January 4th, 2017, 05:52 PM  #1 
Newbie Joined: Jan 2017 From: United States Posts: 4 Thanks: 0  Scalar Line Integral
Hello all, I'm trying to understand the scalar line integral, enough so that I should be able to do it without the formula. But I got kind of confused at the end of the first image (attached). I was thinking that adding up the lengths of all of the tangent vectors would give the total length of the curve C. But then I realized I would be missing the (dt). I did figure it out by looking at the units, but it still feels like I'm missing some part of the big picture. Can somebody please help me? Also, I think that my geometric interpretation is probably wrong. I was thinking that the scalar line integral gave the area between a surface and a curve, (wikipedia called it the area of the "curtain"). But again realizing that f(x,y,z) is not even a surface, and if f(x,y,z)=1 then integrating it over a curve C will just give the length of C. But what about the plane z(x,y)= 1? If C was some curve, wouldn't the line integral of z=1 over C give the area between a plane and a curve? Last edited by skipjack; January 4th, 2017 at 09:53 PM. 
January 5th, 2017, 07:10 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699 
You say, "I'm trying to understand the scalar line integral, enough so that I should be able to do it without the formula". I would recommend NOT worrying about the basic definition in terms of Riemann sums just as you do not use Riemann sums to do integrals in Calculus I. To integrate, say over the curve , , from (0, 0, 0) to (1, 3, 2), write the curve in terms of a single parameter. Here, we we can use x itself as parameter: , , and with t from 0 to 1. Then , , and . . . The path integral is [tex]\int_0^1 3t^3\sqrt{36t^2+ 5}dt[/rtex]. 
January 5th, 2017, 10:28 AM  #3  
Newbie Joined: Jan 2017 From: United States Posts: 4 Thanks: 0  Reply Quote:
 
January 5th, 2017, 01:38 PM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699 
Well, it is good to do it that way once to see how it works. But once you have done that, in one dimension, you shouldn't have to do it again!


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integral, line, scalar 
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