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: Moment of inertia of a spherical segment

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Am I going about this the right way?
    There is a sphere of radius 3 and a region that lies between the planes [tex]z=1[/tex] and [tex]z=2[/tex] and has a density of [tex]cz[/tex]. We are asked to work in cylindrical coordinates.


    2. Relevant equations
    Let [tex]\rho=\sqrt{9-z^2}[/tex], is the following the right formula?

    [tex]I=\displaystyle{\int_R}cz\rho^2\,dV[/tex] where [tex]dV=\rho\,d\theta\,d\rho\,dz[/tex]



    3. The attempt at a solution

    Would this be the integral?

    [tex]I=2\pi c z\displaystyle{\int^2_1}\displaystyle{\int^{\sqrt{9-z^2}}_0}\rho^3\,d\rho\,dz[/tex]

    Thanks
     
  2. jcsd
  3. Sep 1, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Are we to assume that the center of the sphere is at (0, 0, 0)?[/quote]


    2. Relevant equations
    Let [tex]\rho=\sqrt{9-z^2}[/tex], is the following the right formula?[/quote]
    Normally, there is no "[itex]\rho[/itex]" in cylindrical coordinates. Do you mean "[itex]r= \sqrt{9- z^2}[/itex]". Assuming the sphere has center at (0, 0, 0), yes, that is correct.

    [tex]I=\displaystyle{\int_R}cz\rho^2\,dV[/tex] where [tex]dV=\rho\,d\theta\,d\rho\,dz[/tex]



    3. The attempt at a solution

    Would this be the integral?

    [tex]I=2\pi c z\displaystyle{\int^2_1}\displaystyle{\int^{\sqrt{9-z^2}}_0}\rho^3\,d\rho\,dz[/tex]

    Thanks[/QUOTE]
    Well, you don't have the "z" outside the integral: you should have
    [tex]2\pi c\int_{z= 1}^2 z\left(\int_{r= 0}^{9- z^2} r^3 dr\right) dz[/tex]
     
  4. Sep 1, 2010 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Are we to assume that the center of the sphere is at (0, 0, 0)?[/quote]


    Normally, there is no "[itex]\rho[/itex]" in cylindrical coordinates. Do you mean "[itex]r= \sqrt{9- z^2}[/itex]". Assuming the sphere has center at (0, 0, 0), yes, that is correct.

    Well, you can't have the "z" outside the integral: you should have
    [tex]2\pi c\int_{z= 1}^2 z\left(\int_{r= 0}^{9- z^2} r^3 dr\right) dz[/tex]
     
  5. Sep 1, 2010 #4
    Thanks, the z should have been iinside and it is centered at the origin

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