My Math Forum Scalar Line Integral

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 January 4th, 2017, 06:52 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 10:53 PM.
 January 5th, 2017, 08:10 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,492 Thanks: 632 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 $\vec{f}(x, y, z)= xy$ over the curve $y= 3x^2$, $z= 2x$, 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: $x= t$, $y= 3t^2$, and $z= 2t$ with t from 0 to 1. Then $dx= dt$, $dy= 6tdt$, and $dz= 2dt$. $ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{1+ 36t^2+ 4]dt= \sqrt{5+ 36t^2}dt$. $f(x,y,z)= xy= (t)(3t^2)= 3t^3$. The path integral is [tex]\int_0^1 3t^3\sqrt{36t^2+ 5}dt[/rtex].
January 5th, 2017, 11:28 AM   #3
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 Originally Posted by Country Boy 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 $\vec{f}(x, y, z)= xy$ over the curve $y= 3x^2$, $z= 2x$, 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: $x= t$, $y= 3t^2$, and $z= 2t$ with t from 0 to 1. Then $dx= dt$, $dy= 6tdt$, and $dz= 2dt$. $ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{1+ 36t^2+ 4]dt= \sqrt{5+ 36t^2}dt$. $f(x,y,z)= xy= (t)(3t^2)= 3t^3$. The path integral is [tex]\int_0^1 3t^3\sqrt{36t^2+ 5}dt[/rtex].
Thank you for putting it for me in that way. One of my favorite things in math is being able to look at something in multiple different ways. The reason I wanted to do it by the definition is because I spent a good part of my summer trying to figure out what an integral actually was, and so I ended up teaching myself about summations and properties of summations etc, so that I could calculate an integral by actually taking the limit of a sum. But anyways that's why I want to do it this way. But thank you.

 January 5th, 2017, 02:38 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,492 Thanks: 632 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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