1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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 :
    m = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1} kzp^{2}sin\theta dpd\phid\theta

    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


    User Avatar
    Science Advisor

    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].

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook