Polar Plotting Mathematica

[member 428835]

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?

 

[Orodruin]

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.

 

[Dale]

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.

 

[member 428835]

this makes tons of sense, thanks!