What does line integral with respect to arc-length on the cuve C mean?

We're studying some basic vector calculus in my class right now. I can't seem to grasp this.





The first question is just that -


S.c ds (S.c = integral over curve c), where c is a line segment from (0,0) to (0,1). It is obvious that it will be 1. But I don't _really_ understand the concept.

What does line integral with respect to arc-length on the cuve C mean?
♠ indeed example of yours is not much illustrative for peculiarities of linear integration! To feel what is linear integration about you should consider a typical problem from Physics like this one:


♣ let vector F=F(x,y) be a force field of a set of electric charges, acting on a charge q being moved along the path C=C(x,y) from point A=C(x1,y1) to point B=C(x2,y2); Question: what is work done on the charge?


♦ Solution: work =∫(ds·F) from A to B, where (F·ds) is dot product of elementary vector ds along curve C(x,y) by vector F(x,y);


♥ in your example F=1*i, ds=i*dx; A=(0,0), B=(0,1); work=1;
Reply:Perhaps an example may help you. Assume you have a two variable function f(x,y) that takes real values, i.e. at each point (x,y) in the plane the function takes a scalar real value f(x,y). The line integral along a curve or path C in the plane represents the sum of all the values of f(x,y) for all points *on* the curve C, scaled by the arc length along C. This is the limit process of a summation of values of f(x,y) at a finite sequence of points in C scaled by the distance between each successive pairs of points.





A visual image is very good in my experience. Imagine the projection of the C curve onto the surface f(x,y). You will have a curve on f(x,y). Color all the z coordinates of points on this curve. Then you will get a curved 2D surface. The line integral along C is the area of this 2D surface scaled by arc length of the segment of C that you are using in your integral.





So, concerning your example with the curve C being the line between (0,0) and (0,1) that takes a constant value 1 (you only have ds inside the path integral).





Then f(x,y) = 1 is the flat plane at z = 1 spanning all directions in 3D space. C is the segment of line y = 1 between (0,0) and (0,1). This line projects onto the plane z=1 as I mentioned before. All points on this line segment have z value 1, i.e. f(x,y) is 1. Moreover the length in the xy plane has length 1. The area of the square that forms is 1x1, i.e. 1, and this is the value of the line integral.

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