Find the area inside the larger loop and outside the smaller loop of the limacon
r= 1/2 + Cos[x]
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?
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 )
i'd go with the first formula i used, not the difference of two integrals, but the single one
I agree with Galactus . . .
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.