# Polar Coordinate..

• July 21st 2007, 04:28 PM
camherokid
Polar Coordinate..
Find the area inside the larger loop and outside the smaller loop of the limacon
r= 1/2 + Cos[x]
• July 21st 2007, 04:52 PM
galactus
The region between the loops would be:

$2\left[\int_{0}^{\frac{2\pi}{3}}\frac{1}{2}(\frac{1}{2}+c os{\theta})^{2}d{\theta}-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}\right]$
• July 21st 2007, 04:58 PM
Jhevon
Quote:

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

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

could we not use $\int_{ - 2 \pi / 3}^{ 2 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2 ~d \theta$ ?

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

or am i too rusty on polar areas?
• July 21st 2007, 05:52 PM
curvature
the graph
• July 21st 2007, 05:58 PM
camherokid
so what is the answer right?
• July 21st 2007, 06:02 PM
Jhevon
Quote:

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 :D)

i'd go with the first formula i used, not the difference of two integrals, but the single one
• July 21st 2007, 06:09 PM
curvature
Quote:

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 :D)

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.
• July 21st 2007, 06:12 PM
Jhevon
Quote:

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 " $-\int_{\frac{2\pi}{3}}^{\pi}\frac{1}{2}(\frac{1}{2} +cos{\theta})^{2}d{\theta}$" in galactus' solution makes it different from mine. without it, they would be the same
• July 21st 2007, 06:18 PM
curvature
Quote:

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

I have corrcted my last post.
• July 21st 2007, 06:31 PM
curvature
Quote:

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

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

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);
• July 21st 2007, 06:35 PM
curvature
Quote:

Originally Posted by Jhevon
could we not use $\int_{ - 2 \pi / 3}^{ 2 \pi / 3} \frac {1}{2} \left( \frac {1}{2} + \cos \theta \right)^2 ~d \theta$ ?

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

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);
• July 22nd 2007, 06:38 AM
Soroban
I agree with Galactus . . .

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

Check my work . . . please!

I've check and double-checked.
. . If there are errors, I can't find them.

• July 22nd 2007, 07:26 AM
galactus
That's what I got, Soroban.