Surface Integral of quarter-cylinder help

[misterpickle]

Homework Statement

I am taking the surface integral over a quarter cylinder. Everything is fine and I can get the correct answer, it's just a conceptual problem that I need help with.

Homework Equations

The da for the "curved" outer surface is $$da=sd\phi dz\hat{s}$$
The da for the bottom surface is $$da=sdsd\phi (-\hat{s})$$

I understand why the curved da is multiplied by s, since we are integrating over a surface that is projected into 3-space by a distance s.

I do not understand why this s occurs in the da for the bottom (and top) surface. We integrate over a dynamic ds to find the surface integral for this piece...so why multiply the differential area by s?

 

[NateD]

The Attempt at a SolutionI think it is because we have to take into account the distance from the origin. The "ds" of the bottom surface is the arc length, which is defined by s\phi. So when we create a differential area, we must multiply it by s to take that distance into account.