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## Homework Statement

A uniform surface charge lies in the region z = 0 for [tex]x^2 + y^2 > a^2[/tex], and z = [tex]\sqrt{a^2-x^2-y^2}[/tex] for [tex]x^2 + y^2 \leq a^2[/tex]. Find the force on a unit charge placed at the point (0,0,b)

## Homework Equations

dF = G ([tex]\delta[/tex])(vector of point to surface) / (magnitude of surface)^3

[tex]\delta[/tex] is constant because it's a uniform surface.

surface area = [tex]\int\int{\left\|{\frac{\partial f}{\partial x}} \times {\frac{\partial f}{\partial y}}\right\| dx dy}[/tex] = [tex]\int\int{{\sqrt{1 + ({\frac{\partial f}{\partial x}})^2 + ({\frac{\partial f}{\partial y}})^2} dx dy}[/tex]

## The Attempt at a Solution

So basically, I know that this is a hemisphere with radius a centred at the origin, with an infinitesimal surface spanning the rest of the xy-plane. I know how to establish the vector from (0,0,b) to a

*sphere*because a sphere is just radius a all around. I can identify any point on the sphere as (a sin [tex]\varphi[/tex] cos [tex]\theta[/tex], a sin [tex]\varphi[/tex] sin[tex]\theta[/tex], a cos[tex]\varphi[/tex])

But this isn't a sphere; it's a really big sheet with a bump at the origin. I don't know how to identify the vector.

The integral bounds seem simpler, but are just as complicated for me. In cylindricals, for the hemisphere, z goes from 0 to [tex]\sqrt{a^2-r^2}[/tex], r goes from 0 to 1, and [tex]\theta[/tex] goes from 0 to 2 [tex]\pi[/tex]. But isn't this a surface integral? Why do I have three variables? I really have no idea what I'm doing.