Surface Integral of quarter-cylinder

[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 [tex]da=sd\phi dz\hat{s}[/tex]
The da for the bottom surface is [tex]da=sdsd\phi (-\hat{s})[/tex]

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?

 

[lippyka]

The Attempt at a SolutionI think it has something to do with the fact that the top and bottom surfaces of the cylinder are in the x-y plane, so the surface normal is s\hat{s}. But I am still not sure why the s occurs.