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Homework Help: Centroid density function

  1. Nov 18, 2005 #1
    Hi, I am having trouble with the following question. Can someone help me out?

    a) Find the volume of the solid that lies above the cone [tex]\phi = \frac{\pi }{3}[/tex] and below the sphere [tex]\rho = 4\cos \phi [/tex].

    b) Find the centroid of the solid in part (a).

    For the volume I got 10pi which I am fairly sure is correct. I don't know how to find the centroid. There doesn't appear to be a density function. I need it because for instance to find the z coordinate of the centroid I need to calculate [tex]\mathop z\limits^\_ = \frac{{M_{xy} }}{m}[/tex] where [tex]m = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_E^{} {\rho \left( {x,y,z} \right)} } } dV[/tex] and [tex]M_{xy} = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_E^{} {z\rho \left( {x,y,z} \right)} } } dV[/tex].

    I know that if the density function is constant then its value doesn't matter since it cancels in the calculations but this question doesn't mention anything about the density being constant.

    The solution says: By the symmetry of the problem M_(xz) = M_(yz) = 0. Then [tex]M_{xy} = \int_0^{2\pi } {\int_0^{\frac{\pi }{3}} {\int_0^{4\cos \theta } {\rho ^3 \cos \phi \sin \phi d\rho d\phi d\theta } } = ... = 21\pi } [/tex]. Hence [tex]\left( {\mathop x\limits^\_ ,\mathop y\limits^\_ ,\mathop z\limits^\_ } \right) = \left( {0,0,2.1} \right)[/tex].
  2. jcsd
  3. Nov 19, 2005 #2
    The value will "cancel out" due to symmetry arguments in the X-Y plane because there's the same mass distribution evenly around the Z axis. Pick any point (x, y, z) on the graph within the limitations and you will see that there is a point at (-x, -y, z) that counterbalances it. This is established by the cone [tex]\phi = \frac{\pi}{3}[/tex].

    Your prof should have derived for you the volume integral in spherical coordinates and transformations from x, y, z into [tex]\rho , \theta, \phi[/tex]:

    x = [tex]\rho cos(\theta)sin(\phi)[/tex]
    y = [tex]\rho sin(\theta)sin(\phi)[/tex]
    z = [tex]\rho cos(\phi)[/tex]

    So plug all that stuff into [tex]M_{xy} = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_E^{} {z\rho \left( {x,y,z} \right)} } } dV[/tex] where [tex] \rho (x, y, z)[/tex] is assumed to be 1 by lack of an explicit density function.

    where [tex]dV = \rho ^2 sin(\phi)d\rho d\theta d\phi[/tex] in spherical coords. You probably aleady know all this but it really is just "plug and chug"
    Last edited by a moderator: Nov 19, 2005
  4. Nov 19, 2005 #3
    Hmm...it's probably because I don't understand the symmetry all that well. Anyway, could you please explain what a volume integral is? My book gives me a whole bunch of different multiple integrals and usually, the most that is required is a change of coordinates. I'm doing this as self-study so that explains why I'm a bit lacking in the understanding at the moment. But yeah, this is mainly plug and chug.

    Edit: I have the coordinate 'transformation' relations for cylindrical and spherical coordinates.
    Last edited: Nov 19, 2005
  5. Nov 19, 2005 #4
    I'll try to choose my words a bit more carefully. By a volume integral I simply meant an integrating tiny increments of volume across 3 directions, be them x, y and z or [tex]\rho , \theta , \phi[/tex].

    Symmetry is a big time saver in solving center of mass (centroid) questions. Try to visualize the solid in this question. You can see that it is a cone whose apex lies at the origin and extends in the positive z direction at an angle of [tex]\frac{\pi}{3}[/tex] with respect to the z axis, with its end being sort of bulging toward positive z like an icecream cone. Can you convince yourself that no matter what x and y coordinate you choose at a constant z that there is a point that is symmetric about the z axis (point -x, -y) that will counterbalance the mass contained at x, y?

    Consider two equal point masses, one at x = 1 and one at x = -1. What would the center of mass be? It would lie at x = 0 by symmetry arguments, just like the center of mass lies at x = 0 and y = 0 by symmetry arguments in this question.
    Last edited by a moderator: Nov 19, 2005
  6. Nov 20, 2005 #5
    I can see that the x and y coordinates of the centroid would be zero - I've seen situations like this before. What I'm not entirely sure about is how the mass is evenly distributed through the solid. If I am not mistaken then mass = density * volume. The problem seems to be that there is no given density function nor is there any mention of the density being constant. Has constant density just been assumed or is there a way around not having the density function?
  7. Nov 20, 2005 #6


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    Benny, this problem asks for the centroid. That is entirely a mathematical concept, not physics. The centroid is the geometrical center. There is no "density" and no "mass"!
    It is true that you can find the centroid by treating the problem as finding "center of mass" for an object with constant density. What that constant is doesn't matter because, as vsage pointed out, that constant occurs in both the numerator and denominator of a fraction and cancels. If (variable) density is [itex]\rho(x,y,z)[/itex], then the x-coordinate of the center of mass is given by [itex]\left(\int\int\int \rho(x,y,z)dxdydz\right)X= \int\int\int x\rho(x,y,z)dxdydz[/itex] in particular, if [itex]\rho[/itex] is a constant, then
    [tex]X= \frac{\rho\int\int\int xdxdydz}{\rho\int\int\int dxdydz}[/tex]
    which is the same as
    [tex]X= \frac{\int\int\int xdxdydz}{\int\int\int dxdydz}[/tex]. Since that constant cancels anyway, just take it to be 1.
    (Be careful not to confuse the "[itex]\rho(x,y,z)[/itex]" representing the density function with the "[itex]\rho[/itex]" meaning the distance from the origin in spherical coordinates!)
    In this particular problem, you are asking for the centroid of the solid lying between the cone [itex]\phi = \frac{\pi }{3}[/itex] (a cone with vertex at (0,0,0) opening upwards) and the sphere [itex]\rho = 4\cos \phi [/itex], a sphere with radius 4 and center at (0,0,2) so it is tangent to the xy-plane. The region you are looking for looks something like an icecream cone: a cone that goes upward until it gets to that sphere.
    Okay- that means that, for each [itex]\phi[/itex], [itex]\rho[/itex] goes from 0 up to [itex]4\cos\phi[/itex]. Clearly [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] goes from 0 to [itex]\frac{\pi}{3}[/itex]. That is, the volume is given by
    [tex]\int_{\phi=0}^{\frac{\pi}{3}}int_{\theta=0}^{2\pi}\int_{\rho= 0}^{4\cos(\phi)} \rho\sin^2(\phi)d\rhod\thetad\phi[/tex]
    Since [itex]z= \rho cos(\phi)[/itex], the numerator of the fraction will be [tex]\int_{\phi=0}^{\frac{\pi}{3}}\int_{\theta=0}^{2\pi}\int_{\rho= 0}^{4\cos(\phi)} \rho^2 sin^2(\phi)cos(\rho)d\rhod\thetad\phi[/tex]
  8. Nov 20, 2005 #7
    Oh ok, I'll need to go over the definitions. My book doesn't seem to have the formulae for the centroid. My other book should, I'll check it out some time. Thanks for the clarification.
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