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Cylindrical Coordinates Domain

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Let W be the region between the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2


    2. Relevant equations



    3. The attempt at a solution
    for each domain,
    0 ≤ z ≤ 8
    0 ≤ r ≤ 2
    0 ≤ θ ≤ 2[itex]\pi[/itex]

    So...

    [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ
     
  2. jcsd
  3. Nov 15, 2011 #2

    LCKurtz

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    Assuming what you are calculating is the volume, yes.
     
  4. Nov 15, 2011 #3
    wait wait shouldn't it be
    [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ
     
  5. Nov 15, 2011 #4

    LCKurtz

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    No. That is wrong on several levels. If you are doing a triple integral for volume the integrand is always 1dV = 1 rdzdrdθ.

    1. You are missing the dz.
    2. Your z limits are not correct; they would depend on r and maybe θ and you would want the dz on the inside so it is integrated first.
    3. Your integrand was zupper-zlower in your double integral, which was correct, and that is what you should get when you integrate the 1dz in the triple integral.
     
  6. Nov 15, 2011 #5
    oh so this is the final one then

    [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex](8-2[itex]r^{2}[/itex])r dzdrdθ

    Yes I forgot dz accidently but I knew that :p
     
  7. Nov 15, 2011 #6

    LCKurtz

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    Still wrong. Your integrand should be 1dV.
     
  8. Nov 15, 2011 #7
    How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ
     
  9. Nov 15, 2011 #8

    LCKurtz

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    No, not correct.
     
  10. Nov 15, 2011 #9
    but isn't dV = dzdrdθ?
     
  11. Nov 15, 2011 #10

    LCKurtz

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    No, it is r dz dr dθ which you had correct. But now you messed up the limits again.
     
  12. Nov 15, 2011 #11
    ARRRGGGG this thing is driving me nuts
    ok phew
    one more time

    [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex]r dzdrdθ
     
  13. Nov 15, 2011 #12

    LCKurtz

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    Finally. Notice that when you work the dz integral what you have left with is exactly the double integral which you had correct in your initial post.
     
  14. Nov 15, 2011 #13
    Yes i got it
     
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