Volume of sphere cut by two parrallel planes

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Koranzite
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Homework Statement



A sphere of radius R with centre at the origin is cut by two parallel planes at [itex]z=\pm a[/itex], 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



[itex]dV=rdrdθdz[/itex]

The Attempt at a Solution



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

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

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 [itex]\sqrt{R^2-z^2}[/itex]. 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 [itex]\sqrt{R^2-a^2}[/itex]. 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 [itex]\sqrt{R^2-a^2}[/itex].

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.