# limacon area

• Apr 19th 2009, 11:04 AM
silencecloak
limacon area
find the area inside the larger loop and outside the smaller loop

$\displaystyle r = \frac{\sqrt{3}}{2} + cos(\theta)$

I have found the zeros to be at $\displaystyle \frac{5\pi}{6} and \frac{7\pi}{6}$

and have set up my integral as such to exclude the area between those two points

$\displaystyle \frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta + \frac{1}{2} \int_{\frac{7\pi}{6}}^{2\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta$

Resulting in 3.9220114003

This is incorrect, where am I going wrong?

EDIT:

I tried integrating in a different manor

$\displaystyle 2(\frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta - \frac{1}{2} \int_{\frac{5\pi}{6}}^{\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta)$

This was correct, but I do not understand why the first solution didnt work
• Apr 19th 2009, 12:34 PM
skeeter
Quote:

Originally Posted by silencecloak
find the area inside the larger loop and outside the smaller loop

$\displaystyle r = \frac{\sqrt{3}}{2} + cos(\theta)$

I have found the zeros to be at $\displaystyle \frac{5\pi}{6} and \frac{7\pi}{6}$

and have set up my integral as such to exclude the area between those two points

$\displaystyle \frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta + \frac{1}{2} \int_{\frac{7\pi}{6}}^{2\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta$

Resulting in 3.9220114003

This is incorrect, where am I going wrong?

EDIT:

I tried integrating in a different manor

$\displaystyle 2(\frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta - \frac{1}{2} \int_{\frac{5\pi}{6}}^{\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta)$

This was correct, but I do not understand why the first solution didnt work

when you integrate from $\displaystyle 0$ to $\displaystyle \frac{5\pi}{6}$ , the upper half of the inner loop gets included because of the overlap.

same idea for integrating from $\displaystyle \frac{7\pi}{6}$ to $\displaystyle 2\pi$ ... the lower half of the inner loop gets included because of the overlap.