ideasrule said:
Wow, I used the wrong definite integral for csc x.
I don't think that \frac{\rho}{2\pi R}ln\left(\frac{2R-A}{2R+A}\right) can be right either, though, because that equation suggests R should approach 0 as A approaches 0; it should actually approach infinity.
Worse than that, look what happens when you plug in values for A that are fractional values of R (e.g. 0.5R, 0.25R, etc.) Everything turns out negative, although the absolute magnitude increases as the depth of the slab increases--which at least is a good sign.
For completeness, I'll give my
corrected full derivation here.
A circle, centered at the origin, has the following equation:
x^{2}+y^{2}=R^{2}
Solving for y for just the half that is above the origin (the height of our slice at any given point), we have:
y=+\left(R^{2}-x^{2}\right)^{1/2}
For a circular of radius y, the area is:
\pi y^{2}=\pi \left(R^{2}-x^{2}\right)
A circular slab, at any point along x, with thickness dx has a volume of:
\pi \left(R^{2}-x^{2}\right) dx
Assuming we have homogeneous resistivity \rho, the differential resistance (which we'll call dQ because we're already using R) of any of these slabs at any point such that -R<x<R is:
dQ=\frac{\rho \ell}{A}=\frac{\rho}{\pi \left(R^{2}-x^{2}\right)}dx
Instead of finding the resistance of the whole sphere, we'll just take the resistance of half the sphere (to the point x=0) and we'll slice off a slab of thickness A (obviously, \left|A\right|<R) from the rounded face. Consequently, we integrate along the interval -R+A<x<0.
Equivalently, if we sliced off a slab of thickness A, we'd be left with a slab of thickness B (again, \left|B\right|<R). In that case, we'd integrate in the interval between -B<x<0.
The Wikipedia gives the following integral (for \left|x\right| < \left|a\right|):
\int\frac{dx}{x^{2}-a^{2}}=\frac{1}{2a} ln \frac{a-x}{a+x}
http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions
So, we find the resistance Q by integrating the differential resistance dQ:
Q=\int dQ=-\frac{\rho}{\pi} \int \frac{1}{\left(x^{2}-R^{2}\right)}dxWith our expression for a slab of thickness A being sliced off the hemisphere, we do the following integral:
Q=-\frac{\rho}{\pi} \int^{0}_{-R+A} \frac{1}{\left(x^{2}-R^{2}\right)}dx
=-\frac{\rho}{\pi} \frac{1}{2R} ln \left.\frac{R-x}{R+x} \right|^{0}_{x=-R+A}
=-\frac{\rho}{2\pi R}\left(0 - ln \frac{R-(-R+A)}{R+(-R+A)} \right)
=\frac{\rho}{2\pi R} ln \frac{2R-A}{A}With our expression for a remaining slab of thickness B, we do the following integral:
Q=-\frac{\rho}{\pi} \int^{0}_{-B} \frac{1}{\left(x^{2}-R^{2}\right)}dx
=-\frac{\rho}{\pi} \frac{1}{2R} ln \left.\frac{R-x}{R+x} \right|^{0}_{x=-B}
=-\frac{\rho}{2\pi R}\left(0 - ln \frac{R-(-B)}{R+(-B)} \right)
=\frac{\rho}{2\pi R} ln \frac{R+B}{R-B}
EDIT: Which exhibit roughly the correct behaviour as B and A tend to 0 respectively, and whose resistances increase as B increases, and A decreases (i.e. thicker slab = higher resistance)