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anonymous
 5 years ago
find the area of the region inside the lemniscate
r^2 = 2(a^2)(cos2theta) a>0
anonymous
 5 years ago
find the area of the region inside the lemniscate r^2 = 2(a^2)(cos2theta) a>0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The element of area for a lemniscate in polar coordinates is given by,\[dA=\frac{1}{2}[r(\theta)]^2d{\theta}\]For your case, the area will be given by,\[A=\int\limits_{{\theta}_1}^{{\theta}_2}\frac{1}{2}(2a^2\cos(2{\theta}))d{\theta}=\left[ \frac{a^2}{2}\sin(2{\theta}) \right]_{{\theta}_1}^{{\theta}_2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the theta 1 theta 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your limits  you're integrating from one angle to another.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[A=\frac{a^2}{2}\left( \sin(2{\theta}_2)\sin(2{\theta}_1) \right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean, what is the value of theta 1 and theta 2? I am confused on how to find the limits of integration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should be given some information in the question. Usually you're told the angles explicitly. Sometimes you may need to work them out from the geometry of the situation. If the question you've posted is complete, then there's nothing more you can do. This is the formula someone would then use to find a numerical value. Though, like I said, unless you're given the limits, or are asked to find them from other information, what's been posted is as far as you can go.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. Thank you very much :D
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