In this section we are now going to introduce a new kind of
integral. However, before we do that it
is important to note that you will need to remember how to parameterize
equations, or put another way, you will need to be able to write down a set of
parametric equations for a given curve.
You should have seen some of this in your Calculus II course. If you need some review you should go back
and review some of the basics of
parametric equations and curves.
Here are some of the more basic curves that we’ll need to
know how to do as well as limits on the parameter if they are required.
Curve

Parametric Equations

(Ellipse)

CounterClockwise Clockwise

(Circle)

CounterClockwise Clockwise





Line Segment From
to


With the final one we gave both the vector form of the
equation as well as the parametric form and if we need the twodimensional
version then we just drop the z
components. In fact, we will be using
the twodimensional version of this in this section.
For the ellipse and the circle we’ve given two
parameterizations, one tracing out the curve clockwise and the other
counterclockwise. As we’ll eventually
see the direction that the curve is traced out can, on occasion, change the
answer. Also, both of these “start” on
the positive xaxis at .
Now let’s move on to line integrals. In Calculus I we integrated ,
a function of a single variable, over an interval . In this case we were thinking of x as taking all the values in this
interval starting at a and ending at b.
With line integrals we will start with integrating the function ,
a function of two variables, and the values of x and y that we’re going
to use will be the points, ,
that lie on a curve C. Note that this is different from the double
integrals that we were working with in the previous chapter where the points
came out of some twodimensional region.
Let’s start with the curve C that the points come from.
We will assume that the curve is smooth
(defined shortly) and is given by the parametric equations,
We will often want to write the parameterization of the
curve as a vector function. In this case
the curve is given by,
The curve is called smooth
if is continuous and for all t.
We use a ds here
to acknowledge the fact that we are moving along the curve, C, instead of the xaxis (denoted by dx) or
the yaxis (denoted by dy).
Because of the ds this is
sometimes called the line integral of f with respect to arc length.
We’ve seen the notation ds
before. If you recall from Calculus II
when we looked at the arc length of a
curve given by parametric equations we found it to be,
It is no coincidence that we use ds for both of these problems.
The ds is the same for both
the arc length integral and the notation for the line integral.
So, to compute a line integral we will convert everything
over to the parametric equations. The
line integral is then,
Don’t forget to plug the parametric equations into the
function as well.
If we use the vector form of the parameterization we can
simplify the notation up somewhat by noticing that,
where is the magnitude or norm of . Using this notation the line integral
becomes,
Note that as long as the parameterization of the curve C is traced out exactly once as t increases from a to b the value of the
line integral will be independent of the parameterization of the curve.
Let’s take a look at an example of a line integral.
Example 1 Evaluate
where C
is the right half of the circle, . rotated in the counter clockwise direction.
Solution
We first need a parameterization of the circle. This is given by,
We now need a range of t’s
that will give the right half of the circle.
The following range of t’s
will do this.
Now, we need the derivatives of the parametric equations
and let’s compute ds.
The line integral is then,

Next we need to talk about line integrals over piecewise smooth curves. A piecewise smooth curve is any curve that
can be written as the union of a finite number of smooth curves, ,…,
where the end point of is the starting point of . Below is an illustration of a piecewise
smooth curve.
Evaluation of line integrals over piecewise smooth curves is
a relatively simple thing to do. All we
do is evaluate the line integral over each of the pieces and then add them
up. The line integral for some function
over the above piecewise curve would be,
Let’s see an example of this.
Example 2 Evaluate
where C
is the curve shown below.
Solution
So, first we need to parameterize each of the curves.
Now let’s do the line integral over each of these curves.
Finally, the line integral that we were asked to compute
is,

Notice that we put direction arrows on the curve in the
above example. The direction of motion
along a curve may change the value of
the line integral as we will see in the next section. Also note that the curve can be thought of a
curve that takes us from the point to the point . Let’s first see what happens to the line
integral if we change the path between these two points.
So, the previous two examples seem to suggest that if we
change the path between two points then the value of the line integral (with
respect to arc length) will change.
While this will happen fairly regularly we can’t assume that it will
always happen. In a later section we
will investigate this idea in more detail.
Next, let’s see what happens if we change the direction of a
path.
Example 4 Evaluate
where C
is the line segment from to .
Solution
This one isn’t much different, work wise, from the
previous example. Here is the
parameterization of the curve.
for . Remember that we are switching the direction
of the curve and this will also change the parameterization so we can make
sure that we start/end at the proper point.
Here is the line integral.

