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Spherical Coordinates

  1. Nov 27, 2006 #1
    I need to use spherical coordinates to try and find the volume of the region bounded by

    x2 + y2 + z2 = 2 which converts to p=Sqrt.(2) a sphere

    and

    z = x2 + y2, a parabloid which I converted to cot(phi)csc(phi)=p

    I hope the greek letters for these are comonly used
    x=psin(phi)cos(theta)

    so far I have

    V=SSS (p^2)sin(phi) dp d(theta) d(phi)

    i know that theta goes from 0 to 2pi but how do I find the other limits?
     
  2. jcsd
  3. Nov 27, 2006 #2
    Well, think a little about what this shape looks like. It's all in the "upper" half of the universe, so it's not surprising that you can use limits of 0 and pi for phi. As for rho... you'll actually have to divide the integral into two parts: one that starts at 0 and terminates on the sphere, and one that starts at 0 and terminates on the parabloid.

    Make sense?
     
  4. Nov 27, 2006 #3
    The part about spliting it into two intergrals doesnt really make sense to me. I have seen the solution to a similar one with a cone instead of a parabloid and it was done with one triple intergal not two added togreather. That's what you mean right?
     
  5. Nov 27, 2006 #4
    Some, you can do with a single triple integral... this one, you have to divide into two pieces, by phi. Well, if you want to say that the upper limit of rho is min(sphere, parabloid), I suppose that's just one integral. But offhand, it seems to me that you need to find the value of phi where this minimum changes from one surface to the other.

    I'll ask again... you've thought about what this region looks like, right?
     
  6. Nov 27, 2006 #5
    In the case of a cone instead, you never have a case where an upper limit for rho of min(sphere, cone) is ever the cone, so you have a single integral. :smile:

    (Also, I want to make clear that I'm talking about the rho component of the sphere, cone and parabloid here, and I'm too lazy to use proper tex notation and subscripts.)
     
  7. Nov 27, 2006 #6
    When I started out on this problem I attempted to use cylindrical coordinates.

    The set up looked nice but integrated, no so nice

    V=SSS rdzdrd(theta)

    z from r^2 to rt. (2-r^2)
    r from 0 to 1
    and theta from 0 2pi

    but when I got to integrating with respect to r it got weird, mabey a trig sub could be used. I dont know.
    anyways, i have graphed it and know what it looks like.
    I know that they intersect when z = 1

    I just dont see p or rho coming into the picture. I know rho is cot(phi)csc(phi)=rho
     
  8. Nov 27, 2006 #7
    should I stick to cylindrical coordinate for this one?
     
  9. Nov 27, 2006 #8
    No, I think the spherical coordinates are correct.

    So, right, the surfaces intersect at z=1. What phi does this correspond to? (Do you see that for smaller phi, you integrate p (or rho) out to the sphere, but for phi larger than this, you go out to the parabloid?)

    The integral isn't bad, even when the limit is the parabloid. Don't use cot and csc though, just write it as sin and cos. Do theta first, then rho, then phi. The phi integral can be tackled by converting all but one of the cosines into sines, and then substituting u=sin phi.
     
  10. Nov 27, 2006 #9
    Ok, im gonna try and see how that works out.
    Thanks
     
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