# 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.

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 $dy$.

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.
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.