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Volume of a cone

  1. Jan 13, 2006 #1
    Hi,

    I have this problem:

    Compute volume of solid bounded by these planes:

    [tex]
    z = 1
    [/tex]

    [tex]
    z^2 = x^2 + y^2
    [/tex]

    When I draw it, it's cone standing on its top in the origin and cut with the [itex]z = 1[/itex] plane.

    So after converting to cylindrical coordinates:

    [tex]
    x = r\cos \phi
    [/tex]

    [tex]
    y = r\sin \phi
    [/tex]

    [tex]
    z = z
    [/tex]

    [tex]
    |J_{f}(r,\phi,z)| = r
    [/tex]

    I get

    [tex]
    0 \leq z \leq 1
    [/tex]

    [tex]
    0 \leq \phi \leq 2\pi
    [/tex]

    [tex]
    0 \leq r \leq 1
    [/tex]

    And

    [tex]
    V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{1} r\ dr\ dz\ d\phi
    [/tex]

    But I got [itex]\pi[/itex] as a result, which is obviously incorrect :(

    Can you see where I am doing a mistake?

    Thank you!
     
  2. jcsd
  3. Jan 13, 2006 #2

    Dale

    Staff: Mentor

    Your limits of integration on r are wrong. Note that your maximum r changes as a function of z. What you are (correctly) calculating here is the volume of a unit cylinder, not a cone.

    -Dale
     
  4. Jan 13, 2006 #3
    I thought so...anyway, I still can't see what's wrong. When I draw it in x-z 2D plane, I see that the cone is bounded by curve [itex]z = r[/itex] as a "right side" (which is what I got from expressing z from the original equations and using cylindrical coordinates) and [itex]z = -r[/tex] as a "left side". So because [itex]0 \leq z \leq 1[/itex] also [itex]0 \leq r \leq 1[/itex].

    Why isn't it correct?

    EDIT: Oh, I maybe see it now..Gonna try that and possibly write again..
     
  5. Jan 13, 2006 #4

    Dale

    Staff: Mentor

    Well, you probably got it now anyway, but just in case I will tell you before I have to go for the evening. Basically instead of integrating r over 0 to 1 you need to integrate r over 0 to z.
    [tex]
    V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{z} r\ dr\ dz\ d\phi
    [/tex]

    Do you see how this is a cone and the previous integration was a cylinder?

    -Dale
     
  6. Jan 13, 2006 #5

    krab

    User Avatar
    Science Advisor

    Divide the cone into discs of thickness dz. The radius of each is equal to z, so area is piz^2. Now just integrate:

    [tex]\int_0^1\pi z^2dz=\pi/3[/tex]
     
  7. Jan 14, 2006 #6
    Yes, that's exactly how I finally did it. Thank you DaleSpam!

    Thank you Krab, nice approach actually...
     
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