Solving a Polar Plot with r^2 Area Problem: r^2=8cos(2θ)

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EvenSteven
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Area problem regarding [tex]r^2=8cos(2θ)[/tex] and some other curve.

I don't understand how to plot this. I started off with a table of values. I get confused when [tex]θ = π/2[/tex]. I thought it would give [tex]r^2 = -8[/tex]. But looking at mathematica it gives a leminiscate that crosses the origin. How come? Is it because it would give imaginary values? Should I even take the square root? I was doing fine when it was just r = whatever.

Thanks.
 
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EvenSteven said:
Area problem regarding [tex]r^2=8cos(2θ)[/tex] and some other curve.

I don't understand how to plot this. I started off with a table of values. I get confused when [tex]θ = π/2[/tex]. I thought it would give [tex]r^2 = -8[/tex]. But looking at mathematica it gives a leminiscate that crosses the origin. How come? Is it because it would give imaginary values? Should I even take the square root? I was doing fine when it was just r = whatever.

Thanks.

You can't get any point for the curve where r^2 is a negative number. Just plot over values of theta that give you a nonnegative number.
 
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