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Find the area of the region that lies inside the first curve and outside the second curve. r^2 = 8 cos 2theta , r = 2 so this is a leminscate
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4 = 8 cos 2theta 1/2 = cos 2theta theta = pi/6
dude r=2 is a circle
oh right
anyways i set it up but im off by a bit...
\[\int\limits_{0}^{\pi/6} (8\cos2 \theta)^2 - (2)^2 d \theta\]
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@ganeshie8 what am i doing wrong here? i keep getting the wrong answer no matter how i set it up..
\[1/2 \int\limits_{-\pi/6}^{\pi/6} 8\cos 2\theta - 4~ d \theta\]
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