Volume of sphere cut by two parrallel planes

Koranzite
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Homework Statement



A sphere of radius R with centre at the origin is cut by two parallel planes at z=\pm a, where a<R. Write, in cylindrical coordinates, a triple integral which gives the volume of that part of the sphere between the two planes. Evaluate the volume by first performing the r,θ integrals and the the remaining z integral.

Homework Equations



dV=rdrdθdz

The Attempt at a Solution



The main probelm here is the setting up of my integral, as the answer I am getting is independant of R, which is then clearly wrong.

My integral runs from:
r=\sqrt{R^2-a^2} to r=\sqrt{R^2-z^2}
θ=0 to θ=2\pi
z=-a to z=a

I would expect the answer to depend on R, but it keeps cancelling out when I evaluate the r integral. I would be grateful if someone could explain what is wrong with my limits.
 
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I think r should go from 0 to \sqrt{R^2-z^2}. Shouldn't it?
 
Well that is certainly true to get the formula for the volume of the whole sphere, but in this case the minimum value that r takes is \sqrt{R^2-a^2}. Your proposal is one that I have considered, but I don't see how it can be justified.
 
Koranzite said:
Well that is certainly true to get the formula for the volume of the whole sphere, but in this case the minimum value that r takes is \sqrt{R^2-a^2}.

You already have the z limits from -a to a. Doesn't that take care of the a dependency? You may be visualizing the r coordinate wrong. It's the distance from the z axis to the edge of your solid parallel to the x-y plane.
 
Ah yes, just recognised the problem. I was only imagining r as being the distance to the surface from the z axis, neglecting all the interior volume where it can reduce to 0... Thanks!
 
Koranzite said:
Ah yes, just recognised the problem. I was only imagining r as being the distance to the surface from the z axis, neglecting all the interior volume where it can reduce to 0... Thanks!

Right. Your original limits would be for the volume of a sphere with a cylinder cut out of it. Interesting that the R cancels, isn't it? You might not guess that to be true looking a picture of it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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