Line Integrals. where df (x)/dx = f(x) and c is a constant, and the definite integral of f(x) from x = a to x = b as.

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1 Line Integrals What is a line integral? In our integral calculus class ou learned how to perform integrals like b a d f(). () This integral of a single variable is the simplest eample of a line integral. A line integral is just an integral of a function along a path or curve. In this case, the curve is a straight line a segment of the -ais that starts at = a and ends at = b. Just so we re clear on notation, I ll write the indefinite integral of f() with respect to as d f() = F () + c, () where df ()/d = f() and c is a constant, and the definite integral of f() from = a to = b as b a d f() = F () b a = F (b) F (a). (3) The vertical line notation in the second step means evaluate the thing in front at the upper limit, then subtract the thing in front evaluated at the lower limit. Now suppose I gave ou a function of three variables, sa h(,, z), and then described a path P through space that starts at ( i, i, z i ) and snakes around until it ends at ( f, f, z f ). You can integrate h(,, z) along the path P just like ou integrated f() with respect to in (). Here is how we write such an integral P f i dl h(,, z). () What does all this mean? First, the subscript P reminds us that we re integrating along a specific path or curve (I ll use the terms path and curve interchangeabl). Integrating along a different curve ma or ma not give a different answer. Second, the limits i and f on the integral refer to the starting point ( i, i, z i ) and end point ( f, f, z f ), respectivel. Third, dl is the infinitesimal distance from point to point along the path. The reall important thing to notice here is that there is a single integration variable, because we are integrating along a path. You onl need one variable to describe where ou are on a curve, so there is onl one variable to integrate! Over the net few sections we ll unpack all this and figure out how to evaluate integrals like (). Curves Before we can integrate along a curve, we should spend some time reviewing how to describe curves. Suppose ou wanted to describe a curve in the - plane (later on we ll work with three variables, but let s start simple). You ma be able to do this b telling me as a function of, along with the start and end points. For eample, Figure shows the curve () = + 3 from = to = 3. The

2 6 3 - Figure : The curve () = collection of points (, = curve. + 3, from = to = ) with 3 is a perfectl good wa of describing this Of course, this doesn t alwas work. Consider a curve like the one shown in Figure. Do ou think ou could describe it in the same wa as the curve in Figure? It s clear that for this Figure : A parametric curve. curve, but if ou tr to describe the points in the form (, ()) ou run into a problem: there are different points on the curve with the same value of! As an eample, look at values of close to =. Since a function () can t give two different values for a given, ou can t describe this curve like the previous one. Instead, ou need to give a parametric description of the curve, which means epressing both and as functions of a parameter t. The curve in Figure is (t) = cos(t) (t) = sin(t) cos(t) with π t π. (5) You could come up with man other parameterizations of this curve, too there is no unique wa of doing it. The last two eamples are curves in the - plane, but a path through three dimensional space with coordinates (,, z) works the same wa. For eample, ou ma be able to describe a curve b epressing and z as functions of, as (, (), z()). If this isn t possible ou can alwas give a parametric description of the form ((t), (t), z(t)) with ti t tf. For eample, Figure 3 shows the curve ( = cos(t), = sin(t), z = t).

3 z Figure 3: The curve (,, z) = (cos(t), sin(t), t) with t 6π. 3 The infinitesimal displacement vector When ou perform an integral like (), the factor of d in the integrand represents an infinitesimal displacement along the -ais ou are just moving along the -ais, adding up little rectangles with width d and height f(). Naturall, when we integrate along a curve we are interested in the infinitesimal displacement between two nearb points on the curve. Just imagine ou are wrapping the -ais along the curve in question. First, let s consider two infinitesimall close (but otherwise arbitrar) points in the - plane. One of them has coordinates (, ), and the other has coordinates (+d, +d). What is the displacement vector between them? Following the notation used b Griffiths, we ll call this vector d l. The two points are separated b an infinitesimal distance d in the direction, and an infinitesimal distance d in the direction, so the displacement vector is d l = d ˆ + d ŷ. (6) The magnitude of this vector, which we will call dl, is the distance between the two points dl = d l = d l d l = d + d. (7) We haven t said anthing about the points other than the fact that the distance between them is infinitesimal. But what if the were both on a curve whose points are described b (, ())? Then one point is at (, ()) and the other is at ( + d, ( + d)). What is the displacement vector between them in that case? Since is a function of, the displacement in the direction is just d times the derivative of () with respect to d = () d, (8) 3

4 d{ HL d d Figure : Displacement between two infinitesimall close points on a curve. so d ~` is given b () d~` = d + d. (9) This is illustrated in Figure. Notice that the curve looks like a straight line, since we are zooming in on an infinitesimall small patch of the - plane. The magnitude of d~` is s, () d` = d + which is the distance between two infinitesimall close points on the curve. Eample: Epress the distance between two infinitesimall close points on the curve () = 3 in terms of d. d~` = d + d = d + 3 d q d` = d + (3 ). Things work in eactl the same wa if the curve is given in the parametric form ((t), (t)). The displacement between the two points at ((t), (t)) and ((t + dt), (t + dt)) is (t) (t) d~` = dt + dt, () and the distance is s d` = dt +. () What if we are interested in a curve in three dimensions? If it is given in the parametric form ((t), (t), z(t)) then the displacement and distance between two infinitesimall nearb points are (t) (t) z(t) d~` = dt + dt + dt z s z d` = dt + +. (3) ()

5 Eample: What are d l and dl for the curve shown in Figure 3? = sin(t) = cos(t) z = d l = sin(t) dt ˆ + cos(t) dt ŷ + dt ẑ dl = dt sin (t) + cos (t) + = dt. Given a curve, what can we do with d l? Well, dl is the distance between two infinitesimall nearb points on the curve, so integrating dl along the curve should give us the length of the curve. Suppose a and b are two points on the curve. Then the length of the curve between those points, which we ll denote b s(a, b), is s(a, b) = This distance is often referred to as the arc length. b a dl. (5) Eample: Find the length of the curve shown in Figure. The curve in Figure is described b () = + 3, with 3, so the distance between two infinitesimall nearb points on the curve is ( ) dl = d + = d + ( + ) = d + +. Now we want to integrate dl along the curve. Since dl is proportional to d, this just turns into an integral like (). dl = d + + = ( + ) ( log + + ) + + Oka, this is kind of an ugl integral it will actuall come up later this semester and we ll talk about how it is evaluated. To get the length of the curve we need to evaluate the definite integral from = to = 3, which gives 3 d + + = 6.3. (6). In this last eample the integral of dl along the curve becomes an integral with respect to. When the curve is given in parametric form the integral is with respect to the variable that parameterizes the curve. 5

6 Eample: Find the length of the curve shown in Figure 3. We alread saw in a previous eample that dl = dt for this curve. To get the length we need to integrate along the curve from t = to t = 6π 6π This was a little easier than the previous eample. dt = 6π. (7) Paths defined piecewise Sometimes we will encounter paths that are defined piecewise. This means that the are described in terms of a number of curves or straight line segments, with each one beginning where the previous one ended. An eample is shown below, in Figure 5. Piecewise paths are eas to work with we just Figure 5: A path that is defined piecewise. deal with them one piece at a time. Thus, we might have different epressions for things like d l on each part of the curve, but aside from that there aren t an other complications. As an eample, the curve in Figure 5 is a diagonal line from (, ) to (3, 7 ), then a vertical line from (3, 7 ) to (3, ), and finall a horizontal line from (3, ) to ( 3, ). Along the first segment we could describe the path as (, = + ). Thus, d l on this part is given b d l = d ˆ + d ŷ (8) = d ˆ + d ŷ (9) d l = d ˆ + d ŷ, () and its magnitude is dl = d 5/. On the second leg of the path is not changing at all (it s a This makes sense, right? For ever d ou move horizontall, ou also move up a bit verticall, as well. 6

7 vertical line!), so d =. Thus, d~` = d and d` = d on this part. Finall, the third leg of the path is horizontal, so d = there. Now, if we re being careful and following the path as if we were walking along it, we d use d~`3 = d () along the third leg, since we re moving in the direction of decreasing. However, d`3 = d, since the magnitude of a vector is alwas positive. 5 Line Integrals Oka, so we know how to describe curves, and we understand what d` is. How do we evaluate an integral like ()? Well, we actuall just evaluated a few eamples of these integrals the arc-length is what ou get when ou integrate the function around the curve. But let s tr something a little more involved. We ll start with an eample. Suppose we want to integrate the function f (, ) = along the curve (, ) = (cos t, sin t) with π t 3π. That is, we want to evaluate the integral Z f d` f (, ), () P i where P, shown in Figure 6, is a semi-circle that starts at the point (/, / ) and ends at the point ( /, / ). We know how to determine d`, but we still need to work out what to do with Figure 6: The curve (, ) = (cos t, sin t) with π t 3π. the function in the integrand. For this curve, d` is just s q d` = + = dt sin (t) + cos (t) = dt. 7 (3)

