I am very pleased with what I have nutted out (and explained) here, but Alan could you please examine it with a fine toothed combe to check that everything I have said is correct. 
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This question is, in a way, related to finding the volume of a hollow tube.
The normal way to do this is to find the area of the big circle and subtract the area of the little circle. and then times it by the length of the pipe (L)
$$\\V=(\pi b^2-\pi a^2)*h\\
=\pi*(b^2-a^2)*h$$
BUT if the cylinder is made up of an infinite nmber of infintesimally narrow shells there is another way to think about it.
The circumference of a shell is $$2\pi r$$ and the length of the cylinder is L if you flattern this out you will have a rectangle, the area of the rectangle is $$2\pi rL$$
The thinkness of the flatterened of this rectangle is $$dr$$ $$$$(dr is the difference in the radii of the outer circle to the inner circle)
the integral sign $$\int$$ is literally an S it stands to the sum of all the infintesimally small parts. (I would like to word this bit better )
So, if you want the volume of the pipe it will be $$V=L*\int_{inner\; radius}^{outer \;radius} \;2\pi r \;dr$$
NOTE that
$$\\\int_{inner \;radius}^{outer \;radius}\;(2\pi r) dr \\\\
= [\pi r^2]_{inner \;radius}^{outer \;radius}\\\\
=[\pi (outer \;radius)^2]-[\pi (inner \;radius)^2]\\\\
= $Area of the annulus$$$
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NOW - this question is not really about volume - it is about the surface are of an infinite number of infintesimally thin shells (together they make up the volume but think of them as shells)
The resistance is inversely proportional to the surface area that it connects with.
The surface area for any given radius is $$2\pi r *L$$
$$\\R\;\alpha \;\dfrac{1}{2\pi r*L}\\\\
$BUT r is all values from a to b \qquadSO$ \\\\
R\;\alpha \;\displaystyle\int_a^b\dfrac{1}{2\pi r*L}\;dr\\\\
$Adding in the constant which intrinsically belongs to the material we get$\\\\
R= \;\displaystyle\int_a^b\dfrac{\rho}{2\pi r*L}\;dr\\\\$$
NOW is that COMPLETELY CORRECT ? 
I had to think really hard to put that together. 
(I hope I didn't include any more typos and I really hope I know what I am talking about) 
