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

Question

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.

Answer 1

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.