1. ## Polar Coordinate..

Find the area inside the larger loop and outside the smaller loop of the limacon
r= 1/2 + Cos[x]

2. The region between the loops would be:

$\displaystyle 2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}\right]$

3. Originally Posted by galactus
The region between the loops would be:

$\displaystyle 2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}\right]$
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?

4. the graph

5. so what is the answer right?

6. Originally Posted by camherokid
so what is the answer right?
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

7. Originally Posted by Jhevon
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
galactus has used the symmetry of the region, but you haven't. Both solutions are right.

8. Originally Posted by curvature
You have used the symmetry of the region, but galactus hasn't. Both solutions are right.
i don't think so. i think the " $\displaystyle -\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}$" in galactus' solution makes it different from mine. without it, they would be the same

9. Originally Posted by Jhevon
i don't think so. i think the " $\displaystyle -\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}$" in galactus' solution makes it different from mine. without it, they would be the same
I have corrcted my last post.

10. Originally Posted by galactus
The region between the loops would be:

$\displaystyle 2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}\right]$
The above calculation (using symmetry of the region) is verified by the following Maple-generated graph.

A:=plot(cos(theta)+1/2,theta=0..2*Pi/3,coords=polar):
B:=plot(cos(theta)+1/2,theta=2*Pi/3..Pi,coords=polar,color=blue):
display(A,B);

11. Originally Posted by Jhevon
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?
the whole graph

A:=plot(cos(theta)+1/2,theta=-2*Pi/3..2*Pi/3,coords=polar):
B:=plot(cos(theta)+1/2,theta=2*Pi/3..4*Pi/3,coords=polar,color=blue):
display(A,B);

12. I agree with Galactus . . .

Originally Posted by galactus
The region between the loops would be:

$\displaystyle 2\,\left[\frac{1}{2}\int_{0}^{\frac{2\pi}{3}}\left(\frac{1} {2}+\cos{\theta}\right)^{2}d{\theta} \,-\,\frac{1}{2}\int_{\frac{2\pi}{3}}^{\pi}\left(\fra c{1}{2}+\cos{\theta}\right)^{2}d{\theta}\right]$

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.

13. That's what I got, Soroban.

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### lamicon area r= 0.5 cos 2tb

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