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Half-plane in R^3

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Describe in words the surface whose equation is given by theta = pi/4

    2. The attempt at a solution


    It is a fairly simple question, but I'm just trying to understand why this is considered a half-plane that exists for x that is greater than or equal to 0. There are no restriction on r or z. Given this, cannot r be negative? Wouldn't this make a full-plane and let it extend for x less than 0? I just don't quite see why it cannot have the point (-1, pi/4, 0), for example, which would be present for x less than 0. Any feedback is always appreciated!
     
  2. jcsd
  3. Sep 28, 2014 #2

    LCKurtz

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    First of all, you haven't told us what coordinate system you are using. ##\theta = \frac \pi 4## makes sense in 2D polar coordinates, 3D cylindrical coordinates, and in spherical coordinates. In spherical coordinates it might be a plane or cone, depending on what convention you use for ##\theta##.

    But, to answer your question, you seem to understand what is going on perfectly well. Texts are inconsistent about whether or not ##r<0## is used in polar or cylindrical coordinates. If you are using the convention ##r\ge 0## you get a half plane as you say, and if ##r## is allowed to go negative you get the whole plane, as you understand. One problem with disallowing negative values of ##r## is that you don't see all 3 leaves of the rose ##r = \sin(3\theta)## for ##0\le\theta\le \pi##.

    I wouldn't worry too much about this if I were you since you understand it just fine.
     
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