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Calculating mass by double integral

  1. Jul 13, 2009 #1
    1. The problem statement, all variables and given/known data
    The distribution of mass on the hemispherical shell z=(R2 - x2 -y2)1/2 is given by
    [tex]\sigma[/tex]= ([tex]\sigma[/tex]0/R2)*(x2+y2)
    where [tex]\sigma[/tex]0 is constant. Find an expression in terms of [tex]\sigma[/tex]0 and R for the total mass of the shell


    2. Relevant equations
    The mass is given by double integral over hemispherical shell


    3. The attempt at a solution

    [tex]\int\int[/tex][tex]\sigma[/tex]0/R2(x2+y2)dS
    Switch to polar coordinates: x2+y2=r2

    [tex]\int\int[/tex][tex]\sigma[/tex]0/R2*r2*r*dr*d0

    After the iterated integral over region 0->2[tex]\Pi[/tex] and 0->R I get answer:
    Mass = ([tex]\Pi[/tex]*R2*[tex]\sigma[/tex]0)/2

    However the book that I'm studying (Div,Grad,Curl and all that, problem II-6) says that right answer is: (4*[tex]\Pi[/tex]*R2*[tex]\sigma[/tex]0)/3

    What goes wrong?
     
  2. jcsd
  3. Jul 13, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In spherical coordinates, with [itex]\rho= R[/itex] [itex]x= R cos(\theta)sin(\phi)[/itex] and [itex]y= R sin(\theta)sin(\phi)[/itex] so [itex]x^2+ y^2= R^2 sin^2(\phi)[/math] and your density function is given by
    [tex]\frac{\sigma_0}{R}\R^2 sin^2(\phi)[/tex]

    Since the integration is over the upper half spherical shell, the integration is for [itex]\theta[/itex] from 0 to [itex]2\pi[/itex] and for [itex]\phi[/itex] from 0 to [math]\pi/2[/math].

    Finally, the "differential of surface area" for a sphere, of radius R, is [math]R^2 sin(\phi) d\theta d\phi[/math].

    So the mass of the shell is given by
    [tex]\frac{\sigma_0}{R}\R^2 \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2} sin^2(\phi)(R^2 sin^2(\phi) d\theta d\phi[/tex]
    [tex]= \sigma_0\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2} sin^4(\phi)d\theta d\phi[/tex]
    since the "R2" terms cancel out.
     
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