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it has not any special method , just first give theta some value , then do this for r
oh let me explain more first suppose r=1,then for theta between 0 to 2pi what shape do you have? surely a circle with radius r ,
now change r you will have lots of circles
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|dw:1355872768424:dw| I think it'll look like this.... my sketch is ugly, i know
Well shining isn't quite right. Because if you pick an r you're also picking a theta as they are equal...
So as r grows so does theta linearly. So in polar it should look something like this: