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Volume of intersection of spheres

  1. Jul 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the intersection of two spheres of radius 2, give that the center of each sphere lies on the surface of the other.

    3. The attempt at a solution

    I was trying to do this problem with volumes of revolution. I drew two circles, one with a center at -1, the other with a center at 1. I found the point of intersection on the y axis to be + and - root3. I was going to use the washer method, rotating the line y = (4-(x+1)^2)^(1/2) over the x axis but I can't integrate that expression. I'm also unsure if that would even give the correct answer. Any tips on how to approach the problem? This is for a calc 2 class btw, so it shouldn't be especially difficult.
     
  2. jcsd
  3. Jul 29, 2009 #2
    Woops just realized that I would be taking the integral of y^2 so that square root sign doesn't mess up my ability to integrate. Still though, will I get the correct answer if I find that volume and multiply it by two? My concern is that I'm overlooking something about the geometry of the intersection. Like it seems like the volume would be a sum of the area of ellipses, not circles. Yet I don't know how I would end up with an elliptical equation from two intersecting spheres? Help please.
     
  4. Jul 29, 2009 #3

    Dick

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    What range of x are you using when integrating using washers under the curve y=(4-(x+1)^2)^(1/2)? If you've got the right range, yes then you can just double that volume. The cross-sections of your volume perpendicular to the x-axis are circles, not ellipses. I don't think you are missing anything.
     
  5. Jul 29, 2009 #4
    The range I used is just x = 1 to 0...
     
  6. Jul 29, 2009 #5

    Dick

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    I used x=0 to x=1, but yes, that seems fine.
     
  7. Jul 29, 2009 #6
    Did anyone try this and get 8/3 * pi ??
     
  8. Jul 30, 2009 #7

    Dick

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    That's not what I get. Can you show how you got that?
     
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