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Tough Integral problem

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    W is the solid bounded above by [tex]\rho[/tex]=3 and below by[tex]\phi[/tex]=[tex]\pi[/tex]/3, calculate the mass W if the density at each point is directly proportional to the square of its distance above the xy plane.


    3. The attempt at a solution

    I am having a difficult time starting out this problem, I am assuming I need to use spherical coordinates considering the problem is given in terms of rho and phi, however it says bounded above by rho=3 and below by phi =pi/3, I thought phi determined the angle from the z axis to rho??

    Any thoughts direction would be appreciated
     
  2. jcsd
  3. Nov 19, 2008 #2

    Dick

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    They didn't describe that very well, but it looks to me like they want you to integrate rho from 0 to 3 and phi from 0 to pi/3 and over all theta. The region looks like a cone with a spherical cap on it. I think by above and below they mean relative to the z axis.
     
  4. Nov 23, 2008 #3
    I completely stuck
    How do I determine what my density function will be?

    [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] [tex]\delta[/tex](x,y,z)[tex]\rho[/tex]^2sin[tex]\theta[/tex]d[tex]\rho[/tex]d[tex]\phi[/tex]d[tex]\theta[/tex]
    0[tex]\leq[/tex][tex]\rho[/tex][tex]\leq[/tex]3
    0[tex]\leq[/tex][tex]\leq[/tex][tex]\pi[/tex]/3
    0[tex]\leq[/tex][tex]\theta[/tex][tex]\leq[/tex]2pi

    would I use [tex]\rho[/tex]^2??
     
    Last edited: Nov 23, 2008
  5. Nov 23, 2008 #4

    Dick

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    No. They tell you that the density is proportional to z^2. What's z in spherical coordinates?
     
  6. Nov 23, 2008 #5
    z=[tex]\rho[/tex]cos[tex]\phi[/tex]

    z^2=([tex]\rho[/tex]cos[tex]\phi[/tex])^2 ??
     
  7. Nov 23, 2008 #6

    Dick

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    Why the question marks?? Sounds right to me. They only said 'proportional to', so you might want to call it C*z^2, just for generality.
     
  8. Nov 23, 2008 #7
    [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] [tex]\rho[/tex]^2cos^2[tex]\theta[/tex][tex]\rho[/tex]^2sin[tex]\theta[/tex]d[tex]\rho[/tex]d[tex]\phi[/tex]d[tex]\theta[/tex]
    0[tex]\leq[/tex][tex]\rho[/tex][tex]\leq[/tex]3
    0[tex]\leq[/tex][tex]\leq[/tex][tex]\pi[/tex]/3
    0[tex]\leq[/tex][tex]\theta[/tex][tex]\leq[/tex]2pi
     
  9. Nov 23, 2008 #8

    Dick

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    Looks ok so far. But be careful about theta and phi. You said z=r*cos(phI) and then you put in cos(theta). theta and phi tend to get mixed up. There are two different conventions for this. Is theta the polar angle or phi?
     
  10. Nov 23, 2008 #9
    my mistake, should be p^2cos^2(phi) p^2sin(theta)
    can this be broken apart:

    (Integral rho^4 from 0 to 3)*(Integral Cos^2 phi from 0 to pi/3)*(Integral sin theta from 0 to 2pi)

    (1/5rho^5) * (pi/6 + .5sin(pi/3)cos(pi/3)) * (-cos(2pi) - -cos(0))

    (48.6)(.2165)(0)=0 ?????
     
    Last edited: Nov 23, 2008
  11. Nov 23, 2008 #10

    Dick

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    The angle in the z and the angle in the dV part of the integral should both be the same angle (the polar angle), shouldn't they? I'm just worried you are mixing formulas.
     
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