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Finding total mass of hemisphere that has a variable density

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that total mass of half solid hemisphere radius a given as x^2 + y^2 + z^2 = a^2 and z>=0 and having variable density ρ=1+(r^2)z where r is the distance of any point from the origin is given by: M = (2π(a^3))(6+3(a^3)/18 where M is total mass.


    2. Relevant equations
    -


    3. The attempt at a solution
    My first attempt was to find the volume of the hemisphere using spherical polar coordinates and multiply this volume with the density given. Obviously this was wrong. I then tried integrating again but this time i integrated 1+(x^2 + y^2 + z^2)z
    I replaced the r^2 in the density equation with a^2. I still ended up with the wrong answer. Have i applied the right method the second time round and maybe just made a mistake when integrating, or am i approaching this question the wrong way? Thanks.
     
  2. jcsd
  3. May 9, 2009 #2

    gabbagabbahey

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    Of course you ended up with the wrong answer; not every point inside the hemisphere is a distance [itex]a[/itex] from the origin...only the points on the outer shell of the hemisphere are that far away.

    Try integrating the density over the volume of the sphere using spherical polar coordinates. What is [itex]z[/itex] in spherical polars?...What are your integration limits for [itex]r[/itex], [itex]\theta[/itex] and [itex]\phi[/itex]? What is the infinitesimal volume element (usually denoted [itex]dV[/itex]) in spherical polars?...
     
  4. May 9, 2009 #3
    i see! i got the answer, thank you.
     
  5. May 9, 2009 #4
    I think something is wrong in your solution

    My first thoughts:
    since m=[itex]\rho[/itex]V

    [itex]\Rightarrow[/itex] dm=Vd[itex]\rho[/itex]+[itex]\rho[/itex]dV

    but in spherical : dV=r2sin([itex]\theta[/itex])dr d[itex]\theta[/itex] d[itex]\phi[/itex]
    and z=r cos([itex]\theta[/itex])
    [itex]\Rightarrow[/itex] dz=cos([itex]\theta[/itex]) dr - r sin([itex]\theta[/itex])d[itex]\theta[/itex]

    hence, d[itex]\rho[/itex]=2 r z dr + r2 dz and substitute for dz

    then substitute again in dm expression and perform the integration from
    r: 0[itex]\rightarrow[/itex]a
    [itex]\phi[/itex]: 0[itex]\rightarrow[/itex] 2 pi
    [itex]\theta[/itex]: 0 [itex]\rightarrow[/itex] pi/2

    solve it and tell me about the result :smile:
     
    Last edited: May 9, 2009
  6. May 9, 2009 #5

    gabbagabbahey

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    Huh?!:confused:

    [itex]m=\rho V[/itex] is only valid for regions of uniform charge density (in which case [itex]d \rho[/itex] would be zero)

    More generally, [tex]m=\int_{\mathcal{V}} \rho(\vec{r})dV[/tex].
     
  7. May 10, 2009 #6
    you're true ....

    so [itex]dm=\rho dV[/itex] only
    and the term [itex]Vd\rho[/itex] is wrong ...
     
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