evaluation form of the Fundamental Theorem of Calculus: the line integral of a gradient depends only on the values of the potential at the two endpoints, not on the route between them. Computing work then requires no parametrization at all — only $f$ at $A$ and $B$.

Path independence, closed loops, and the component test

The theorem has three equivalent consequences that together characterize conservative fields on a suitable domain.

Since a gradient integral is $f(B)-f(A)$ regardless of $C$, conservative fields are path independent. The converse holds on connected domains: a path-independent field is conservative. A closed curve returns to its start, so a path-independent integral around any closed loop is zero.

To test whether a given field is conservative without hunting for a potential, compare cross-partials. If $\vec{F}=\nabla f$ then $P=f_x$ and $Q=f_y$, so $P_y=f_{xy}$ and $Q_x=f_{yx}$, equal by Clairaut's Theorem. This gives a necessary condition, and on a simply connected domain it is sufficient.

Simply connected means the domain has no holes, so any closed loop can be shrunk to a point. The condition can fail on a domain with a hole — the field $\vec{F}=\frac{1}{x^2+y^2}⟨-y,x⟩$ satisfies $P_y=Q_x$ everywhere it is defined yet is not conservative on the punctured plane, because a loop around the origin cannot be contracted.

The four statements are equivalent for a $C^1$ field on a simply connected domain:

Recovering the potential

When the component test passes, build $f$ by partial integration. From $f_x=P$, integrate in $x$ treating $y$ as constant, picking up an unknown function of $y$; then differentiate in $y$ and match $f_y=Q$ to pin it down.

The quantity behind the component test

The component test sets $Q_x-P_y$ to zero. That same quantity measures a field's tendency to circulate, and integrating it over a plane region recovers the loop integral of $\vec{F}$ around the region's boundary. That identity is Green's Theorem, which turns a line integral around a closed curve into a double integral and accounts for why $P_y=Q_x$ detects conservativeness.¹