I am asked to find the work of $f(x, y, z) = (x, z, 2y)$ through the curve given by the intersection of two surfaces. I have been doing a series of exercises on this and my question has simply to do with the parametrization of the curve.
The two surfaces are:
$\{(x, y, z) \in R^3 : x = y^2 + z^2\}$ and
$\{(x, y, z) \in R^3 : x + 2y = 3\}$
Although I managed to calculate a function $g$ such that $g(\alpha) = (3-2\alpha, \alpha, \sqrt{3 - \alpha^2 - 2\alpha})$ gives me points on both those surfaces, I am pretty sure there is a nicer parametrization for proceeding to calculate the integral, involving modified polar coordinates. Even with this one, I could only find that $\alpha \le \frac{3}{2}$, leaving me wondering what the lower bound for $\alpha$ is.
The intersection of the two surfaces is given by the equation : $$y^2+z^2-x=3-2y-x$$ or, $$z^2+y^2+2y-3=z^2+(y+1)^2=4$$ which is the equation of a circle in the yz-plane. We have the parametrization $$z=2cost$$, $$y=-1+2sint$$ and $$x=5-4sint$$
Finaly : $\gamma(t)=(5-4sint,-1+2sint,2cost)$ for $t\in[0,2\pi]$.