# Limacon, Inner Loop Area

• April 2nd 2008, 03:10 PM
Del
Limacon, Inner Loop Area
Find the area inside the loop of the following limacon: r = 7 − 14 sin (θ).

• April 2nd 2008, 03:26 PM
TheEmptySet
Quote:

Originally Posted by Del
Find the area inside the loop of the following limacon: r = 7 − 14 sin (θ).

to find the limits of integration we set r =0

$0=7-14\sin(\theta) \iff \sin(\theta)=\frac{1}{2}$

so we get $\frac{\pi}{6} \mbox{ and } \frac{5 \pi}{6}$

$\frac{1}{2}\int_{\pi/6}^{5\pi/6}(7-14\sin(\theta))^2d\theta$

After integrating I get $49\pi-\frac{147\sqrt{3}}{2} \approx 26.63$
• April 2nd 2008, 03:28 PM
Jhevon
Quote:

Originally Posted by Del
Find the area inside the loop of the following limacon: r = 7 − 14 sin (θ).

did you draw the graph?

now that we have drawn it, we see that the inner loop happens when the graph goes to the origin, that is, when $r = 0$. so let's solve for that:

$\Rightarrow 0 = 7 - 14 \sin \theta$

$\Rightarrow \sin \theta = \frac 12$

$\Rightarrow \theta = \frac {\pi}6,~\frac {5 \pi}6$ for $0 \le \theta \le 2 \pi$

thus, the area is given by: $A = \frac 12 \int_{\pi /6}^{5 \pi /6}r^2~d \theta = \frac 12 \int_{\pi / 6}^{5 \pi / 6}(7 - 14 \sin \theta)^2 ~d \theta$

EDIT: Geez, too late. Thanks a lot EmptySet!
• April 2nd 2008, 03:49 PM
galactus
We can factor out the 7 and write it as $7(1-2sin{\theta})$

$1-2sin{\theta}=0, \;\ {\theta}=\frac{\pi}{6}, \;\ {\theta}=\frac{5\pi}{6}$

Then, we get $\frac{49}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}( 1-2sin{\theta})^{2}d{\theta}$

$\frac{49}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\ left[4sin^{2}{\theta}-4sin{\theta}+1\right]d{\theta}$

Oops, beat to the punch. At least we agree.
• April 2nd 2008, 03:51 PM
Jhevon
Quote:

Originally Posted by galactus
EDIT ... Well, we agree. That's good.

Yes, the one benefit to posting after others have posted.
• April 2nd 2008, 03:53 PM
galactus
I didn't just copy what you all did. What would be the point in that.

I was in the middle of it when you fellas posted.
• April 2nd 2008, 03:58 PM
Jhevon
Quote:

Originally Posted by galactus
I didn't just copy what you all did. What would be the point in that.

I was in the middle of it when you fellas posted.

of course not. no one suggested that.

but it's good that we all agree. our answers confirm one another. i mean, what are the chances we all got it wrong ... (Thinking)
• April 2nd 2008, 04:17 PM
galactus
I'm sorry, I misunderstood what you meant.:o
• April 2nd 2008, 04:18 PM
Jhevon
Quote:

Originally Posted by galactus
I'm sorry, I misunderstood what you meant.:o

that's alright. no hard feelings whatsoever (Handshake)