Volume of sphere cut by two parrallel planes

In summary, the student is trying to solve a problem involving the integration of a triple integral, but is having difficulty due to incorrect limits.
  • #1
Koranzite
6
0

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 independant 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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  • #2
I think r should go from 0 to [itex]\sqrt{R^2-z^2}[/itex]. Shouldn't it?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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!
 
  • #6
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.
 

What is the formula for finding the volume of a sphere cut by two parallel planes?

The formula for finding the volume of a sphere cut by two parallel planes is V = (2/3)πr3 - (1/3)πh2(3r - h), where r is the radius of the sphere and h is the distance between the two parallel planes.

Can the volume of a sphere cut by two parallel planes be negative?

No, the volume of a sphere cut by two parallel planes cannot be negative. It is always a positive value since volume is a measure of the amount of space an object occupies.

How do you find the height of the cut portion of a sphere?

The height of the cut portion of a sphere can be found by using the formula h = 2r - (V/(πr2)), where r is the radius of the sphere and V is the volume of the cut portion.

What is the relationship between the volume of a sphere and the distance between the two parallel planes?

The volume of a sphere cut by two parallel planes is directly proportional to the distance between the planes. As the distance increases, the volume of the cut portion also increases. Similarly, as the distance decreases, the volume of the cut portion decreases.

Can the volume of a sphere cut by two parallel planes be greater than the volume of the original sphere?

Yes, it is possible for the volume of a sphere cut by two parallel planes to be greater than the volume of the original sphere. This can happen if the distance between the two planes is greater than half of the sphere's radius.

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