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: 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