Vector line integrals

A vector field is defined as a set of vectors associated with each point in space. For instance, the velocity ${\bf v}({\bf r})$ in a moving liquid (e.g., a whirlpool) constitutes a vector field. By analogy, a scalar field is a set of scalars associated with each point in space. An example of a scalar field is the temperature distribution $T({\bf r})$ in a furnace.

Consider a general vector field ${\bf A}({\bf r})$. Let $d{\bf l} = (dx,dy,dz)$ be the vector element of line length. Vector line integrals often arise as

$$\int_P^Q {\bf A}\cdot d{\bf l} = \int_P^Q (A_x dx+A_y dy + A_z dz).$$ (72)

For instance, if ${\bf A}$ is a force then the line integral is the work done in going from $P$ to $Q$.

As an example, consider the work done in a repulsive, inverse-square, central field, ${\bf F} = - {\bf r}/ \vert r^3\vert$. The element of work done is $dW={\bf F}\cdot d{\bf l}$. Take $P=(\infty, 0, 0)$ and $Q=(a,0,0)$. Route 1 is along the $x$-axis, so

$$W = \int_{\infty}^a \left(-\frac{1}{x^2}\right) dx = \left[\frac{1}{x}\right]_{\infty}^a =\frac{1}{a}.$$ (73)

The second route is, firstly, around a large circle ($r=$ constant) to the point ($a$, $\infty$, 0), and then parallel to the $y$-axis. In the first, part no work is done, since ${\bf F}$ is perpendicular to $d{\bf l}$. In the second part,

$$W = \int_{\infty}^0 \frac{-y dy}{(a^2 + y^2)^{3/2}} = \left[\frac{1}{(y^2+a^2)^{1/2}} \right]_0^\infty = \frac{1}{a}.$$ (74)

In this case, the integral is independent of the path. However, not all vector line integrals are path independent.