Volume of two pieces of a sphere cut by a plane

ParoxysmX
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Homework Statement



Consider the unit sphere x^{2} + y^{2} + z^{2} = 1

Find the volume of the two pieces of the sphere when the sphere is cut by a plane at z=a.

The Attempt at a Solution



My interpretation is that a is a point on the z-axis that the plane cuts at. So the height of the segment is r-(r-a). After that, I'm not sure how to proceed. Should you somehow integrate the volume of the segment between r=1 and r=a?
 
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Volume or area? You say volume in the title and area in the question.
 
LastOneStanding said:
Volume or area? You say volume in the title and area in the question.

Volume. Edited accordingly.
 
ParoxysmX said:
My interpretation is that a is a point on the z-axis that the plane cuts at. So the height of the segment is r-(r-a). After that, I'm not sure how to proceed. Should you somehow integrate the volume of the segment between r=1 and r=a?

Well, you will have ##a<1## (since it's a unit sphere) so you will want to go the other way around; but, yes, what you are proposing should work. If you give a try and get stuck, show your work in lots of detail and we'll be able to help more.
 
The height of the segment is r-a. Your expression of the height, r - (r - a), equates to just a.
 
Obviously a must be between -1 and 1. It is sufficient to assume that 0\le a\le 1 and calculate the volume above z= a and below the sphere. (Of course, the volume of the entire sphere is (4/3)\pi so the volume low z= a is (4/3)\pi minus the volume above. And if a< 0, just flip it over.)

In cylindrical coordinates the base is given by r^2+ a^2= 1 so the cover that base r goes from 0 to 1- a^2 and \theta from 0 to 2\pi. For each r and \theta, the height is z- a= \sqrt{1- r^2}- a
 
Ok I believe I have that one figured out. The next problem now is to calculate the surface area of those segments using spherical coordinates. I'm told the formula S = \int^{b}_{\phi=a}\int^{d}_{\theta=c} sin\theta d\theta d\phi should be used. What are a, b, c and d here?
 
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