Find the line integral of $\int y dx + z dy + x dz$ along the curve
given by the intersection of $x+y=2$ and $x^2+y^2+z^2=2(x+y)$, in
clockwise direction.
I think I'm not getting the correct parametrization from this, what I did was:
From $x+y=2 \to y=2-x \to x^2+(2-x)^2+z^2=4$
Completing the square I got $(x-1)^2+z^2=2$, a circle of radius $\sqrt2$ and center $(1, 0)$, so the natural parametrization I thought of was $(1+\sqrt2 \cos t, 1-\sqrt2 \cos t, \sqrt 2 \sin t)$.
But this might be wrong right? I think this parametrization is in counter clockwise direction right?
Also, the solutions say the parametrization should be $(1+ \cos t, 1- \cos t, \sqrt 2 \sin t)$, and I don't know how they got that (which is also in counter clockwise direction I think ?)
Best Answer
We have, $$x+y=2 \implies x^2+y^2 = 4 -2xy$$
Then, $$x^2+y^2+z^2=2(x+y) \implies z^2 = 2xy \implies (\sqrt2 x-\sqrt2)^2 + z^2 = 2$$
From this we get the parametrisation, $$(x,y,z) = (1+ \cos t, 1-\cos t, \sqrt2 \sin t)$$
Also, a parametrisation is neither clockwise or counterclockwise. The direction is required for loop circumvention, as in whether you go across the loop in a clockwise or anticlockwise manner.