What do the differentials $dx$ and $dy$ do in line integrals (path integrals)? I think that using the arc length differential $ds$ in a line integral finds the area like in https://en.wikipedia.org/wiki/File:Line_integral_of_scalar_field.gif, but I cannot imagine what the area using $dx$ or $dy$ is. 
Sometimes the question asks "Evaluate 
a) $$\int_C xy^2\,dx$$ b) $$\int_C xy^2\,dy$$ c) $$\int_C xy^2\,ds$$
where the path $C$ is defined by $x=4\cos(t)$, $y=4\sin(t)$, $0\leq t\leq \pi/2$."
I know how to evaluate these, but what do they mean geometrically? Specifically, a) and b) since their differentials are $dx$ and $dy$.
 A: The integrals in (a) and (b) are components of the line integral of a vector field.  For example, the line integral of the vector field $F(x,y)=(xy^2,x^2y)$ can be written as
$$
  \int_C F(\mathbf{r}) \cdot \,d\mathbf{r} = \int_C xy^2 \,dx + x^2y \,dy
$$
where 
$$
  \int_C xy^2 \,dx = \int_a^b x(t)y(t)^2 x'(t) \,dt.
$$
The Wikipedia "Line integral" article has a gif depicting the line integral of a vector field, but there really isn't any geometric meaning to the individual components of this integral.  
A: This question is essentially the same as the one here. Let me point you to the second answer given there by bfhaha (and then expand on it): the integral $\int_C xy^2 dx$ is the "net area" of the projection of the ribbon above the curve $C$ onto the $xz$-plane, and similarly $\int_C xy^2 dy$ is the "net area" of the projection of the ribbon above the curve $C$ onto the $yz$-plane. Here is the picture of one of these "projected line integrals" that is given in that answer.
I put "net area" in quotes because we have to be careful: these integrals takes in to account the "folds" obtained when we project the ribbon. Consider in the picture linked above the projection onto the $yz$-plane involved with $\int_C xy^2 dy$. If we imagine creating this projection starting from the endpoint of $C$ closest to the origin, then we begin by moving in the positive $y$ direction, but then turn around and move in the negative $y$ direction, then once more turn and move in the positive $y$ direction. When we move in the positive $y$ direction, the positive area from our projected ribbon is added, but when we move in the negative $y$ direction, the positive area we're projecting is then subtracted, and finally when we turn again and start moving in the positive $y$ direction once more, the positive area we're projecting is added. 
