Find the area inside the larger loop and outside the smaller loop of the limacon

r= 1/2 + Cos[x]

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- Jul 21st 2007, 04:28 PMcamherokidPolar Coordinate..
Find the area inside the larger loop and outside the smaller loop of the limacon

r= 1/2 + Cos[x] - Jul 21st 2007, 04:52 PMgalactus
The region between the loops would be:

$\displaystyle 2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}\right]$ - Jul 21st 2007, 04:58 PMJhevon
could we not use $\displaystyle \int_{ - 2 \pi / 3}^{ 2 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2 ~d \theta$ ?

or, the equivalent answer would be, i think, $\displaystyle \int_{0}^{2 \pi} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2~d \theta - \int_{2 \pi / 3}^{4 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2~d \theta$

or am i too rusty on polar areas? - Jul 21st 2007, 05:52 PMcurvature
the graph

- Jul 21st 2007, 05:58 PMcamherokid
so what is the answer right?

- Jul 21st 2007, 06:02 PMJhevon
I never calculated the answer, I assume you are capable of doing the integration on your own, so I don't know what answer you got or whether or not it is correct.

I do believe galactus' solution would give a different answer from mine though, but i think i'm right (i may be wrong :D)

i'd go with the first formula i used, not the difference of two integrals, but the single one - Jul 21st 2007, 06:09 PMcurvature
- Jul 21st 2007, 06:12 PMJhevon
- Jul 21st 2007, 06:18 PMcurvature
- Jul 21st 2007, 06:31 PMcurvature
- Jul 21st 2007, 06:35 PMcurvature
- Jul 22nd 2007, 06:38 AMSoroban
I agree with Galactus . . .

Quote:

Originally Posted by**galactus**

I would use: .$\displaystyle \int_0^{\frac{2\pi}{3}}\left(\frac{1}{2}+\cos\thet a\right)^2 d\theta \:-\:\int_{\pi}^{\frac{4\pi}{3}}\left(\frac{1}{2} + \cos\theta\right)^2d\theta$

The integral is: .$\displaystyle \int\left(\frac{1}{4} + \cos\theta + \cos^2\theta\right)d\theta \;=\;\int\left(\frac{1}{4} + \cos\theta + \frac{1 + \cos2\theta}{2}\right)d\theta$

. . $\displaystyle = \;\int\left(\frac{3}{4} + \cos\theta + \frac{1}{2}\cos2\theta\right)d\theta \;= \;\frac{3}{4}\theta + \sin\theta + \frac{1}{4}\sin2\theta$

The "outer" area is: .$\displaystyle \frac{3}{4}\theta + \sin\theta + \frac{1}{4}\sin2\theta\,\bigg|^{\frac{2\pi}{3}}_0$

. . $\displaystyle = \;\left[\frac{3}{4}\!\cdot\!\frac{2\pi}{3} + \sin\frac{2\pi}{3} + \frac{1}{4}\sin\frac{4\pi}{3}\right] - \left[\frac{3}{4}\!\cdot\!0 + \sin 0 + \frac{1}{4}\sin0\cos0\right]$

. . $\displaystyle = \;\left[\frac{\pi}{2} + \frac{\sqrt{3}}{2} + \frac{1}{4}\left(-\frac{\sqrt{3}}{2}\right)\right] - \left[0+ 0 + 0\right] \;\;=\;\;\frac{\pi}{2} + \frac{3\sqrt{3}}{8}$

The "inner" area is: .$\displaystyle \frac{3}{4}\theta + \sin\theta + \frac{1}{4}\sin2\theta\,\bigg|^{\frac{4\pi}{3}}_{\ pi}$

. . $\displaystyle = \;\left[\frac{3}{4}\!\cdot\!\frac{4\pi}{3} + \sin\frac{4\pi}{3} + \frac{1}{4}\sin\frac{8\pi}{3}\right] - \left[\frac{3}{4}\!\cdot\!\pi + \sin\pi + \frac{1}{4}\sin2\pi\right] $

. . $\displaystyle = \;\left[\pi - \frac{\sqrt{3}}{2} + \frac{1}{4}\cdot\frac{\sqrt{3}}{2}\right] - \left[\frac{3}{4}\pi + 0 + 0\right] \;\;=\;\;\frac{\pi}{4} - \frac{3\sqrt{3}}{8}$

The desired area is: .$\displaystyle \left(\frac{\pi}{2} + \frac{3\sqrt{3}}{8}\right) - \left(\frac{\pi}{4} - \frac{3\sqrt{3}}{8}\right) \;\;= \;\;{\bf{\color{blue}\boxed{\frac{\pi + 3\sqrt{3}}{4}}}}$

Check my work . . .*please!*

I've check and double-checked.

. . If there are errors, I can't find them.

- Jul 22nd 2007, 07:26 AMgalactus
That's what I got, Soroban.