Arc length of curve of intersection between cylinder and sphere 
Given the sphere $x^2+y^2+z^2 = \frac{1}{8}$ and the cylinder $8x^2+10z^2=1$, find the arc length of the curve of intersection between the two.

I tried parametrizing the cylinder (the task specifies this as a hint). My attempt:
$$x(t) = \frac{1}{\sqrt{8}}  \sin(t)$$
$$z(t) = \frac{1}{\sqrt{10}} \cos(t)$$
Plugging this into $x^2+y^2+z^2 = \frac{1}{8}$, I solve for $y$ to get 
$$y = \sqrt{\frac{\cos(2t)+1}{4\sqrt{5}}}$$
I then tried integrating $|x(t), y(t), z(t)|$ from $0$ to $2\pi$
with no luck. I suspect my parametrization is wrong as my expression for $y$ looks rather ugly. Any ideas?

Apparently I was right.
If we define $r(t) = (x, y, z)$ where
$$x = \frac{1}{\sqrt{8}}sin(t)$$
$$z = \frac{1}{\sqrt{10}}cos(t)$$
$$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$
we find 
$$|r(t)| = \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} = \frac{1}{2\sqrt{2}}$$
$$\int_{0}^{2\pi} |r(t)|  dt = \int_{0}^{2\pi} \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} dt $$ 
$$ = \int_{0}^{2\pi} \frac{1}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$
The correct answer is $\pi\sqrt{2}$, I suspect the reason lies in the fact that $$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$, meaning we must also take into account the negative value here (seems we must multiply our integral by 2). I used wolfram to my advantage to figure out the integral, my approach by hand was a failure as it got quite messy. 
This is from my calculus 2 class, I'm not familiar with elliptic integrals yet, but I thank you for the assistance Satish. 
