Vector line integrals

In the previous section we looked at evaluating line integrals of scalar functions; here, we are going to evaluate line integrals of vector fields.

Consider first a vector field representing force, denoted by $\boldsymbol{F}(\boldsymbol{r})$, being applied over a displacement $d \boldsymbol{r}$; note that now the differential is a vector. The work done is calculated by the product of force and distance but only the force in the direction of displacement does any work:

where $\theta$ is the angle between $\boldsymbol{F}$ and $d \boldsymbol{r}$. This leads us to an important difference between vector and scalar line integrals: the vector line integrals depend on a direction along the curve.

Now, suppose we have a particle moving along a curved path $\mathcal{C}$ and the particle is acted on by a force $\boldsymbol{F}(\boldsymbol{r})$. We want to determine the total amount of work done as the particle moves along $\mathcal{C}$. To calculate this, we can divide $\mathcal{C}$ into infinitesimal pieces and 'sum' the work contributions over all pieces. We call this the line integral of $\boldsymbol{F}$ along the curve $\mathcal{C}$ and define it as,

To relate the vector line integral to the scalar line integral ideas above, we re-write the former as follows in terms of the arc length $s$ :

where $\boldsymbol{r}(t)$ is a parametrisation for the path $\mathcal{C}$ while $\boldsymbol{T}$ is the unit tangent vector, defined as:

here, we assume that $\boldsymbol{r}(t)$ is a regular parametrisation, meaning that $\boldsymbol{r}^{\prime}(t) \neq$ $0, \forall t \in[a, b]$. This ensures that $\boldsymbol{T}$ does not become undefined anywhere in the integration interval. Also, it is interesting to note that since $\boldsymbol{F} \cdot \boldsymbol{T}$ is the tangential component of $\boldsymbol{T}$, it follows from Eq. (2.53) that the line integral of vector $\boldsymbol{F}$ along a curve $\mathcal{C}$ is the integral of the tangential component of $\boldsymbol{F}$.

Using the arc length differential as defined in Eq. (2.41), we write $\boldsymbol{F}(\boldsymbol{r}) \cdot \boldsymbol{T} d s$ as follows:

We now observe that all terms on the RHS of Eq. (2.55) are given in terms of $t$.

At this point, we need to discuss another important difference between line integrals of scalar and vector fields: the orientation of the parametrisation $\boldsymbol{r}(t)$. Recall that this is defined as a 'specified orientation along a path', say from a point $A$ to a point $B$. We refer to this direction as the positive direction and the curve $\mathcal{C}$ itself is the oriented curve. The opposite direction is known as the negative direction. Now, if we reverse the direction of the path then the positive direction becomes the one moving from point $B$ to $A$. We use these definitions below to give the theorem associated with the evaluation of vector line integrals:

Computing a vector line integral

Let $\boldsymbol{r}(t)$ be a regular parametrisation for an oriented curve $\mathcal{C}$ for $a \leq t \leq b$. Then,

Next, we look at an example with a vector field $\boldsymbol{F}$ and a three-dimensional, smooth curve given by a parametrisation $\boldsymbol{r}(t)$.

Example 2.7 Evaluate $\int_{\mathcal{C}} \boldsymbol{F} \cdot d \boldsymbol{r}$ where $\boldsymbol{F}=\left(8 x^{2} y z, 5 z,-4 x y\right)$ and $\mathcal{C}$ is the curve defined by $\boldsymbol{r}(t)=\left(t, t^{2}, t^{3}\right)$ where $t \in[0,1]$.

Solution:

We first need to evaluate $\boldsymbol{F}$ along the curve:

this gives us the first term in the integrand on the RHS of Eq. (2.56). Next, we need $\boldsymbol{r}^{\prime}(t)$ :

The integrand in Eq. (2.56) therefore becomes:

With $\boldsymbol{F}(\boldsymbol{r}(t)) \cdot \boldsymbol{r}^{\prime}(t)$ expressed in terms of $t$, the line integral is:

Properties:

(a) If $\mathcal{C}$ is piecewise smooth then we can break it down into several, say $n$, concatenated oriented curves. We define the union of the paths as done in Subsec. 2.3.1 and the line integral is given by

(b) The line integral of vector fields does depend on orientation. Consider a curve connecting two points $A$ and $B$. As previously mentioned, there are two possible orientations: the oriented curve, which we denote as $\mathcal{C}$ and the curve in the opposite direction, $-\mathcal{C}$. The unit tangent vector changes sign from $\boldsymbol{T}$ to $-\boldsymbol{T}$ when we change orientation. It follows that the tangential component of $\boldsymbol{F}$ and the line integral also change sign:

(c) For a smooth curve, $\mathcal{C}$ and two vector fields, $\boldsymbol{F}$ and $\boldsymbol{G}$, the following linearity properties hold:

where $k$ is a constant.

(d) If $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are different parametrisations of the same curve with the same orientation, then: