1. ME 2304: 3D Geometry & Vector Calculus
Line Integrals
Dr. Faraz Junejo

2. Line Integral
In this lecture, we define an integral that is similar to a single integral except that, instead of integrating over an interval [a, b], we integrate over a curve C.
Such integrals are called line integrals.

3. Line Integral
In mathematics, a line integral (sometimes called a path integral, contour integral, or curve integral) is an integral where the function to be integrated is evaluated along a curve.
The function to be integrated may be a scalar field or a vector field.

4. Line Integrals (Contd.)
Consider the following problem:
A piece of string, corresponding to a curve C, lies in the xy-plane. The mass per unit length of the string is f(x,y). What is the total mass of the string?
The formula for the mass is:
The integral above is called a line integral of f(x,y) along C.

5. Line Integrals with Respect to Arc Length
We use a ds here to acknowledge the fact that we are moving along the curve, C, instead of the x-axis (denoted by dx) or the y-axis (denoted by dy).
Because of the ds this is sometimes called the line integral of f with respect to arc length.

6. Line Integrals with Respect to Arc Length
Question: how do we actually evaluate the above integral?
The strategy is:
(1) parameterize the curve C,
(2) cut up the curve C into infinitesimal pieces i.e. small pieces,
(3) determine the mass of each infinitesimal piece,
(4) integrate to determine the total mass.

7. Arc Length
We’ve seen the notation ds before. If you recall from Calculus I course, 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.

8. Computing 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.

9. Computing Line Integral
If we use the vector form of the parameterization we can simplify the notation up by noticing that,
Using this notation the line integral becomes,

10. Special Case
In the special case where C is the line segment that joins (a, 0) to (b, 0), using x as the parameter, we can write the parametric equations of C as:
x = x
y = 0
a ≤ x ≤ b

11. Special Case Line Integral formula then becomes
So, the line integral reduces to an ordinary single integral in this case.

12. Line Integrals
Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area.

13. Line Integrals
In fact, if f(x, y) ≥ 0, represents the area of one side of the “fence” or “curtain” shown here, whose:
Base is C.
Height above the point (x, y) is f(x, y).

14. Example: 1

15. Example: 1 (contd.)

16. Exercise: 1 x = cos t y = sin t Evaluate
where C is the upper half of the unit circle x2 + y2 = 1
To use Line Integral Formula, we first need parametric equations to represent C.
Recall that the unit circle can be parametrized by means of the equations
x = cos t y = sin t

17. Exercise: 1 (contd.)
Also, the upper half of the circle is described by the parameter interval 0 ≤ t ≤ π

18. Exercise: 1 (contd.)
So, using Line integral Formula gives:

19. Exercise: 2 Evaluate Where, C is the upper right quarter of a circle
x2 + y2 = 16, rotated in counterclockwise direction.
Answer:
256/3

20. Piecewise smooth Curves

21. Piecewise smooth Curves
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,

22. Example: 2

23. Example: 2 (contd.)
At first we need to parameterize each of the curves, i.e.

24. Example: 2 (contd.)

25. Example: 2 (contd.)
Notice that we put direction arrows on the curve in this example.
The direction of motion along a curve may change the value of the line integral as we will see in the next example.

26. Example: 2 (contd.)
Also note that the curve in this example can be thought of a curve that takes us from the point (-2,-1) to the point (1, 2) .
Let’s first see what happens to the line integral if we change the path between these two points.

27. Example: 3 vector form of the equation of a line
we know that the line segment start at (-2,-1) and ending at (1, 2) is given by,

28. Example: 3 (contd.)

29. Summary: Example: 2 & 3
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.

30. Example: 4

31. Example: 4 (contd.)
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 (discussed in Exercise: 2 later on) in which this won’t be the case.

32. Example: 4 (contd.)
We will see more examples of this in next sections so don’t get it into your head that changing the direction will never change the value of the line integral.

33. Fact: Curve Orientation
Let’s suppose that the curve C has the parameterization
x = h(t ) , y = g (t )
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 (i.e. drawn) out as t increases.

34. Fact (Contd.)
Finally, let -C 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 -C is the same curve as C except the direction has been reversed.

35. For instance, here Fact (Contd.)
The initial point A corresponds to the parameter value.
The terminal point B corresponds to t = b.
We then have the following fact about line integrals with respect to arc length.

36. Exercise: 1
Evaluate
where C consists of the arc C1 of the parabola y = x2 from (0, 0) to (1, 1) followed by the vertical line segment C2 from (1, 1) to (1, 2).

37. Exercise: 1 (Contd.) The curve is shown here.
C1 is the graph of a function of x, as y = x2
So, we can choose t as the parameter.
Then, the equations for C1 become: x = t y = t ≤ t ≤ 1

38. Exercise: 1 (Contd.)
Therefore,

39. Exercise: 1 (Contd.) On C2, we choose y as the parameter.
So, the equations of C2 are: x = y = t ≤ t ≤ 2 and

40. Exercise: 1 (Contd.)
Thus,

41. Line Integrals w.r.t. x & y
The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t:
x = x(t)
y = y(t)
dx = x’(t) dt
dy = y’(t) dt

42. Line Integrals w.r.t. x & y (contd.)

43. Exercise: 2 Evaluate Where, C is the upper right quarter of a circle
x2 + y2 = 16, rotated in counterclockwise direction.

44. Exercise: 2 (contd.)

45. ABBREVIATING
It frequently happens that line integrals with respect to x and y occur together.
When this happens, it’s customary to abbreviate by writing

46. Line Integrals
When we are setting up a line integral, sometimes, the most difficult thing is to think of a parametric representation for a curve whose geometric description is given.
In particular, we often need to parametrize a line segment.

47. VECTOR REPRESENTATION
So, it’s useful to remember that a vector representation of the line segment that starts at r0 and ends at r1 is given by:
r(t) = (1 – t)r0 + t r ≤ t ≤ 1
It is simplification of vector equation of line

48. Exercise: 3 Evaluate where
C = C1 is the line segment from (–5, 3) to (0, 2)
C = C2 is the arc of the parabola x = 4 – y2 from (–5, 3) to (0, 2).
In effect,
C2 = - C1

49. Exercise: 3(a) A parametric representation for the line segment is:
x = 5t – 5 y = 5t – ≤ t ≤ 1
Use Vector Equation of line with r0 = <–5, 3> and
r1 = <0, 2>.

50. Exercise: 3(a) contd.
Then, dx = 5 dt, dy = 5 dt, and Formulas give:

51. Exercise: 3(b) The parabola is given as a function of y.
So, let’s take y as the parameter and write C2 as:
x = 4 – y2 y = y –3 ≤ y ≤ 2

52. Exercise: 3(b) contd.
Then, dx = –2y dy and, by employing Formulas, we have:

53. Exercise: 3 (Summary)
Notice that we got different answers in parts a and b of Exercise 3 although the two curves had the same endpoints.
Thus, in general, as discussed earlier, the value of a line integral depends not just on the endpoints of the curve but also on the path.

54. Exercise: 3 (Summary)
Notice also that the answers in Exercise 3 depend on the direction, or orientation, of the curve.
If –C1 denotes the line segment from (0, 2) to (–5, –3), you can verify, using the parametrization x = –5t y = 2 – 5t 0 ≤ t ≤ 1 that gives

55. Arc length w.r.t. x & y
If –C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A in the previous figure), we have:

56. Example: 5

57. Example: 5 (contd.)

58. Example: 5a (contd.)

59. Example: 5b (contd.)

60. Example: 5b (contd.)
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.
As mentioned earlier 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.

61. Example: 5c (contd.)

62. Line Integrals in 3D

63. Example: 6
Here is a sketch of the helix.

64. Example: 6 (contd.)

65. Exercise: 1
Evaluate
where C is the circular helix given by the equations x = cos t y = sin t z = t 0 ≤ t ≤ 2π

66. Exercise: 1 (contd.)