If you want to add up something along a curve, you need to compute a line integral. Common examples are the length of a curve, how much a wire weighs, and how much work is done when moving along a particular path.

We start with the last of these, namely finding the work $W$ done by a force $\FF$ in moving a particle along a curve $C$. We begin with the relationship $$\hbox{work} = \hbox{force} \times \hbox{distance}$$ Suppose you take a small step $d\rr$ along the curve. How much work was done? Since only the component along the curve matters, we need to take the dot product of $\FF$ with $d\rr$. Adding this up along the curve yields $$W = \Lint \FF\cdot d\rr$$

So how do you evaluate such an integral? If you have an explicit parameterization $\rr=\rr(u)$, you can differentiate it in order to determine \begin{equation} d\rr = {d\rr\over du}\,du \label{drparam} \end{equation} thus turning your line integral into an ordinary integral with respect to $u$.

This is essentially the strategy used in most texts: the distance you go is $ds=|d\rr|$, the direction you are going is given by the unit tangent vector $\TT$, the component of $\FF$ in your direction is $\FF\cdot\TT$, and so the work done when moving a small distance is $\FF\cdot\TT\,ds$. But $$\TT \, ds = {d\rr\over ds} \, ds = d\rr $$ so you wind up with the same integral as before. This is nice in theory, but in practice it is not always easy to write things explicitly in terms of the arclength $s$; it's usually easier to work with some other parameterization, which brings us back to (\ref{drparam}).

But there is another strategy for doing line integrals, especially if you are not handed an explicit parameterization of the curve. Recall from (1) of Section 2 that the vector differential $d\rr$ is given by $$d\rr = dx\,\ii + dy\,\jj + dz\,\kk$$ If you like, you can think of this as taking the differential of both sides of the expression for the position vector (in rectangular coordinates!), namely $$\rr = x\,\ii + y\,\jj + z\,\kk$$

But what if you don't have an explicit parameterization? Use what you know!

Suppose you want to find the work done by the force $\FF=x\,\ii+y\,\jj$ when moving along a given curve $C$. Curves can be specified in several different ways; let us consider some examples, all of which start at $(1,0)$ and end at $(0,1)$.

1. Consider first the parametric curve $\rr=(1-u^2)\,\ii+u\,\jj$. It is straightforward to compute $$ d\rr = (-2u\,\ii + \jj) \, du $$ Since a curve is described by a single parameter, in this case $u$, we write everything in terms of that parameter, obtaining \begin{eqnarray*} \Lint \FF\cdot d\rr &=& \int_0^1 \left( (1-u^2)\,\ii + u\,\jj \right) \cdot (-2u\,\ii + \jj) \> du \\ &=& \int_0^1 \left( (1-u^2)(-2u)+u \right) du = \int_0^1 (2u^3-u) \,du = 0 \end{eqnarray*}
2. In physical applications, one is rarely given an explicit parameterization of the curve, but rather some other description. For instance, the curve just discussed might have been defined by the equation $x=1-y^2$. Finding the differential of both sides of this expression yields $dx=-2y\,dy$, and substituting into (1) of Section 2 leads to $$ d\rr = (-2y\,\ii + \jj) \,dy $$ The computation is exactly the same as before, using $y$ instead of $u$.
3. Alternatively, one can solve for $y$, obtaining $y=\sqrt{1-x}$, then compute $dy$ in terms of $dx$, then substitute into (1) of Section 2 obtaining $$ d\rr = dx\,\ii - {{dx}\over{2\sqrt{1-x}}}\,\jj $$ so that 1) $$ \Lint\FF\cdot d\rr = \int_1^0 \left(x-{1\over2}\right)\,dx = 0 $$

It is important to realize that all of these methods work.

Consider now the problem of trying to find the length of a quarter of a circle. What do you know? In polar coordinates, a circle is given by $r=\hbox{constant}$, so that $dr=0$. Inserting this fact into the expression (4) of Section 2 for arclength in polar coordinates, one immediately obtains $$ ds^2 = r^2 \, d\phi^2 $$ and finally \begin{equation} \Lint \, ds = \int_0^{{\pi}\over{2}} r\,d\phi = {\pi r\over{2}} \label{Answer} \end{equation}

But what if you didn't remember (4) of Section 2? The calculation is not much harder in rectangular coordinates: You know that $x=r\cos\phi$ and $y=r\sin\phi$, with $r=\hbox{constant}$, so that $dx=-r\sin\phi\,d\phi$ and $dy=r\cos\phi\,d\phi$. Inserting this into (2) of Section 2 again leads to (\ref{Answer}).

