Vector Calculus: Line Integrals

This is section three of my series on vector calculus.  As always, if you are able to follow along and have questions pop up, or a section is totally unclear, drop a comment or email to me.  I’ve got the two exams coming up on Wednesday and Thursday.  I could theoretically bomb the crap out of the Wednesday one, I just have to pass the one on Thursday (matches lowest exam grade with the final).

Now we come to the part of multi-variable that always confuses me, the line integral.  Line integrals, as you may suspect, attempt to measure the effect of a vector field over a line.  That line may be simple, such as a straight line or curve, or it may be complex, making closed shapes or strange curves with quick changes in direction.  To be more specific, we’ll be looking at line integrals over continuous lines that may or may not be differentiable.  This means we may have something that isn’t a smooth and pretty curve.  The following is such an example (assuming the whole curve is C, and each number is a piece of the whole curve):

A Smooth Piecewise Curve

A Smooth Piecewise Curve

As you can see, the line is continuous as it does not have any breaks.  You can also see it is not differentiable because there are extreme sharp curves.  In the case of the curve C above, we would actually take an integral over each piece of the curve and add them together.  We’ll discuss what it means to take that integral in a second, but first I want to mention orientation.

In this case, we would say the curve is orientated from C1 to C2 to C3 to C4.  Orientation in this means if we were to draw this curve from scratch, we would start in the 2nd quadrant, begin drawing C1 into the first quadrant and continue to on C2.  Orientation is just the direction we are headed in.  If we were to switch the orientation and start at the end of C4 and trace it the other way, our integral essentially stays the same, we just have to multiply it by negative 1.  The absolute value of the integral would stay the same, just the change in direction makes it negative.

Arc Length as a Line Integral

The easiest way to start with a line integral is to think about it with no vector field. In line integrals, the idea is to find the value of the vector field on tiny parts of the line, and then add them all up (integrals are nothing but sums of things over small sections).  In this case, the small sections are lengths of the curve.  Therefore, by taking the line integral with no vector field, all you are taking is the arc length of the curve, something you learn how to do much earlier in the class. Remember that we will need to treat the curve as a parametric equation for this.  So let’s take a look at the basic arc length integral:

Arc Length = ∫ ds = ∫ √ ( x'(t)2 + y'(t)2 ) dt

Sorry for not using formula writing software and pasting pictures to make the equation look clearer.  Deal with it.  For those of you who might be confused, that is the integral of ds (length of a small segment) is equal to the integral with respect to t of the square root of the derivative of a function x of t squared plus the derivative of y of t with respect to t squared.

Line Integrals of Functions

When looking at line integrals of function (the next step up in difficulty), we have to add the vector field into the integral.  This is done easily by inserting it before ds or the square root of the second equation.  For this example, we’ll be using the function f(x,y) with both x and y being functions of t (due to parametrization).  So the equation becomes:

∫ f(x,y) ds = ∫ f(x(t),y(t)) * √ ( x'(t)2 + y'(t)2 ) dt

This just becomes a regular integral where we take the sum of the value of a function at each particular part of a line.  These steps are easy, but we need to look at them in terms of vector fields, not just functions.  For this we need to make our integral in terms of two vectors.  The equation for a vector field line integral in its most basic form would look like this:

f • dr = ∫ fr‘(t) dt

In this equation, r(t) is the line we are taking the integral over and f is the vector field function.  Let’s compare the terms of our line integral of a function and our line integrals of a vector function so you can see where the vector function line integral originates from.  First, you’ll notice a version of f appears in both integrals in the same spot.  This should be slightly intuitive as we are just replacing the function with a vector.  They both are what we are integrating over the line, so it should make sense they stay in the same position in our integral.  Something you should notice though is the vector field f will need to be parametrized in order to integrate in respect to t.  This is easy though,  as you replace any x, y or z variable in the vector f with the corresponding x- ,y- , or z-components of r(t).  For example, if f=<xy,y,z> and you want it over the line r(t)=<1+t, 2+3t, 0>, then the parametrization of f would be f(r(t)) = <(1+t)(2+3t), 2+3t, 0>.

The second items, the radical and r‘(t) are one in the same.  If you remember back to parametric equations, a line can be defined by a function r(t).  When we wanted to find the arc length (on for integrals, several small arc lengths), we took the derivative of r(t) and multiplied it by small amounts of t (dt) to approximate small lengths of the curve.  If you remember, the norm of the derivative of r(t) is our radical.

Fundamental Theorem of Application

Last time, we discussed how to find potential functions.  This time, we’ll discuss another practical use of those potential functions.  The first thing to remember is the fundamental theorem of calculus.  It says that if you take the definite integral of the derivative of a function,  you can use the function and input the end points to find the answer.  In equation form, it is the same as saying this:

ab F ‘(x) dx = F(b) – F(a)

This also works for vector field integrals.  Say we are taking the line integral from point a to point b:

ab F • dr

And let’s also say that f is a potential function for F.  If that is the case, we can skip doing the full integral, and instead, just put in the necessary values of a and b into the function f as such:

ab F • dr = f(b)-f(a)

This is much easier than finding the parametric equation of r(t), taking the dot product, and finally integrating.  Remember, this only works for conservative vector fields.

Another note is that this fundamental theorem to find the integral is independent of path.  The curve r(t) could travel in any direction for any length, but all that matters is where it starts and where it ends.  Expounding on that idea, this also says that if a line r(t) creates a closed loop, then the integral over that loop in a conservative vector field will be zero as the line starts and stops at the same point (provided no overlap of the line).

Next Time

In the next post, we’ll discuss Green’s Theorem and the relationship between a line integral over a simple closed loop and the area bound inside of it.  After that it’s onto surfaces and their integrals.  As always, if you are able to follow along and found a point to question, drop me a line.  The better I can express this stuff, the better I’ll understand it, the better I’ll do on my two exams this week.


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