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Cone edge

  1. Nov 26, 2005 #1
    Hi ,
    I dont know how to get the edge of the cone in cylindrical coordinates.

    For example, we have a cone starting at the origin, of heigth 2 and the top is a circle of radius 1 (center at the origin).

    the edge of the cone is z=2r. but I dont know how they find it.
    Please can someone help me?

    Thank you

  2. jcsd
  3. Nov 26, 2005 #2


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    The phrasing "edge of the cone" is a little confusing! I would have interpreted it to mean the circle where the slant side meets the base but it seems clear that you are talking about the slant (curved) side itself.

    Look at the cone "from the side". That is, imagine that you are looking along the y-axis so that positive z is up and positive x is to the right.
    The line forming the slant side of the cone, from that perspective, passes through (0,0,0) and (1, 0, 2). Ignoring the y-coordinate, that is (0,0) to (1, 2). What is the equation of a line passing through those coordinates? (Remember to use z instead of y.)
    Now, remember that the cone is formed by that line rotating around the z-axis. "x" is really the straight line distance from the z-axis to the point: that's r.
  4. Nov 26, 2005 #3
    Thank you
    In fact, I took this explample for a book.
    for the same cone, we want to calculate the volume.
    I set the integral
    (|(a,b) means integral from a to b)
    for theta : |(0,2*pi)
    for r: |(0,1)
    for z: |(X,2) I dont know how to find the lower bound X
  5. Nov 26, 2005 #4


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    IF you are doing it in that order: the [itex]\theta[/itex] integration is the outer integration from [itex]\theta= 0[/itex] to [itex]2\pi[/itex] and then next from r= 0 to r=1, then inner most integral is from the line z= 2r up to 2:
    [tex]\int_{\theta = 0}^{2\pi}\int_{r= 0}^1\int_{z=2r}^2 rdzdrd\theta[/tex]

    I, personally would be inclined to change the order:
    [tex]\int_{\theta= 0}^{2\pi}\int_{z= 0}^2\int_{r= 0}^{\frac{r}{2}} rdrdzd\theta[/tex].

    They should give exactly the same answer.
    Last edited by a moderator: Nov 26, 2005
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