Interpreting Line Integrals with respect to $x$ or $y$ 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 $ds$.

What I'm wondering is how do I interpret line integrals with respect to $x$ or $y$ geometrically?
 A: 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. $(2)$.
A: 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 $f(x,y)\not= 1$). 

$\int_C 1\,dx$ 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.
A: 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.
A: The geometric interpretation of line integral with respect to $x$ is as the following figure. I can't give a geometrical interpretation of the line integral with respect to $y$ in this case because the direction of $y$ back and forth when $t$ increase.

A: Wouldn't a geometric representation of a line integral in respect to x be taking the surface whose area you would have calculated when taking the line integral in respect to arc length:
http://i.stack.imgur.com/8LVrc.png  (the same image in another answer here)
and then projecting that surface onto the XZ plane. The area of that projection would essentially be the geometric representation of the line integral in respect to x. A line integral in respect to y would be the projection of that same surface onto the YZ plane. The projections would be like viewing that surface in the image above just through the XZ plane and then through the YZ plane. Is this a correct interpretation? Sorry, I am new here so I couldn't put the image in my post, so I just put the link. 
