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Mass of a sphere with a vertical bound

  1. Mar 17, 2010 #1
    Find the total mass that occupies a solid region D bounded by a sphere of radius 3
    centered at the origin and z = 1 if the density of the function is (x, y, z) = 1/1+x^2+y^2+z^2 .








    I would like to be able to do this problem using spherical coordinates but I am unsure about how this z=1 will affect my bounds of integration for both phi and rho.
     
  2. jcsd
  3. Mar 17, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Harlow! Welcome to PF! :smile:

    (have a theta: θ and a phi: φ and a rho: ρ :wink:)

    Personally, I'd use Cartesian coordinates or cylindrical coordinates, but if you must use spherical coordinate, draw a radius at angle θ to the z-axis, and see where the limits of r are along that radius. :smile:
     
  4. Mar 17, 2010 #3
    Thanks for the reply. To be honest I hadn't even really considered using Cartesian coordinates. We have been covering spherical and cylindrical coordinates so I figured that the problem was probably designed for one of those two coordinate systems. Cartesian looks easier for this problem. Heres what I have for the bounds x: -sqrt(9-y^2-z^2) to sqrt(9-y^2-z^2), y: -sqrt(9-z^2) to sqrt(9-z^2) and z:-3 to 1.

    My only issue is with the integral, I have the first integral set up as follows: 1/(1+x^2+y^2+z^2)dx evaluated from -sqrt(9-y^2-z^2) and sqrt(9-y^2-z^2). does this look correct? How can I do this integral?
     
    Last edited: Mar 17, 2010
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