Compute $\int_{\Gamma}\omega$ where $\omega=(y-2z)dx+(x-z)dy+(2x-y)dz$ and $\Gamma$ is the intersection between: $x^2+y^2+z^2=r^2$ and $x-y+z=0$
My attempt:
$\Gamma$ is some kind of ellipse in the plane $x-y+z = 0$. Taking $z$ from the second equation and substituting in the first we get:
$$x^2+y^2+(x-y)^2=r^2$$
How do you continue from here with the parametrization? Is there a general approach to this kind of things?
Calculated it further and got: $\displaystyle x^2-xy+y^2=\frac {r^2}2$. How do I transform it in the form of a ellipse?
$\Gamma$ is the intersection between a sphere centred on the origin and a plane passing through the origin, thus a circle centred on the origin. The normal vector of this circle is $(1,-1,1)$, so find two unit vectors perpendicular to each other and to $(1,-1,1)$. One possible choice is $\frac{\sqrt2}2(1,1,0)$ and $\sqrt{\frac23}\left(-\frac12,\frac12,1\right)$.
The two new vectors are an orthonormal basis for the plane the circle lies in, so we can use the ordinary parametrisation of the circle, just with different basis vectors: $$\sqrt{\frac23}\left(-\frac12,\frac12,1\right)r\cos t+\frac{\sqrt2}2\left(1,1,0\right)r\sin t$$ $$=r\left(-\frac{\sqrt2}{2\sqrt3}\cos t+\frac{\sqrt2}2\sin t,\frac{\sqrt2}{2\sqrt3}\cos t+\frac{\sqrt2}2\sin t,\sqrt{\frac23}\cos t\right)$$ Of course, $t$ runs from 0 to $2\pi$.