Polar Plotting Mathematica

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joshmccraney
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Why is it when I plot a circle from ##\theta \in [\pi/2,-\pi/2]## I get the right side of the unit circle:

PolarPlot[1, {\[Theta], \[Pi]/2, -\[Pi]/2}]

? Shouldn't I get the left side?
 

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Orodruin
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No. It is perfectly in order. ##\theta = 0## is the positive ##x##-axis and is part of your plot. ##\theta = \pi## is the negative ##x##-axis and is not part of your plot.
 
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Dale
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Why is it when I plot a circle from ##\theta \in [\pi/2,-\pi/2]## I get the right side of the unit circle:

PolarPlot[1, {\[Theta], \[Pi]/2, -\[Pi]/2}]

? Shouldn't I get the left side?
Note that ##\theta=0## is in the range from ##\pi/2## to ##-\pi/2##, so you expect to get the side of the circle containing 0.

What you want is the range from ##\pi/2## to ##3\pi/2## which does not contain 0.
 
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joshmccraney
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this makes tons of sense, thanks!
 

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