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I must find the area of the region that lies inside the first curve and outside the second curve for:

$ r^2 = 8cos2\theta, r = 2$

How do I find $a$ and $b$? My solution manual tells me that the curves intersect when $cos2\theta = 1/2$ but I don't see where that came from. I tried setting it equal to each other, but keep getting $1/4$.

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Did you notice the $r^2$ in the first curve? –  Aryabhata Jan 9 '11 at 21:51

2 Answers 2

up vote 4 down vote accepted

The curves will intersect if $8\cos2\theta$ is positive (since it equals $r^2$) and equals 4 (because $r=2$ so $r^2=2^2=4$).

[I emphasize that it must be positive, because for example $r=8\cos2\theta$ and $r=2$ intersect whenever $8\cos2\theta=2$ and also when $8\cos2\theta=-2$.]

Anyway, $8\cos2\theta=4$ gives us $\cos2\theta=4/8=1/2$. This gives us the needed values of $\theta$.

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Thank you Andres. –  ShrimpCrackers Jan 9 '11 at 22:29
    
Andres, or to anyone else, I also have a question related to this problem. When I enter this equation into my ti-83, the leminiscate never completely closes its loop. Any suggestions how to do so? –  ShrimpCrackers Jan 10 '11 at 1:23

my answer for this is like this,,

since r^2=8cosθ , a graph of this is lemniscate! *symmetric wrt x-axis.

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This answer can use some cleanup (MathJax) and clarification. Regards –  Amzoti Aug 4 '13 at 1:36

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