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Surface Integral of quarter-cylinder

by misterpickle
Tags: cylinder, electrodynamics, griffiths
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Sep30-10, 08:56 PM
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 [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 why multiply the differential area by s?
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