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Finding mass from mass density of lamina

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    wlNsyAz.png

    2. Relevant equations
    m = ∫∫sρ(x,y,z)dS

    3. The attempt at a solution
    I used polar coordinates for this (is that necessary here)? I found zx and zy, took the cross product of those, and found it to be √(2). So dS = √2 dA.
    √2 02pi25r2drdθ = 78pi√2
     
  2. jcsd
  3. Apr 25, 2016 #2

    Simon Bridge

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    Polar coordinates are entirely appropriate: try cylindrical-polar and exploit the symmetry.
    Don't know what you mean by ##z_x## and ##z_y## or the role you used ##dS## and ##dA## for.
    You have to spell out your reasoning - different courses can use different conventions.

    Since it is a lamina - the mass density is probably an area rather than a volume density; so ##dm = \rho\;d\!A## (check).
    Here dA is an element of area lying on the surface of the cone and dm comes from: ##m=\int_{Area}\;dm##

    To select a strategy you can feel confident with, 1st figure out what ##\rho## depends on.
    Does it depend on r? z? ##\theta##? All of them? None of them? Which?
    Then a good choice for the geometry of dA may present itself.
     
  4. Apr 25, 2016 #3
    Doesn't it depend on all of them?
     
  5. Apr 25, 2016 #4
    Okay I'm thinking about dA and would it be dzrdrdθ? The boundaries for z would be from z=2 to z=5, θ would be from 0 to 2pi, but what about r?
     
  6. Apr 25, 2016 #5

    Simon Bridge

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    Back up a bit.
    You have ##\rho(x,y,z)=\sqrt{x^2+y^2}## in post #1.
    Notice how that equation does not have a z in it? That means that the value of ##\rho## does not depend on the value of ##z##.
    Can you write that equation in terms of the cylindrical-polar coordinates?

    Do the same for the equation of the cone.
     
  7. Apr 25, 2016 #6
    So it only depends on r and θ, correct?
    p(x,y,z) = r
    z = 6-r
     
  8. Apr 25, 2016 #7

    Simon Bridge

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    OK - so you have ##\rho = r## ... look at what you wrote: is there a ##\theta## in that equation?

    Note: ##\rho(x,y,z)=r## is a bit misleading because x,y,z belong to rectangular coords while the r is cylindrical. You could write ##\rho(r,\theta,z)=r##
     
  9. Apr 25, 2016 #8
    No... so it only depends on r?
    I am getting confused on what to use now...
     
  10. Apr 25, 2016 #9

    Simon Bridge

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    You are doing fine. The density only depends on the distance from the z-axis.
    This means you could use a shortcut ... you only need dA to be the area of the surface of the cone between r and r+dr.
    Technically you could also make dA the area of the cone between z and z+dz (sketch).

    If the density depended on the angle as well, then you'd need dA to be the area of the cone between r and r+dr which is also between ##\theta## and ##\theta + d\theta##. You can still do it that way if it helps you think about it.
     
    Last edited: Apr 25, 2016
  11. Apr 25, 2016 #10
    I think I see what you are saying... So since this is not dependent on z or theta, would the integrand just be rdr?
     
  12. Apr 25, 2016 #11

    Simon Bridge

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    No. the integrand is ##\rho\; d\!A##
    The area on the cone between r and r+dr is circular so you'd expect to see a ##\pi## in there someplace right?
    Did you draw a sketch?
     
  13. Apr 25, 2016 #12
    A cone with its center at (0,0,-6), but between the planes z=2 and z=5, is that correct?
     
  14. Apr 25, 2016 #13
    But pdA = rdr? Actually, would dA = dzrdr?
     
  15. Apr 25, 2016 #14
    Can anyone confirm I am thinking correctly? This homework is due tonight. ):
     
  16. Apr 25, 2016 #15

    vela

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    If I understand what you did, you found ##dS = \sqrt{2}\,dx\,dy## since you differentiated with respect to ##x## and ##y##. Now if you want to use polar coordinates, you need to find the equivalent of ##dx\,dy## in terms of polar coordinates. Do you know what that is?

    The other thing you need to do is figure out the bounds of integral. I've attached a top-down view of the lamina, so you can see its projection onto the xy-plane. Untitled-1.png
     
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