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Setting up some triple integrals

  1. Feb 10, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    I want to know if I've gone about setting up these integrals in these questions properly before I evaluate them.

    (i). Find the mass of the cylinder [itex]S: 0 ≤ z ≤ h, x^2 + y^2 ≤ a^2[/itex] if the density at the point (x,y,z) is [itex]δ = 5z^4 + 6(x^2 - y^2)^2[/itex].

    (ii). Evaluate the integral of [itex]f(x,y,z) = (x^2 + y^2 + z^2)^{1/2}[/itex] over the region D which is above the cone [itex]z^2 = 3(x^2 + y^2)[/itex] and between the spheres [itex]x^2 + y^2 + z^2 = 1[/itex] and [itex]x^2 + y^2 + z^2 = 9[/itex].

    2. Relevant equations

    Cylindrical polars and spherical polars.

    3. The attempt at a solution

    (i). Switching to cylindrical polars, x=rcosθ, y=rsinθ and z=z.

    So [itex]S → S^{'}: 0 ≤ z ≤ h, -a ≤ r ≤ a, 0 ≤ \theta ≤ 2 \pi[/itex]
    and the density becomes [itex]δ = 5z^4 + 6r^4cos^2(2θ)[/itex].
    The Jacobian of the cylindrical polars is just r so our integral over S' becomes :

    [itex]\int_{0}^{2π} \int_{-a}^{a} \int_{0}^{h} 5rz^4 + 6r^5cos^2(2θ) dzdθdr[/itex]

    (ii). Switching to spherical polars, x=ρcosθsinφ, y=ρsinθsinφ and z=ρcosφ and we also know that [itex]x^2+y^2+z^2 = ρ^2[/itex].

    So f(x,y,z) now becomes [itex](ρ^2)^{1/2} = ρ[/itex]
    The two spheres yield [itex]1 ≤ ρ ≤ 3[/itex]
    θ is as is in cylindrical, so [itex]0 ≤ θ ≤ 2π[/itex]
    Also [itex]0 ≤ φ ≤ π[/itex]

    The purpose of the cone is not yet clear to me.
     
  2. jcsd
  3. Feb 10, 2013 #2

    Mark44

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    The cone is part of the definition of the region, which looks sort of like a tapered cork or stopper.
     
  4. Feb 10, 2013 #3

    Zondrina

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    I'll take it (i) was good to go then.

    For (ii) does the cone not impact some of the information for the integral?

    Also, a tapered cork? Looks more like an ice cream cone with a bite in it I find.

    EDIT : If anyone could help me verify the second one it would be greatly appreciated. I don't set this up often so it's hard to tell if I'm doing it correctly.
     
    Last edited: Feb 10, 2013
  5. Feb 10, 2013 #4

    Dick

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    Did you sketch a graph of the region? If you know it's an ice cream cone with a bite, you probably did. Doesn't being above the cone imply some restriction on ##\phi##? What's the angle of the cone?
     
  6. Feb 10, 2013 #5

    Zondrina

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    Well when I drew it out, I had a bigger sphere with a smaller sphere inside of it. Then the cones would come in and cut the region created by the boundaries of the spheres; both in the positive and negative portions, but we only care about the positive portion since we want 'above' the cone ( I'm guessing this is what is intended since in the question it said 'which is above the cone' ).

    That portion of the spheres that the cone cuts into looks like someone took a bite out of the bottom of the cone.

    Hmm so would I not need to find out when the cone intersects the first sphere and then similarly the second?

    Just a thought. Not sure if it has merit.
     
  7. Feb 10, 2013 #6

    Dick

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    You aren't listening. Doesn't the angle of the cone have something to do with the range of the polar angle ϕ?
     
  8. Feb 10, 2013 #7

    Zondrina

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    Yes it does, there are two rays of the cone which go off at an angle.
     
  9. Feb 10, 2013 #8

    Dick

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    So ϕ doesn't go from 0 to pi, yes?
     
  10. Feb 10, 2013 #9

    Zondrina

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    Indeed... by the looks of it by the way I'm picturing it, it would go from [itex]-π/2[/itex] along to [itex]π/2[/itex].
     
  11. Feb 10, 2013 #10

    Dick

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    Does not. What's the angle of the cone from the z-axis? It's not pi/2.
     
  12. Feb 11, 2013 #11

    Zondrina

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    Ohhh wait so phi is the angle from the z axis to the ray of the cone... so I should...

    Switch the cone to spherical and then calculate a value for phi I believe?

    So in spherical the cone becomes :

    [itex]ρ^2cos^2 \phi = 3ρ^2sin^2φ[/itex]
    [itex]cos^2 \phi = 3sin^2φ[/itex]
    [itex]tan^2 \phi = 1/3[/itex]
    [itex]\phi = arctan( \sqrt{1/3} )[/itex]

    So pi/6? So I would take my bounds from 0 to pi/6?
     
    Last edited: Feb 11, 2013
  13. Feb 11, 2013 #12

    Dick

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    Now you've got it.
     
  14. Feb 11, 2013 #13

    Zondrina

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    Ah I see now. So my final integral for (ii) would be :

    [itex]\int_{1}^{3} \int_{0}^{2π} \int_{0}^{π/6} ρ d \phi dθdρ[/itex]

    and then I've got two integrals to evaluate.

    Thank you for helping me understand what's going on. I should feel comfortable setting these up from now on.
     
  15. Feb 11, 2013 #14
    Zondrina: check the volume element in polar coordinates.
     
  16. Feb 11, 2013 #15

    Zondrina

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    What do you mean check the volume element?
     
  17. Feb 11, 2013 #16

    Dick

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    When you are integrating in polar coordinates, you don't just write dϕdθdρ after the function you are integrating. There's the jacobian part of the volume element as well.
     
  18. Feb 11, 2013 #17

    Zondrina

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    *facepalm.jpg*

    Totally forgot about the jacobian there for a moment. The jacobian is ρ2sin(phi) which would change my integral a bit.

    Thanks for reminding me guys :)
     
  19. Feb 12, 2013 #18
    Yes, the determinant of the Jacobian matrix can be seen as the scale factor of the unit volume when changing coordinates.
     
  20. Feb 13, 2013 #19

    Zondrina

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    Sorry for bumping this, but I evaluated these integrals awhile ago and something has been bothering me.

    For the cylindrical integral in (i), I get the mass to be 0 which does not make any sense at all?

    Where might I have gone wrong? I'm guessing it has something to do with one of my limits. Also, I re-arranged my integral like this :

    [itex]\int_{-a}^{a} \int_{0}^{2 \pi} \int_{0}^{h} 5z^4r + 6r^5cos^2(2 \theta) dzd \theta dr[/itex]
     
  21. Feb 13, 2013 #20

    Dick

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    Why are you integrating over negative values of r?
     
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