Volume of sphere cut by two parrallel planes

Click For Summary

Homework Help Overview

The problem involves calculating the volume of a sphere of radius R that is intersected by two parallel planes at z = ±a, where a is less than R. The task is to set up a triple integral in cylindrical coordinates to find the volume of the portion of the sphere between these planes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the limits of integration for the cylindrical coordinates, particularly the range for r and z. There is a focus on ensuring the integral reflects the geometry of the sphere and the planes correctly.

Discussion Status

Some participants have offered insights into the limits of integration, with one suggesting a reconsideration of the minimum value for r. There is acknowledgment of the complexity in visualizing the geometry and how it affects the integral setup. The discussion is ongoing, with participants exploring different interpretations of the problem.

Contextual Notes

There is a concern regarding the dependency of the final volume on the radius R, as some calculations appear to lead to results that do not include R. This raises questions about the setup of the integral and the assumptions made about the geometry involved.

Koranzite
Messages
5
Reaction score
0

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 independent 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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K