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A line integral (with respect to arc length) can be interpreted geometrically as an area - The area between C and f(x,y) along C as in the picture. You sum up the areas of all the infinitesimally small 'rectangles' formed by f(x,y) and ds.

What Im wondering is how do I interpret line integrals with respect to x or y geometrically?

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You rather integrate with respect to the arclength, that's why you use $dS$. In turn, this will depend on $dx$ and $dy$ for cartesian cordinates or on $d\theta$ and $dr$ for polar coordinates, i.e. (and don't take this strictly) $$ds^2 = dx^2+dy^2$$ $$ds^2 = dr^2+r^2d\theta^2$$ –  Pedro Tamaroff Mar 4 '12 at 19:44
    
You get the "algebraic sum" of the projected areas. E.g., if $C$ is a circle in the $(x,y)$-plane you get zero. –  Christian Blatter Mar 4 '12 at 21:07
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2 Answers

This is not a really detailed answer, however:

I think that pulling back the integral to the $x$ or $y$ axis is geometrically unnatural (and you have to decide how you want to do this, since generically $C$ will not be a graph over either axis), so I wouldn't expect a really good geometric interpretation. But you could do something like the following. Break up $C$ into pieces $C_i$ so that each $C_i$ is a graph over either the $x$ or $y$ axis, and then you can use the chain/substitution rule to literally write the line integral as an integral w.r.t. $x$ or $y$ for each $C_i$, and this makes the relationship explicit. This construction gives you (if $C_i$ is a graph of $r(x)$, for instance) something like the area under the curve $f(x, r(x))\sqrt{1 + r'(x)^2}$, so it doesn't seem very natural.

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So if there is no geometric interpretation of line integrals wrt to x or y, what exactly am i calculating when i perform this operation...what is the purpose of line integrals wrt to x and y? –  Jim_CS Mar 4 '12 at 21:17
    
I guess I would turn it around and ask you, what do you mean by line integral w.r.t. x or w.r.t y? Whenever I do a line integral over a curve $C$, I usually want to think of $C$ as a parametrized curve, that is $C$ is the image of a function $$ s \to (f(s), g(s))$$ in which case it makes sense to use the line element $ds$ instead of trying to relate things back to the $x$ or $y$ axis. If I want $C$ to lie in the $x, y$ plane I can set $f(s) = x$ and $g(s) = y$ but under this setup it isn't clear that I should get anything meaningful if i try to integrate over the $x$ or $y$ axis. –  treble Mar 4 '12 at 21:22
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By w.r.t. to x or y I mean using dx instead of ds. I am reading Chapter 16 of Calculus Early Transcendentals and it features these line integrals w.r.t. to x and y but it gives no information as to their meaning. –  Jim_CS Mar 4 '12 at 21:38
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I took a calculus class from that book a long time ago, but I no longer have a copy and therefore I would need to see the page of text to which you are referring in order to figure out what is happening. But if I had to guess I would say they probably mean to consider $s = x$ so that $C$ is given by the parametric equations $(x, g(x))$. In this case you may use the definition of line integral to re-write the line integral as an ordinary single-variable calculus integral. –  treble Mar 4 '12 at 21:43
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I doubt if this will be helpful since it's such a late reply. however, I do want to share my thoughts on this. So usually the line integral of a graph over a curve C is ∫P(x,y)dx+Q(x,y)dy (over C), but if you are only talking about ∫P(x,y)dx, it is simply an integral of the graph over x, which in graphic sense will be the integral of the graph of f(x,y) over C after projecting to xz plane. Not sure if this is what u r looking for :)

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The question was about a line integral but it seems you are considering it as a Riemann integral. –  Ittay Weiss Nov 12 '12 at 4:52
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