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

1. anonymous

The 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,$A=\int\limits_{{\theta}_1}^{{\theta}_2}\frac{1}{2}(2a^2\cos(2{\theta}))d{\theta}=\left[ \frac{a^2}{2}\sin(2{\theta}) \right]_{{\theta}_1}^{{\theta}_2}$

2. anonymous

what is the theta 1 theta 2?

3. anonymous

Your limits - you're integrating from one angle to another.

4. anonymous

$A=\frac{a^2}{2}\left( \sin(2{\theta}_2)-\sin(2{\theta}_1) \right)$

5. anonymous

i mean, what is the value of theta 1 and theta 2? I am confused on how to find the limits of integration.

6. anonymous

You 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.

7. anonymous

okay. Thank you very much :D

8. anonymous

No probs.