Q:

Solve the integral $ \int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$

Solve the integral $ \int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$.
I tried to apply a 2 parametrizations:
1)
$x= \sqrt{\frac{2t^2}{-3t-\frac{a^2}{2}}}$
$y= \sqrt{-\frac{3}{2}t-\frac{a^2}{4}}$
$z= \sqrt{-6t-a^2}$
From this (using $\int{f \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}}$ i got something like $\int f(t^2) \sqrt{\frac{p(t^2)}{g^3(t)}}dt$
which I don't know how to solve.
2)
Moving to spherical coordinates (r, fi, theta): 
$r=a$
and there is an equation with fi and theta (from $x+y+z=0)$ which seems too complex do deal with.
Now I hope to find the parametrization that will be more fitting.

A:

I think $\int{x^2}d l$=$\int{y^2}d l$=$\int{z^2}d l$ by the symmetry of $L$，so
$$\int{y^2}d l=\frac{1}{3}\int{(x^2+y^2+z^2)}d l=\frac{1}{3}\int{a^2}d l$$
And 
$$x=\frac{\sqrt{2}}{2}a\cos{t}-\frac{\sqrt{6}}{6}a\sin{t},\,\,y=\frac{\sqrt{6}}{3}a\sin{t},\,\,z=-\frac{\sqrt{2}}{2}a\cos{t}-\frac{\sqrt{6}}{6}a\sin{t}$$
where $t\in[0,2\pi]$ will be a good parametrization.

