User:Egm6341.s10.team3.heejun/HW5 update

=  Compare the result of the given circumference formular with the complete elliptic integral of the second kind method =

Problem Statement
$$ \displaystyle Compare\, the\, results\, i)\, and\, ii)\, $$

$$ \displaystyle i)\, Evaluate\, the\, given\, circumference\, formular\, with\, Comp. Trap,\, Romberg\, and\, Clencurt\, $$

$$ \int_{\theta=0}^{\theta=2\pi} dl = C $$

$$ \displaystyle Where\,\,\,\, \begin{align} dl &=d\theta [r^2+(\frac{dr}{d\theta})^2]^(\frac{1}{2})\\ r(\theta) &= \frac{1-e^2}{1-ecos(\theta)} \\ e &= sin(\frac{\pi}{12}) \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg25.djvu|p.25-2]] and [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle ii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle C = 4a E (e) $$

$$ \displaystyle E(e) = \int_{0}^{\pi /2} [1-e^2sin^2(\theta)]^{(\frac{1}{2})} d\theta $$

Ref: Team 4 in HW 4 [|problem_11] and Wikipedia[|Elliptic_integral]

Solution
$$ \displaystyle i)\, Quad\, function\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Quad\, $$

$$ \displaystyle Therefore\,\,\, \int_{\theta=0}^{\theta=2\pi} dl = 6.176601938988738 $$

$$ \displaystyle ii)\, Comp.\, Trap.\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Comp.\, Trap.\, $$

$$ \displaystyle Therefore\,\,\, dl = 6.176601987658693 $$

$$ \displaystyle iii)\, Romberg\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Romberg\, $$

$$ \displaystyle Therefore\,\,\, dl = 6.176601987700596 $$

$$ \displaystyle iiii)\, Clencurt\, $$

$$ \displaystyle The\, above\, input\, is\, under\, BUSY\, on\, Matlab.\,\, Acually\, I\, have\, no\, idea\, for\, it.\, I\, may\, approach\, with\, other\, ways\,$$

$$ \displaystyle iiiii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle Therefore\,\,\, C = 4aE(e) = 6.176601933080789\,\,\, when\, a=1\, $$

$$ \displaystyle Therefore\, $$