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Problem determining p in triple integral

  1. Mar 29, 2009 #1
    The question states:

    Find the center of mass of the solid that is bounded by the hemisphere z = sqrt(21 - x ^2 - y^2) and the plane z = 0 if the density at a point P is directly proportional to the distance from the xy-plane.

    I know that the integral is setup :
    [itex]
    m = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1} kzp^{2}sin\theta dpd\phid\theta
    [/itex]

    How ever I do not see how p = 1 for this equation...

    i know x^2 + y^2 + z^2 = p^2

    any help would be appreciated... for all my other problems i would isolate x^2 + y^2 + z^2 and determine my p, however, for this problem, that isnt working...

    thanks a lot
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2009 #2

    HallsofIvy

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    So "p" (actually the Greek letter "rho": [itex]\rho[/itex]) is the radius of the sphere? If [itex]z= \sqrt{21- x^2- y^2}[/itex] then squareing both sides gives [itex]z^2= 21- x^2- y^2[/itex] or [itex]x^2+ y^2+ z^2= 21[/itex], a sphere of radius 21, not 1. Of course in the orginal form z cannot be negative so that is the upper hemisphere. If what you have is correct then the integral you want should be
    [tex]\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2}\int_{\rho= 0}^{21} k\rho cos(\phi) \rho^2 sin(\phi) d\rho d\phi d\theta [/tex]
    Notice that I have made two corrections: in spherical coordinates, [itex]z= \rho cos(\phi)[/itex] and the differential involves [itex]sin^2(\phi)[/itex], not [itex]sin^2(\theta)[/itex]. But it is true that the radius of the sphere, as given, is 21, not 1.

    I wonder if there isn't a typo here and the formula was intended to be [itex]z= \sqrt{1- x^2- y^2}[/itex].

     
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