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

enter image description here

What I'm 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

3 Answers 3

Let's compare the definitions of these three related, but distinct concepts. Let $C$ be a parametrized curve with respect to the parameter $t\in[a,b]$. Then

\begin{equation}\tag{1} \int_C f(x,y)\,ds := \int_a^b f(x(t),y(t))\,\color{blue}{\sqrt{[x'(t)]^2+[y'(t)]^2}}\,dt \end{equation} whereas \begin{align} \int_C f(x,y)\,dx &:= \int_a^b f(x(t),y(t))\,\color{red}{x'(t)}\,dt,\tag{2}\\ \int_C f(x,y)\,dy &:= \int_a^b f(x(t),y(t))\,\color{green}{y'(t)}\,dt.\tag{3} \end{align}

You seem to understand the geometric interpretation of (1): it is the area of the "fence" built along the curve $C$ whose height along any point $(x,y)$ on $C$ is given by $f(x,y)$. Alternatively, focus on the multiplier in blue in (1): we are weighting the integrand $f(x(t),y(t))$ by the length of the velocity vector along $C$.

On the other hand, in (2), we are weighting the integrand by only the $x$ component of the velocity vector.

In (3), we are weighting the integrand by only the $y$ component of the velocity vector.

As a simple example, consider $f(x,y)=1$.

\begin{align} \int_C 1\,ds&=\int_a^b \sqrt{[x'(t)]^2+[y'(t)]^2}\,dt =\text{length of }C\\ \int_C 1\,dx&=\int_a^b x'(t)\,dt =x(b)-x(a)=\text{net displacement in $x$ direction as $C$ is traversed}\\ \int_C 1\,dy&=\int_a^b y'(t)\,dt =y(b)-y(a)=\text{net displacement in $y$ direction as $C$ is traversed}. \end{align}

Draw a simple example of something like an $S$ shaped curve for $C$ and look at the three quantities above in that setting.

Edit: Here is an admittedly crude graphical interpretation of what (2) and (3) mean in the particular case when $f(x,y)=1$ (and I realize that in the picture $f(x,y)\not= 1$).

enter image description here

$\int_C 1\,dx$ corresponds to the dark red line on the $x$ axis while $\int_C 1\,dy$ corresponds to the dark blue line on the $y$ axis.

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This diagram came from James Stewart's Calculus. One example from the text was $\int_C 2x\,ds$, where $C$ goes from $(0, 0)$ to $(1,1)$ along $y = x^2$ and from $(1, 1)$ to $(1, 2)$ along the vertical line segment. With the parametrization $x = x$, $y = x^2$ in the parabola and $x = 1$, $y = y$ in the line segment, you will get $\int_C f(x,y)\,ds = \int_{0}^{1} 2x\sqrt{1+4x^2}\,dx + \int_{1}^{2} 2\,dy$. I'm not seeing this as displacements in the x- or y- direction. Rather, as Mr. Treble indicated below, the parametrization just "decomposed" as such. Please kindly comment, TY! –  Andy Tam Nov 11 at 4:01
    
I said that was the interpretation when $f(x,y)=1$. –  JohnD Nov 11 at 4:07

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
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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

Given a function $(x,y)\mapsto z=f(x,y)$ and a curve $$\gamma:\quad s\mapsto{\bf z}(s)=\bigl(x(s),y(s)\bigr)\qquad(a\leq s\leq b)$$ parametrized with respect to arc length, the integral $$\int_\gamma f(x,y)\>ds:=\int_a^b f\bigl(x(s),y(s)\bigr)\>ds\tag{0}$$ can be interpreted in various ways. You have chosen to interpret it as surfacr area of a "Christo curtain" $S$ displayed along $\gamma$ and having height $f(x,y)$ at the point $(x,y)\in\gamma$.

Now you want an interpretation of the integral

$$\int_\gamma f(x,y)\>dx:=\int_a^b f\bigl(x(s),y(s)\bigr)\>\dot x(s)\>ds\tag{1}$$ in a similar vein.

Since $\gamma$ is parametrized with respect to arc length one has $$\dot {\bf z}(s)=\bigl(\cos\theta(s),\sin\theta(s)\bigr)\ ,$$ where $\theta(s)$ is the angle between the positive $x$-axis and the tangent vector $\dot{\bf z}(s)$. So we can replace $(1)$ by $$\int_\gamma f(x,y)\>dx=\int_a^b f\bigl(x(s),y(s)\bigr)\>\cos\theta(s)\>ds\tag{2}$$ In $(2)$ an "infinitesimal curtain element" no longer weighs in with its area $dS=f\bigl(x(s),y(s)\bigr)\>ds$ as in $(0)$ but with the area $dS'$ of the shadow (or projection) of this element onto the $(x,z)$-plane. Therefore one is tempted to say that $(1)$ represents the total area of the projected curtain.

But there is more to it: Note that $\cos\theta(s)$ has a sign. When $\cos\theta(s)$ is positive then the curtain has its good side towards the $x$-axis, and when $\cos\theta(s)$ is negative its backside. The latter parts of the shadow are counted negative. Similarly, if the same part of the $(x,z)$-plane is "shadowed" several times by successive pleats of the curtain, all these "coverings" are summed up with their proper sign in $(1)$, resp. $(2)$.

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