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Mass of half ball

  1. Mar 13, 2016 #1

    RaulTheUCSCSlug

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    1. The problem statement, all variables and given/known data
    If the density of the half-ball x ^2 + y ^2 + z ^2 ≤ 4 ; z ≥ 0 is given by δ(x, y, z) = ( x^ 2 + y^ 2 + z ^2)^(1/2) find its mass.

    2. Relevant equations
    ∫∫F⋅ds
    ∫∫FoΦ(u,v)||Tu,Tv||dudv
    ∫∫FoΦ(u,v)⋅(Tu,Tv)dudv


    3. The attempt at a solution
    For the last problem I was asked to find the density of the shell and used the equation
    ∫∫FoΦ(u,v)||Tu,Tv||dudv, but now I don't know if I use the same thing or not. I know how to parametrize it and know that it will be a ball of radius 2 but don't know if I use the same equation or not.
     
    Last edited: Mar 13, 2016
  2. jcsd
  3. Mar 13, 2016 #2

    BvU

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    No idea what your T stand for. Normally mass is ##\int \rho \, dV## and I'm missing that relevant equation...
     
  4. Mar 13, 2016 #3

    RaulTheUCSCSlug

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    Tu stands for Φ(u,v) derivative with respect to u and Tv is Φ(u,v) with respect to v.
     
  5. Mar 13, 2016 #4

    RaulTheUCSCSlug

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    What I did in the previous problem was parameterize the surface, but that was asking for mass of shell, now it is mass of ball. So am I supposed to use volume then divide by the density? ? ?
     
  6. Mar 13, 2016 #5

    BvU

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    Displaces the unknown to u and v. Why not use the simple equation ?
     
  7. Mar 13, 2016 #6

    RaulTheUCSCSlug

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    Oops it's supposed to be x^2+y^2+z^2=/< 4, so it would be best to convert to spherical coordinates.
     
  8. Mar 13, 2016 #7

    BvU

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    Yep.
     
  9. Mar 13, 2016 #8

    RaulTheUCSCSlug

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    doesnt answer my question though.
     
  10. Mar 13, 2016 #9

    RaulTheUCSCSlug

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    That was a typ-o from copying and pasting, I am still stuck on what to do...
     
  11. Mar 13, 2016 #10

    Mark44

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    This problem is a natural for spherical coordinates, as the density varies with the radius, and the object is half a sphere.
     
  12. Mar 13, 2016 #11

    Ray Vickson

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    What does not answer which question?
     
  13. Mar 13, 2016 #12

    RaulTheUCSCSlug

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    Yes, I know, there was a typo before, but the typo is no longer there, I already switched to spherical coordinates, hence the equation that I am using since I am perimeterizing.
     
  14. Mar 13, 2016 #13

    RaulTheUCSCSlug

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    ^^^^
     
  15. Mar 13, 2016 #14

    RaulTheUCSCSlug

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    Okay, I know to use spherical coordinates. I just am not sure which integral to use. That is my confusion.
     
  16. Mar 13, 2016 #15

    RaulTheUCSCSlug

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    How does this question differ from when they ask for the mass of a spherical shell. Because I can do the shell fine, but I just don't know how it jumps to volume... or when the object is full.
     
  17. Mar 13, 2016 #16

    Mark44

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    It wouldn't be any of the three you wrote in your relevant equations. As BvU wrote, the basic integral is ##\int \rho dV##, which in this problem turns out to be a triple integral in spherical coordinates. For the shape of the object and the density function, I think you might be able to get away with a double integral, although I haven't set up the problem.

    In any case, the typical volume element would be a hemispherical shell, sort of like a layer in an onion. At each point in a given shell, all the points are the same distance from the origin, hence the density is the same for each point.
     
  18. Mar 13, 2016 #17

    RaulTheUCSCSlug

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    so triple integral from the bounds being the same as if i was finding half a sphere (obviously) times the Jacobian integrated with the density function in terms of spherical coordinates. That would be this equations ∫∫FoΦ(u,v)||Tu,Tv||dudv as this part ∫∫FoΦ(u,v)(((((||Tu,Tv||)))))dudv is the Jacobian.

    So I'm not sure why you are all saying that the equations are wrong or unfamiliar? But if this is the case, which I do not think, where do we account for the change of radius from 0 to 2. We wouldn't account for it at all.
     
  19. Mar 13, 2016 #18

    BvU

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    There is no Jacobian. And why don't you start writing out ##dV## and ##\rho## ( your ##\delta## ) in spherical coordinates ? Who knows, the angular part can be integrated easily and you are left with one simple integral in ##r##....
     
  20. Mar 13, 2016 #19

    RaulTheUCSCSlug

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    So just take the volume integral with the density formula and that will account for the change in density. But why is there no jacobian if we are changing coordinate systems we must account for this somehow...
     
  21. Mar 13, 2016 #20

    Mark44

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    You aren't really changing coordinate systems, since you aren't starting from an iterated integral ##\int \int \int f(x, y, z) dx~dy~dz##, and then changing to a different integral with a different volume element.
     
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