But what if you didn't even remember how to parameterize a circle, or, equivalently, how to use polar coordinates? Well, you still know that $x^2+y^2=r^2=\hbox{constant}$, so that $2x\,dx+2y\,dy=0$. Solving for $dy$ and inserting this into (2) of Section 2 yields $$ ds^2 = \left(1+{{x^2}\over{y^2}}\right) dx^2 = {{r^2}\over{r^2-x^2}} \> dx^2 $$ This leads to the (improper!) integral $$ \int_0^r {{dx}\over\sqrt{1-{{x^2}\over{r^2}}}} $$ which is, of course, most easily computed via a trig substitution — or numerically — yielding the same answer.

The point is that using what you know will always yield correct answers — eventually.

A similar procedure can be used to determine the mass of a wire in the shape of the curve $C$ if one knows the linear mass density $\lambda$ (mass per unit length). In this case, the mass of a small piece of the wire is $\lambda\,ds$, so the integral becomes $$m = \Lint \lambda \, ds$$

Another standard application of this type of line integral is to find the {\it center of mass} of a wire. This is done by averaging the values of the coordinates, weighted by the density $\lambda$ as follows: $$\bar{x} = {1\over m} \Lint x\lambda \, ds$$ where $m$ was defined earlier in the lesson. Similar formulas hold for $\bar{y}$ and $\bar{z}$; the center of mass is then the point $(\bar{x},\bar{y},\bar{z})$.

In a “use what you know” strategy, you may not be sure when to stop! A common error is to substitute for $dy$ in terms of $dx$ in $d\rr$, but to forget to substitute for $y$ in terms of $x$ in $\FF$. The rule of thumb is that you shouldn't start integrating until you have the integral in terms of a single parameter — including correctly determining the limits in terms of that parameter. Curves are one-dimensional!

Here is another example. Suppose you want to integrate $\DS\Lint\FF\cdot d\rr$, where $\FF=y\,\ii$, and $C$ is the line segment from $(1,0)$ to $(0,-1)$. Start with expression (1) of Section 2 for $d\rr$. What do you know? Well, the slope of the line segment is clearly $+1$, and its $y$-intercept is $-1$, so the equation of the line is $y=x-1$. Taking the differential of both sides, $dy=dx$, so that $d\rr=dx\,\ii+dx\,\jj$. Finally, note that $x$ runs from 1 to 0! Thus, $$\Lint\FF\cdot d\rr = \int_1^0 (x-1)\,\ii \cdot (\ii+\jj)\>dx = \int_1^0 (x-1) \,dx = \left( { x^2\over2} - x \right) \Bigg|_1^0 = {1\over2} $$ Yes, of course, this particular curve is easy to parameterize, but this is not always the case. Note also how easy it was to get the limits right, and thus get the correct sign, simply by always starting with expression (1) of Section 2 for $d\rr$, and integrating from your starting point to your final point.

This is important: Vector line integrals of the form $\DS\Lint\FF\cdot d\rr$ are directed integrals; the sign of the answer depends on which way you traverse the curve. You will obtain the correct sign automatically if you integrate from the beginning point to the final point, without putting in any artificial signs. As in the last example, this may result in an integral which goes from a larger value of the integration variable to a smaller one.

This is not the case for line integrals with respect to arclength. Since $ds=|d\rr|$, such integrals do not depend on which way the curve is traversed. Standard examples are arclength and mass, which must be positive! Other times you may have to think about it to get the sign right; the total charge on a wire could be positive or negative. One way to keep this straight is to remember that $ds=|d\rr|$, which has an absolute value in it, which requires care with signs.

We summarize this discussion by writing $$\Int_{-C} f \, ds = + \Lint f \, ds \qquad\hbox{but}\qquad \Int_{-C} \FF\cdot d\rr = - \Lint \FF\cdot d\rr $$ where $-C$ denotes the reversed curve. You will always get the sign right for vector line integrals if you are careful to put the limits in correctly (from starting point to ending point), whereas you may have to think about signs for the scalar line integrals.

GOALS