Here is my task:
Calculate directly and using Stokes theorem $\int_C y^2 dx+x \, dy+z \, dz$, if $C$ is intersection line of surfaces $x^2+y^2=x+y$ and $2(x^2+y^2)=z$, orientated in positive direction viewed from point $(0;0;2R)$.
I did it using Stokes theorem, but I don't know how to do it directly. Any idea? Result is $0$.
On
Notice $$\int_C y^2 dx + x dy + z dz = \int_C \left[((x^2 + y^2) - (x+y)) dx + d\left( \frac{x^2}{2} + xy - \frac{x^3}{3} + \frac{z^2}{2}\right)\right] $$ The first term vanishes because $x^2 + y^2 = x + y$ on $C$, the second term vanishes because it is a total differential and $C$ is a closed curve.
One way to do it is to use the parametrization $$x=\frac{1}{2}+\frac{\sqrt{2}}{2}\cos\theta, y=\frac{1}{2}+\frac{\sqrt{2}}{2}\sin\theta, 0\leq \theta \leq 2\pi$$ Then computing $z$ using the second equation gives you $$z=2(1+\frac{\sqrt{2}}{2}(\cos\theta+\sin\theta))$$
Plugging all these into the line integral will give you a single integral with respect to $\theta$.