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

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The discussion focuses on the area problem related to the polar plot of the equation r^2 = 8cos(2θ). A participant expresses confusion about plotting the curve, particularly when θ = π/2, which leads to r^2 = -8, resulting in imaginary values. It is clarified that points cannot be plotted where r^2 is negative, and one should only consider values of θ that yield nonnegative results. The mention of a leminiscate shape indicates that the curve crosses the origin, which is a characteristic of this polar equation. Understanding the restrictions on r^2 is crucial for accurate plotting and analysis.
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Area problem regarding r^2=8cos(2θ) and some other curve.

I don't understand how to plot this. I started off with a table of values. I get confused when θ = π/2. I thought it would give r^2 = -8. 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 r^2=8cos(2θ) and some other curve.

I don't understand how to plot this. I started off with a table of values. I get confused when θ = π/2. I thought it would give r^2 = -8. 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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