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3D Surfaces - Equation Formed When Rotating a 2D Line About an Axis?

  1. Sep 10, 2007 #1

    dt_

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    Hi everyone, I'm pretty new to Physics Forums but it seems like a fairly friendly community. :)

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

    Determine the equation of the surface formed when the line x=3y is rotated about the x-axis.

    2. Relevant equations

    x=3y is the given line.


    3. The attempt at a solution

    First I write it in terms of x because it's simpler: [tex]y = \frac{1}{3} x[/tex]

    The slope is 1/3, thus, and you have a diagonal line that passes through the origin in a 2-D graph with the X-Y plane.

    Now, if you rotate this about the X-axis, you see you get a sort of cone. rather, two cones, one for each side of the y-axis ; these two cones have their tops(tips) facing each other.

    How, though, can I determine an equation for the cone? I know there is a generic equation that involves x,y,z variables and a,b,c constants (I think it's something like.. [tex](x-a)^{2} + (y-b)^{2} = (z-c)^{2}[/tex] )

    but what do I plug in for the variables and constants? I think I need to substitute [tex]\frac{x}{3}[/tex] for [tex] x [/tex] , or maybe with one of the other variables (y or z) but I'm not sure where and how.

    If anyone could help me on this I would very much appreciate it. Thank you!! :)
     
  2. jcsd
  3. Sep 10, 2007 #2

    Dick

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    You are SOO dumb! Oh, wait, we're friendly! :) Just kidding. The absolute value of y at any point x determines the radius of the circle in the y-z plane with center at y=0, z=0, right? So what's the equation of such a circle in y-z with radius |x/3|?
     
  4. Sep 10, 2007 #3

    dt_

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    [tex]| \frac{x}{3} | ^{2} = \frac{x^{2}}{9} [/tex]

    so the r^2 in x^2 + y^2 = r^2 is equal to (x^2) / 9

    and..

    the equation of the circle is
    y^2 + z^2 = (x^2 / 9)


    but that's a.. cone? i think?

    which would make sense

    does that seem right?
     
  5. Sep 11, 2007 #4

    Dick

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    That looks good to me.
     
  6. Sep 11, 2007 #5

    HallsofIvy

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    A cone, in the mathematical sense, has two parts. Each one is a "nappe" of the cone.
     
  7. Sep 11, 2007 #6

    dt_

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    Thanks Dick. :)

    HallsOfIvy: Not quite getting that, but do you think you can check my arithmetic and see if I've worked out the solution correctly?
     
  8. Sep 11, 2007 #7

    Dick

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    I'm betting Halls already checked the math. He tends to catch small errors. As far as the "nappe" goes he's just telling that full geometry of your equation looks like a 'double cone' (one for x>0 and one for x<0), but it's still ok to call it a 'cone'. "nappe" is generic name for one of these two parts.
     
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