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Finding the volume between two spheres

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the volume outside the sphere x2 + y2 + z2 = 1/2 and inside the sphere x2 + y2 + z2 = z

    2. The attempt at a solution

    I've gotten as far as to visually seeing that's there's two spheres and determining that the radius of the first sphere is 1/√2. However, I'm still not clear as to how to approach this problem in order to find the bounds since the second equation is only equal to z. I think this is using spherical coordinates since the problem literally utilizes two spheres. My approach was to bring over the z, factor out the z and solve for it, leaving me with z = 0 and z = 1.

    Any help in clearing up this confusion is greatly appreciated.
     
  2. jcsd
  3. Dec 1, 2011 #2

    Filip Larsen

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    Perhaps you can write up an expression for the volume inside of each of the two spheres? If so, can you then think of a way to calculate the volume between the two concentric spheres from the two volumes?
     
  4. Dec 1, 2011 #3
    Would that then involve the subtraction of the volume of the first sphere from the second sphere?
     
  5. Dec 1, 2011 #4
    For the 2nd sphere, you could fiddle with the algebra, but it's easier to make a sketch. (Note that the points (0,0,0), (±1,0,1), and (0,±1,1) all satisfy the equation.) From the sketch, the 2nd sphere obviously has a radius of 1 and is centered at (0,0,1). (The point (0,0,1) also satisfies the equation, but it's degenerate.)

    Therefore, the spheres intersect on a plane parallel to the XY plane. Find the Z coordinate of that plane and then use the formula for the volume V of a spherical segment,
    [tex]V = \pi * h^2 * (r - h/3)[/tex]
    where r is the radius of the sphere and h is the thickness of the segment to find the volumes of the two segments. Add 'em up, and there you are!

    (Alternatively, you could solve a triple integral, but that seems like overkill.)
     
  6. Dec 1, 2011 #5
    I appreciate the help but I need to strictly use integrals, even if that involves triple integrals
     
  7. Dec 1, 2011 #6

    Office_Shredder

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    If you're having trouble with the second equation, being a bit weird, this is how you should re-write it:
    [tex] x^2+y^2+z^2-z=0[/tex]
    [tex]x^2+y^2+(z-1/2)^2-1/4=0[/tex]
    by completing the square with the z's

    Then you need to find a way to describe the intersection of the two spheres with a couple of equations, and write down an integral (probably in polar coordinates)
     
  8. Dec 1, 2011 #7

    Filip Larsen

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    Indeed, yes, but I most admit I didn't catch before that you have z on both sides of the equation for the last sphere which makes the spheres overlapping and the problem a bit more difficult.

    In order to make the equation for the last sphere on a form that resembles the equation of a sphere you need to bring the right hand side z over to the other side and then try if you can bring the resulting z2-z into the form (z-z0)2 - r2, where z0 and r are two new constants. After that you can then move r2 over to the other side as the square of the radius of the sphere.

    However, all this doesn't help that much since you now end up with two overlapping spheres where you cannot simply subtract the volumes like I suggested in my first post. The only way forward I can see is to make a sketch of the spheres in a x-z diagram and then use this a as guide to find the resulting volume by integrating the difference in z between the two spheres upper half-sphere for relevant x and y.
     
  9. Dec 1, 2011 #8
    I used this method and set y = 0 for equations. I then solved for x and set both equations equal to each other as follows:

    [tex]x^2+z^2=1/2[/tex]
    [tex]x^2+(z-1/2)^2=1/4[/tex]
    [tex]0=1/2-z^2[/tex]
    [tex]0=1/4-(z-1/2)^2[/tex]

    I got z to equal 1/2 and plugged it back into those 2 equations I got when I set y = 0 and got 1/2 for the first one and ±1/2 for the second one. Am I on the right track? I'm guessing the 1/2's represent the z bounds
     
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