Find given line integral without any fancy theorems(Stokes)
$$\int_L (y-z)dx+(z-x)dy + (x-y)dz$$ where L is circle given by $x^2+y^2+z^2=a^2, y\cos\alpha= x\sin\alpha$, $0<\alpha<\pi$
What I am thinking to do is to switch to spherical coordinates. In that case we get $x=r\cos(\theta)\cos\alpha,y=r\sin\theta \sin\alpha,z=r\cos\theta.$ From here I am unsure how to continue. I have $\theta$ and $\alpha$ should I write this in form of double integrals?
As I wrote in comment, our line is a circle in plane $y=x\tan \alpha$. Assume $\phi$ is an angle formed by the radius of a circle with a straight line and counted in the direction of movement counterclockwise, when viewed from the side of positive $x$. Then parametric representation of line, for $\phi\in [0, 2\pi]$, will be $$\begin{cases}x=a\cos \alpha \cos \phi, \\ y = a\sin \alpha \cos \phi, \\ z = a \sin \phi \end{cases}$$ Now using fact, that $(y-z)dx+(z-x)dy + (x-y)dz=a^2(\cos \alpha - \sin \alpha)d\phi$ and by definition of line integral(II) we have our integral equal $$\int\limits_{0}^{2\pi}a^2(\cos \alpha - \sin \alpha)d\phi = 2\sqrt{2}\pi a^2\sin\left(\frac{\pi}{4}- \alpha\right)$$