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Working on Triple Integral

  1. Nov 4, 2011 #1
    Hi everyone, I have problems solving triple integral question like this:

    Find the volume of the solid bounded below by the cone [itex]\varphi[/itex]=[itex]\frac{\pi}{6}[/itex] and above by the plane z=a.

    I can do simple triple integration questions,but can some please give me some guidance on how to solve triple integrals which requires us to convert to other coordinates(spherical,cartesian)?I have problems finding the new limits after I convert.

    Appreciate any help and guidance.Thanks in advance!
  2. jcsd
  3. Nov 4, 2011 #2
    My opinion :
    It seems for me that the other bounds is z = 0, maybe I do not understand but mentioned bounds is not quite enough for finite volume of the body.
  4. Nov 4, 2011 #3

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    Welcome to PF, violette! :smile:

    The coordinates you're looking for are cylinder coordinates (r, theta, z).
    What do you think the limits are for z?
    What would r on the cone be for a specific z?
  5. Nov 4, 2011 #4
    I think maybe something like this, but do not sure

    For any z the cone has a distinct radius of the cross-section, then R ( z ) = z * tg ( pi/6 )
    A) cartesian system:

    Integral_z ( from 0 to a )
    { Integral_x ( from 0 to z*tg( pi/6 ) )
    { Integral_y ( from 0 to sqrt[ (z*tg( pi/6 ) )^2 - x^2 ] )

    B) cylindrical system:

    Integral_z ( from 0 to a )
    { Integral_angle ( from 0 to 2*pi )
    { Integral_R ( from 0 to z*tg( pi/6 ) )
    R*(dz)*(d angle)*(d R)

    C) The volume of a cone ( 1/3 )*H*pi*R^2 = ( 1/3 )*a*pi*(a*tg( pi/6 ))^2 = 22.34
  6. Nov 4, 2011 #5
    Hi thanks for the replies =)
    but may I know why do we need to change it to cylindrical coords?I thought spherical would be better?

    for my spherical coords, i got:

    0 ≤ z ≤ a
    0 ≤ θ ≤ 2∏
    0 ≤ ρ ≤ (2a)/√3

    can i also ask,is θ always between 0 and 2∏?How do we determine whether it is or not?
  7. Nov 5, 2011 #6

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    Your z and θ are right! :smile:

    But I'm afraid your ρ is not quite right.
    As you have it now, it is a constant, meaning you're describing a cylinder.
    But it is supposed to be a cone.
    ρ should be dependent on z.

    And what you have there is cylindrical! ;)
    Cylindrical is polar with an added z coordinate.
    Just like polar, the range of θ is 0 to 2∏.

    In spherical you have 2 angle θ and φ.
    That works like longitude and latitude on earth.
    One goes all the way around (longitude), and the other has range of 0 to ∏ (compare to latitude).
  8. Nov 6, 2011 #7
    Ah I see,so for cone, ρ will be in terms of z all the time?
  9. Nov 6, 2011 #8

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    Btw, I see that you have multiplied by 2/√3.
    How is your phi defined?
    However it is, this is not the right multiplier.
  10. Nov 6, 2011 #9
    Why is that so?
    cosine [itex]\frac{\pi}{6}[/itex]=[itex]\frac{\sqrt{3}}{2}[/itex],hence my [itex]\rho[/itex] is 2/[itex]\sqrt{3}[/itex]z?
  11. Nov 6, 2011 #10

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    The cosine gives the ratio with the hypotenuse, but that one doesn't play.
    z and ρ are aligned with the adjacent and the opposite sides.
  12. Nov 6, 2011 #11
    oh no,sorry i think i did it wrongly.
    Im trying to convert to spherical coordinates.
    0 ≤ θ ≤ 2[itex]\pi[/itex]
    0 ≤ ρ ≤ [itex]\frac{2a}{\sqrt{3}}[/itex]
    0 ≤ [itex]\phi[/itex] ≤ [itex]\frac{\pi}{6}[/itex]

    Does it look right now?
  13. Nov 6, 2011 #12

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    Oh, okay, then the multiplier is good and the ranges too!

    But... :uhh: spherically the bottom (top?) of the cone would be part of the sphere (round).
    But your problem says it's a plane at z=a... :confused:
  14. Nov 6, 2011 #13
    oh man!u mean to use spherical coordinates,the surface has to be round?!
    i never knew that =S
  15. Nov 6, 2011 #14

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    Well, at ρ=constant, you're tracing a sphere.

    Actually, you can trace a plane with a ρ that depends on φ:
    [tex]\rho = {a \over \cos \phi}[/tex]
    But to integrate that tends to get ugly! :yuck:
  16. Nov 6, 2011 #15
    Ah I see..thanks so much =D
  17. Nov 9, 2011 #16
    I have a similar problem, where you have to find the area cut between the cylinder x^2+y^2≤1 and 0≤ρ≤a, im so lost its not even funny. I think we are supposed to use spherical to get ρ as variable to a, and 0 ≤ θ ≤ 2π, and 0 ≤ ϕ ≤ (the intersection of sphere and cylinder), and thats all ive got right now, this setup seems to stem from actual constants, but it does not make logical sense how to make ρ change from the edge of the cylinder to a. Should I repost this on another thread? Or can you guys maybe help me?
  18. Nov 9, 2011 #17


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  19. Nov 9, 2011 #18

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    Welcome to PF, GENHEN! :smile:

    It seems okay to use spherical.
    In your case you should still use the full range 0 ≤ ϕ ≤ π.
    But your ρ should depend on ϕ.
    [tex]\rho = \left[\begin{matrix}
    a, & \text{ if } \sin \phi < {1\over a}, \\
    {1 \over \sin \phi}, & \text{ otherwise}.
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