So, it looks like when we switch the direction of the curve
the line integral (with respect to arc length) will not change. This will always be true for these kinds of
line integrals. However, there are other
kinds of line integrals in which this won’t be the case. We will see more examples of this in the next
couple of sections so don’t get it into your head that changing the direction
will never change the value of the line integral.
Before working another example let’s formalize this idea up
somewhat. Let’s suppose that the curve C has the parameterization ,
. Let’s also suppose that the initial point on
the curve is A and the final point on
the curve is B. The parameterization ,
will then determine an orientation for the curve where the positive direction is the
direction that is traced out as t
increases. Finally, let be the curve with the same points as C, however in this case the curve has B as the initial point and A as the final point, again t is increasing as we traverse this
curve. In other words, given a curve C, the curve is the same curve as C except the direction has been reversed.
We then have the following fact about line integrals with
respect to arc length.
Fact
So, for a line integral with respect to arc length we can change
the direction of the curve and not change the value of the integral. This is a useful fact to remember as some
line integrals will be easier in one direction than the other.
Now, let’s work another example
Example 5 Evaluate
for each of the following curves.
(a) [Solution]
(b) :
The line segment from to . [Solution]
(c) :
The line segment from to . [Solution]
Solution
Before working any of these line integrals let’s notice
that all of these curves are paths that connect the points and . Also notice that and so by the fact above these two should
give the same answer.
Here is a sketch of the three curves and note that the
curves illustrating and have been separated a little to show that
they are separate curves in some way even though they are the same line.
(a)
Here is a parameterization for this curve.
Here is the line integral.
[Return to Problems]
(b) : The line segment from to
.
There are two parameterizations that we could use here for
this curve. The first is to use the
formula we used in the previous couple of examples. That parameterization is,
for .
Sometimes we have no choice but to use this
parameterization. However, in this
case there is a second (probably) easier parameterization. The second one uses the fact that we are
really just graphing a portion of the line . Using this the parameterization is,
This will be a much easier parameterization to use so we
will use this. Here is the line
integral for this curve.
Note that this time, unlike the line integral we worked
with in Examples 2, 3, and 4 we got the same value for the integral despite
the fact that the path is different.
This will happen on occasion.
We should also not expect this integral to be the same for all paths
between these two points. At this point
all we know is that for these two paths the line integral will have the same
value. It is completely possible that
there is another path between these two points that will give a different
value for the line integral.
[Return to Problems]
(c) : The line segment from to
.
Now, according to our fact above we really don’t need to
do anything here since we know that . The fact tells us that this line integral
should be the same as the second part (i.e.
zero). However, let’s verify that,
plus there is a point we need to make here about the parameterization.
Here is the parameterization for this curve.
for .
Note that this time we can’t use the second
parameterization that we used in part (b) since we need to move from right to
left as the parameter increases and the second parameterization used in the
previous part will move in the opposite direction.
Here is the line integral for this curve.
Sure enough we got the same answer as the second part.
[Return to Problems]

To this point in this section we’ve only looked at line
integrals over a twodimensional curve.
However, there is no reason to restrict ourselves like that. We can do line integrals over threedimensional
curves as well.
Let’s suppose that the threedimensional curve C is given by the parameterization,
then the line integral is given by,
Note that often when dealing with threedimensional space
the parameterization will be given as a vector function.
Notice that we changed up the notation for the
parameterization a little. Since we
rarely use the function names we simply kept the x, y, and z and added on the part to denote that they may be functions of
the parameter.
Also notice that, as with twodimensional curves, we have,
and the line integral can again be written as,
So, outside of the addition of a third parametric equation
line integrals in threedimensional space work the same as those in
twodimensional space. Let’s work a
quick example.
Example 6 Evaluate
where C
is the helix given by, ,
.
Solution
Note that we first saw the vector equation for a helix
back in the Vector Functions section. Here is a quick sketch of the helix.
Here is the line integral.
You were able to do that integral right? It required integration by parts.

So, as we can see there really isn’t too much difference
between two and threedimensional line integrals.