How can I find the area under a polar curve with the equation r^2 = 4cos(2θ)?

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SUMMARY

The discussion focuses on calculating the area under the polar curve defined by the equation r² = 4cos(2θ). The integral setup is correctly established as (1/2)∫ from 0 to π of 4cos(2θ) dθ. The user initially encounters issues with the limits of integration, mistakenly believing they should be from 0 to π, which results in an area of zero. Upon further reflection, the user realizes that they are only calculating the area of one loop of the curve, leading to the conclusion that the correct area is 4.

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johnhuntsman
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r2 = 4cos(2θ)

First I graph it. Then I set up the integral.

_____π
(1 / 2)∫ 4cos(2θ) dθ
_____0

________π
= [sin(2θ)]
________0

I thought the limits ought to be π and 0, but that comes out to zero. I pick other limits and they come out to 0. My graph matches the one in the back of the book. I know I've integrated correctly. I think the problem is the limits I'm picking somehow. The correct answer is 4.

[Edit] I also picked (π / 4) and -(π / 4) as limits, but that certainly didn't get me 4. [Edit]
 
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Forget it I see what I'm doing. I'm only finding the area of one loop. My bad. Feel free to delete this post if you're a mod.
 

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