I'm having trouble understanding the idea of a surface integral of a real valued function $f$. I've read some of the other answers here on Stack Exchange, but they seem to be focused on the surface integral of a vector field.
If I have a surface $S$, and I'm taking the surface integral $\int_S f dA$, my understanding is that I'm finding the "volume" underneath the the surface S, analogous to how a line integral of a real valued function is finding the area underneath the graph of $z=f(x,y)$ when we traverse through a curve $C$ on the $x,y$ plane.
But I am having trouble visualizing this. It seems like I can't visualize in a 3-d plot, since $f(x,y)$ would take 3 dimensions to plot, but the surface also lives in 3-d space, so I'm not sure what it means geometrically to take the surface integral of a real valued function. Does anyone have any analogies to help withe intuition here?
Analogies:
Line Integral over scalar function: Walking along a path in the x-y plane, graphing f(x,y) on the z-axis. The total area underneath the "fence" carved out by the the path, and f(x,y).
Line Integral over vector field: Walking along a path in the x-y plane, and being pushed around by a mysterious force at each point. The total amount of "work" exerted on me as I walk along the curve.
Surface Integral over vector field: Placing a parachute (surface) in a region with lots of turbulence, such that the force acting on the parachute at each point is different.
Questions: What's the relevant one for the surface integrals over a real valued function, similar to the ones I've provided above? What's the geometric intuition? (i.e. what volume am I calculating)?
What's keeping you from visualizing a volume on a surface embedded in 3d? For instance, imagine a perfect sphere centered at the core of our earth whose surface is at sea level. And imagine the surface of earth with all its mountains and deep sea trenches as the plot of a function on this perfect sphere. The surface integral of this function would be the signed volume between earth's surface and this perfect sphere (not exactly, since the surface integral doesn't take into account the effect that volume doesn't scale linearly with the radius, but it's approximately true as long as earth's surface remains close to that of the perfect sphere).