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Two spheres mass of inertia multivariable calculus problem

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Let R be the solid region that is bounded by two spheres x^2 + y^2 + z^2=1 and x^2 + y^2 + z^2=2. Determine the moment of inertia of R around the x-axis if the mass density per unit volume of R is u=sqrt(x^2 + y^2 + z^2).

    2. Relevant equations

    Moment of Inertia around the x-axis: triple integral (y^2+z^2)*u(x,y,z) dxdydz

    using (p,@,theta) for spherical coordinates as it will be faster to type, sorry for any confusion.

    spherical coordinates:
    x=psin(@)cos(theta)
    y=psin(@)sin(theta)
    z=pcos(@)

    3. The attempt at a solution

    Using the Moment of Inertia around the x-axis equation, i first got the equation:

    triple integral (y^2+z^2)*sqrt(x^2 + y^2 + z^2) dxdydz

    now converting to spherical coordinates i got:

    triple intergal (p^2*sin^2(@)*sin^2(theta) + p^2*cos^2(@))*sqrt(p^2*sin^2(@)*cos^2(theta) + p^2*sin^2(@)*sin^2(theta) + p^2*cos^2(@)) dpd@dtheta

    After some simplification:

    triple intergal p^3(sin^2(@)sin^2(theta) + cos^2(@)) dpd@dtheta

    where 1<p<sqrt(2) and 0<@<pi and 0<theta<2pi

    the 1<p<sqrt(2) is because of the two radius and the other two because its a sphere and its centred at the origin, i know 0<theta<2pi is correct but i'm not sure if 0<@<pi or 0<@<2pi.

    solving this integral i got 9*pi^2/4 which i'm not sure is correct as i'm not sure if i'm even able to do it this way, any tips or help would be much appreciated !
     
  2. jcsd
  3. Oct 2, 2014 #2

    pasmith

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    You have forgotten that [itex]dx\,dy\,dz = \rho^2 \sin \phi\,d\rho\,d\theta\,d\phi[/itex]. The region you are integrating over is [itex]\{1 \leq \rho \leq \sqrt{2}, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi\}[/itex].
     
  4. Oct 2, 2014 #3

    vela

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    Because of spherical symmetry in the problem, the moment of inertia about the x-axis is equal to the moment of inertia about the z-axis. In the latter calculation, the integral is a little simpler.
     
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