8 Now, we are integrating this function along the curve, so we re onl concerned with the values the function f(, ) takes for points on the curve. In other words, we need to evaluate the function for points (, ) = (cos t, sin t): So the integral () is f(cos t, sin t) = (cos t) sin t. () 3π π dt (cos t) sin t. (5) This integral can be worked out using the change of variable u = cos t, and we get 3π π dt (cos t) sin t = 3π 3 (cos t)3 = 3 π ( ) 3 ( 3 ( ) 3 ) (6) (7) = 3. (8) That s all there is to it ou work out the appropriate dl, evaluate the rest of the integrand for points on the curve, and then perform the integral. It isn t too hard to see what an integral like () means. First, let s plot the path in the - plane, with a third ais that we ll use to indicate the value of the function we re integrating (Figure 7). Net,. f, Figure 7: The curve (, ) = (cos t, sin t) with π t 3π. we ll add the function f(, ) = to the plot, along with a black curve that shows the value of the function at the points along the curve (Figure 8). Finall, let s take the function out of the plot but leave the black curve that shows its value along the path we re interested in. Filling in the area between our path and values of the function along the path makes the meaning of an integral like () clear: it gives the area of the shaded region in Figure 9. So ou see, there isn t much difference between the simple integral () and the integral along a path in (). The line integral ma be a bit harder to visualize especiall if the path winds around in three dimensions, so that we can t make a plot like Figures 8 or 9 but it still means basicall the same thing. 8

9 -..5 f H,L Figure 8: The function f (, ) =, and its values on the curve we re integrating along f H,L Figure 9: The integral () gives the area of the shaded region between the two curves. 6 Vector Fields Now we understand (at least in principle) how to evaluate an integral like (): describe the curve, determine d`, evaluate the rest of the integrand for points on the curve, and then perform the resulting integral. But we will encounter other line integrals which do not take quite the same form as (). These integrals involve vector fields. Recall that a vector field is just a vector with components that are functions (a plain old function is a scalar ). For eample, if we re sticking to the - plane, an eample of a vector field is ~ =. V (9) The -component is the function, and the -component is the function. This vector field is shown in Figure, with the direction and size of each arrow indicating the direction and magnitude, respectivel, of the vector field at that point. 9

10 Figure : The vector field V = ˆ ŷ. Given a vector field like V, we are often interested in the integral P f i d l V. (3) If ou take a minute to think about what the integrand means, ou ll see that this integral isn t reall that different than (). We understand what d l is it s the infinitesimal displacement vector along the path and we know how to take its dot product with the vector V. So, to evaluate (3) we take the integrand d l V = d V + d V, (3) work out what each part is on the curve, and perform the integral (if this were a three-dimensional curve and vector field, there would also be a dz V z term). Let s work out a few eamples. Eample: Evaluate the integral (3) for the vector field V = ˆ ŷ, along the curve (, + 3) shown in Figure. First let s work out d l V : d l V = d d (3) Along the curve we re interested in, = + 3, d = ( + ) d, and (3) becomes ( ) d l V ( ) = + 3 d + 3 ( + ) d ( = ) d.

11 The curve in Figure runs from = to = 3, so the integral is 3 d ( ) = ( ) = 6. Eample: Evaluate the integral (3) for the vector field V = ˆ + z ŷ ẑ, along the curve (cos t, sin t, t) with t 6π (shown in Figure 3). The infinitesimal displacement vector along the curve is and the vector field V is The integrand is The curve runs from t = to t = 6π, so d l = ( sin t ˆ + cos t ŷ + ẑ) dt V = (cos t) ˆ + (sin t) t ŷ (sin t)(cos t) ẑ. d l V = ( sin t cos t + t sin t cos t sin t cos t ) dt. 6π dt ( sin t cos t + t sin t cos t sin t cos t ) = 3π. (33) There are obviousl a few steps missing in that last integral!