Find the area of the region between the loops of the limacon

hank

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Sep 13, 2006
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Find the area of the region between the loops of the limacon r = (1/2) = cos@.

I just need help setting it up.

I know I want to take the region of the bigger loop and subtract the region of the inner loop from it.

Bigger loop area = .5 S (.5 + cos@)^2 d@ limits: 0, pi
Inner loop area = .5 S(.5 + cos@)^2 d@ limits: 2pi/3, 4pi/3

I suspect my setup is wrong. Can anyone verify for me what it should be?
 
\(\displaystyle \L\\2\left[\int_{0}^{\frac{2{\pi}}{3}}\frac{1}{2}(\frac{1}{2}+cos({\theta}))^{2}d{\theta}-\int_{\frac{2{\pi}}{3}}^{\pi}\frac{1}{2}(\frac{1}{2}+cos({\theta}))^{2}d{\theta}\right]\)
 
Re: Find the area of the region between the loops of the lim

Hello, Hank!

A number of errors . . .


Find the area of the region between the loops of the limacon: \(\displaystyle \,r\:=\:\frac{1}{2}\,+\,\cos\theta\)

I know I want to take the region of the bigger loop
and subtract the region of the inner loop from it.

Bigger loop area = \(\displaystyle \L\frac{1}{2}\int^{\;\;\;\pi}_0 (\frac{1}{2}\,+\,cos\theta)^2\,d\theta\;\)

Inner loop area = \(\displaystyle \L\frac{1}{2}\int^{\;\;\;\frac{4\pi}{3}}_{\frac{2\pi}{3}}(\frac{1}{2}\,+\,\cos\theta)^2\,d\theta\)

Your "Bigger loop area" has only half of the outer loop.
. . And it doesn't go to \(\displaystyle \pi\)

Your "Inner loop area" has all of the inner loop.


If you want to work with the entire limacon . . .

. . The outer area is: \(\displaystyle \L\:\frac{1}{2}\int^{\;\;\;\frac{2\pi}{3}}_0(\frac{1}{2}\,+\,\cos\theta)^2\,d\theta\;\) plus \(\displaystyle \L\;\frac{1}{2}\int^{\;\;\;2\pi}_{\frac{4\pi}{3}}(\frac{1}{2}\,+\,\cos\theta)^2\,d\theta\)

. . The inner area is: \(\displaystyle \L\:\frac{1}{2}\int^{\;\;\;\frac{4\pi}{3}}_{\frac{2\pi}{3}}(\frac{1}{2}\,+\,\cos\theta)^2\,d\theta\)


If you want to work with the upper half of the region (and multiply by 2),
. . Galactus has the correct set-up.

 
Cool.

Can I pull the 1/2 out of the integrals which then cancels out the 2 in front?
 
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