Mass of a sphere with a vertical bound

  • Thread starter Harlow
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  • #1
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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.
 

Answers and Replies

  • #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:
 
  • #3
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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:

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