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Scetching the graph of a region in polar co-ords

  1. Oct 8, 2008 #1
    right my problem is express the region given by x^2+y^2<x in plane polar co-ords, hence sketch the region.
    so i end up getting r<cos£ and £ lies between 0 and pi/2 and 3pi/2 and 2pi (if theres a better way or writing this please show me) i can show more working if required but i think that should suffice, the part i'm stuck on is the drawing of the graph, never been good at showing things graphically so could someone please scetch it out for me and work through what comes from where? thanks in advance
     
  2. jcsd
  3. Oct 9, 2008 #2
    (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle radius r centred at (a,b). So your equation will be a circle radius root x centred at the origin. The inequality indicates that you want the region enclosed by the circle INCLUDING the boundary.

    Alternatively you can express x and y as rcos(£) and rsin(£) respectively and rewrite the equation as rcos^2() + rsin^2(£) <= r^2
     
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