Limacon, Inner Loop Area

**Del** · April 2nd 2008, 03:10 PM

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

Please help!

**TheEmptySet** · April 2nd 2008, 03:26 PM

Originally Posted by Del

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

Please help!

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 $49\pi-\frac{147\sqrt{3}}{2} \approx 26.63$

**Jhevon** · April 2nd 2008, 03:28 PM

Originally Posted by Del

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

Please help!

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!

**galactus** · April 2nd 2008, 03:49 PM

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.

**Jhevon** · April 2nd 2008, 03:51 PM

Originally Posted by galactus

EDIT ... Well, we agree. That's good.

Yes, the one benefit to posting after others have posted.

**galactus** · April 2nd 2008, 03:53 PM

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.

**Jhevon** · April 2nd 2008, 03:58 PM

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 ...

**galactus** · April 2nd 2008, 04:17 PM

I'm sorry, I misunderstood what you meant.

**Jhevon** · April 2nd 2008, 04:18 PM

Originally Posted by galactus

I'm sorry, I misunderstood what you meant.

that's alright. no hard feelings whatsoever

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