# Surface Integral of quarter-cylinder

by misterpickle
Tags: cylinder, electrodynamics, griffiths
 P: 12 1. The problem statement, all variables and given/known data 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. 2. Relevant 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?