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Sketching surfaces described in cylindrical coordinates

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    The surface is described by the equation ## (r-2)^2 + z^2 = 1 ## in cylindrical coordinates. Assume ## r ≥ 0 ##.
    a) Sketch the intersection of this surface with the half plane ## θ= π/2 ##

    2. Relevant equations
    ## r= psin φ ##
    ## p^2 = r^2 + z^2 ##

    3. The attempt at a solution
    First off, I'm new to cylindrical coordinates, so I don't really know how to work with equations given in these coordinates. Following the few examples given in my textbook, I tried to convert this equation into another involving Cartesian or spherical coordinates. I converted to spherical ones. The result I got was:
    ## p^2 =4p sinφ -3 ##
    The term ## 4p sinφ## is giving me a problem because I don't know how to express the equation in terms of p only. I can't factorise, etc.

    Am I starting in the right way at all? Any hints on how to proceed? Thanks.
     
  2. jcsd
  3. Sep 16, 2016 #2

    andrewkirk

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    Before you think about the half-plane, work out what shape the surface described by the equation is.
    Think about its horizontal cross sections. To get a horizontal cross section we fix ##z##. What two-dimensional shape does the equation describe if ##z## is fixed rather than a variable? Bear in mind that it will be the set of all points with polar coordinates ##(r,\theta)## that obey that equation.

    Once you've figured that out, try to work out how that 2D shape of the horizontal cross section changes as we change ##z##. Are there values of ##z## for which there is no cross-section - ie for which the surface does not intersect that horizontal plane?

    The nature of the surface should soon become apparent.
     
  4. Sep 16, 2016 #3

    LCKurtz

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    Another thing you might think about -- when ##\theta = \frac \pi 2## that is in the ##yz## plane. In that plane the ##r## variable is in the ##y## direction. If you think of ##r## as ##y## in that plane, does that help?
     
  5. Sep 16, 2016 #4

    andrewkirk

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    I realised that the problem can be solved without knowing the original shape. LCKurtz has hinted at this. Do it his way. That will be quicker.
     
  6. Sep 17, 2016 #5
    Oh, do you mean ## (y-2)^2 + z^2 = 1 ##? Then, it will be a circle with centre ##(0, 2,0)## right, with radius of 1?
     
  7. Sep 17, 2016 #6

    LCKurtz

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    Yes. And that answers your specific question in the OP. But another thing you might notice is that your cylindrical equation is independent of ##\theta##. That means your graph has the same shape in the ##zr## plane plane no matter what ##\theta## is, not just when ##\theta = \frac \pi 2## and ##r## is in the ##y## direction. So do you see what the surface looks like?
     
  8. Sep 17, 2016 #7
    I have trouble going on with the method you suggested. If say, I fix ##z## as zero, then I would have the ##xy## plane. Then the equation becomes ##(r-2)^2=1##, which I still don't understand. What does ##(r-n_{1})^2 = n_{2}## mean? It isn't the same as in Cartesian coordinates, right?
     
  9. Sep 17, 2016 #8
    Oh, a donut shaped graph? A ##zr## plane is something like a ##(r,θ)## plane that spans the ##z## axis?
     
  10. Sep 17, 2016 #9

    LCKurtz

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    Yes. It's called a torus.

    Beats me. I don't know what you mean.
     
  11. Sep 17, 2016 #10
    What do you mean by a ##zr## plane then? Is it like a ##(r, \theta)## plane that forms a cylinder abt the ##z## axis?
     
  12. Sep 17, 2016 #11

    LCKurtz

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    To say a plane "forms a cylinder about the ##z## axis" is meaningless to me.

    On a 3d axis, draw a radius in the xy plane at some angle ##\theta##. That radius line and the z axis form part of a vertical plane. You can locate any point in that plane by knowing z and r. That's what I mean by the zr plane.
     
  13. Sep 17, 2016 #12
    Okay, understood. Thanks!
     
  14. Sep 17, 2016 #13

    vela

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    That equation has solutions ##r=1## and ##r=3##. What shape do those represent?
     
  15. Sep 17, 2016 #14
    A parabola?
     
  16. Sep 17, 2016 #15

    vela

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    You're just guessing, aren't you?
     
  17. Sep 17, 2016 #16
    Yes, I honestly don't know how to predict the shape just from identifying the roots of the equation. Any hints?

    Anyway, after thinking for some time, it occurred to me to treat the ##zr## plane just as I would an ##xy## plane. So, in this case, the variable ##r## behaves like ##x##, the difference is that ##\theta## is variable. Hence, the equation of a circle applies here, and since equation is independent of ##\theta##, a torus shaped is formed when all the circles are combined.
     
  18. Sep 17, 2016 #17

    andrewkirk

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    @toforfiltum are you familiar with the concept of a surface of revolution? That is what is happening here. The curve is a circle in the ##r,z## half-plane, and the surface is the set of points 'swept out' by rotating the circle around the ##z## axis - ie varying the angle ##\theta##.
     
  19. Sep 17, 2016 #18
    Yes, I do think I have learnt about surface of revolution, but only in Cartesian coordinates. But, many thanks anyway. I understand the graph formed now.
     
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